Chapter 8 - Continuous-Time Fourier Transform (CTFT)

08 Chapter Map - Continuous-Time Fourier Transform (CTFT)

Chapter 8 Overview & Map of Content (MOC)

Rayleigh’s energy theorem, Gaussian self-transforms, modulation shifting, and low-pass RC network analysis.

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📚 Study Notes Index

Read in order — each note assumes the previous one.

#	Note	What it covers
8.00	8.00 Continuous-Time Fourier Transform CTFT Compact Review	CTFT Compact Review, Fourier Transform Formula Sheet
8.01	8.01 Foundation of the Continuous-Time Fourier Transform CTF	CTFT Foundation, Derivation of Fourier Transform, Fourier Integral, Dirichlet Conditions
8.02	8.02 Properties of the Continuous-Time Fourier Transform	CTFT Properties, Fourier Transform Theorems, Modulation Property, Scaling Property
8.03	8.03 Rayleighs Energy Theorem and Spectral Density	Rayleigh’s Theorem, Energy Spectral Density, ESD, Power Spectral Density, PSD
8.04	8.04 CTFT Pairs for Singularity and Common Functions	Fourier Transform Pairs, Common CTFT Pairs, Gaussian Self-Transform, Signum Transform
8.05	8.05 Time-Domain Convolution and Multiplication Properties	Convolution Property, Multiplication Property, Spectral Multiplication, Spectral Convolution
8.06	8.06 Frequency Spectra Phase Spectra and Modulation	Frequency Spectra, Phase Spectra, Modulation Theorem, Amplitude Modulation
8.07	8.07 CTFT System Analysis of Continuous Networks	Circuit Fourier Analysis, Frequency response of RC networks, LPF CTFT

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🎯 Exam Weight

ECE 2107 Exam Relevance

Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.

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🔗 Related Resources

• Course Teaching Plan: ECE 2107 - Signals and Systems
• Previous chapter: 07 Chapter Map - Continuous-Time Fourier Series (CTFS)
• Next chapter: 09 Chapter Map - Laplace Transform & s-Domain Circuit Applications

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Chapter 8: Continuous-Time Fourier Transform (CTFT) - Compact Review

7.00 Continuous-Time Fourier Series - Compact Review | 9.00 Laplace Transform & s-Domain Circuit Applications - Compact Review

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8.01 Foundation of the CTFT & Existence Conditions

*(Target: Theory Descriptive / Mathematical Proof / 10-Mark Derivation)*

• Concept: Represents aperiodic, transient signals in the continuous frequency domain by modeling them as periodic signals whose fundamental period approaches infinity ($T_{0}o\infty$).
• The CTFT limiting derivation steps:
  1. Express periodic counterpart $ildex(t)$ with fundamental period $T_0$ and spacing \omega_0 = rac{2\pi}{T_0} using the complex exponential Fourier series: ilde{x}(t) = \sum_{n=-\infty}^{\infty} C_n e^{j n \omega_0 t} \quad ext{where} \quad C_n = rac{1}{T_0} \int_{-T_0/2}^{T_0/2} ilde{x}(t) e^{-j n \omega_0 t} \, dt \quad ext{[221, 236, 240]}
  2. Substitute $C_n$ back into the series, scale terms with rac{\omega_0}{2\pi} = rac{1}{T_0}, and set the limit as $T_{0}o\infty$ (where fundamental frequency spacing ${\omega }_{0}od\omega$ and discrete index $n{\omega }_{0}o\omega$ becomes a continuous variable) (derivation):
• Forward Fourier Transform: $F(j\omega )={\int }_{-\infty }^{\infty }f(t)e^{-j\omega t}dtext[124,225,240,344]$
• Inverse Fourier Transform: f(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} F(j\omega) e^{j\omega t} \, d\omega \quad ext{[124, 225, 240, 345]}
• Dirichlet Convergence Conditions:
  • Condition 1 (Absolute Integrability): ${\int }_{-\infty }^{\infty }|f(t)|dt<\infty$ {prevents infinite growth}.
  • Condition 2 (Finite Extremas): Finite number of maxima and minima within any finite interval.
  • Condition 3 (Finite Discontinuities): Finite number of discontinuities within any finite interval, with all jump heights bounded.
• Gibbs Phenomenon in CTFT: Truncating high frequencies causes a permanent 9% overshoot at sharp discontinuities when reconstructing $f(t)$ via a finite frequency band.

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8.02 Master CTFT Transform Pairs

*(Target: Numerical Solving / Rapid Lookups)*

Signal $f(t)$	Fourier Transform $F(j\omega )$	Notes / Physical Interpretation
Unit Impulse $\delta (t)$	$1$	Infinite, flat spectrum; contributes equally to all frequencies.
DC Constant $1$	$2\pi \delta (\omega )$	Spectral impulse located exclusively at DC frequency ($\omega =0$).
Complex Exponential $e^{j{\omega }_{0}t}$	$2\pi \delta (\omega -{\omega }_0)$	Pure phase phasor; a single positive spectral line.
Cosine Wave $\cos ({\omega }_{0}t)$	$\pi \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0)
ight]$	Two real, symmetrical line impulses at $\pm {\omega }_0$.
Sine Wave $\sin ({\omega }_{0}t)$	$j\pi \left[ \delta(\omega + \omega_0) - \delta(\omega - \omega_0)
ight]$	Symmetrical imaginary impulses pointing in opposite directions.
Signum Function $extsgn(t)$	rac{2}{j\omega}	Model for a polarity flip; pure imaginary, odd spectrum.
Unit Step $u(t)$	rac{1}{j\omega} + \pi \delta(\omega)	Integrates $\delta (t)$; includes DC impulse ($\pi \delta (\omega )$).
Causal Exponential $e^{-at}u(t)$	rac{1}{a + j\omega} \quad (\Re e\{a\} > 0)	Single-sided decaying transient.
Double Exponential e^{-a\lvert t ert}	rac{2a}{a^2 + \omega^2} \quad (\Re e\{a\} > 0)	Symmetrical infinite-duration decay; yields real Lorentian curve.
Causal Ramp Transient $t{e}^{-at}u(t)$	rac{1}{(a + j\omega)^2} \quad (\Re e\{a\} > 0)	Evaluated using s-domain/frequency differentiation.
Rectangular Gate Pulse $ ext{rect}\left(rac{t}{ au}
ight)$	$ au ext{sa}\left(rac{\omega au}{2}
ight) = au ext{sinc}\left(rac{\omega au}{2\pi}
ight)$	Flat pulse of width $au$ centered at $t=0$; sinc spectrum.
Ideal Low-Pass Filter $rac{W}{\pi} ext{sinc}\left(rac{Wt}{\pi}
ight)$	$ ext{rect}\left(rac{\omega}{2W}
ight)$	Infinitely long sinc impulse response; brick-wall spectrum of width $2W$.
Normalized Gaussian Pulse $e^{-\pi t^2}$	e^{-\pi f^2} = e^{-rac{\omega^2}{4\pi}}	Symmetrical bell curve; transforms exactly into itself.
Delayed Impulse $\delta (t-t_0)$	$e^{-j\omega t_0}$	Pure linear phase delay; constant unit magnitude.

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8.03 Mathematical Properties of the CTFT

*(Target: Theory Descriptive / Mathematical Proof / Numerical Solving)*

• Linearity:

ight} = a_1 X_1(j\omega) + a_2 X_2(j\omega) \quad ext{[124, 126, 226]}$$

• Time Shifting:

ight} = X(j\omega) e^{-j\omega t_0} \quad ext{[124, 126, 226, 346]}$$ Physical Interpretation: Delaying a signal in time shifts its phase linearly with frequency, keeping the magnitude spectrum unchanged.

• Frequency Shifting (Modulation):

ight} = X(j(\omega - \omega_0)) \quad ext{[124, 126, 226, 347]}$$

• Time Scaling:

ight} = rac{1}{\lvert a ert} X\left(jrac{\omega}{a} ight) \quad ext{[124, 126, 226, 348]}$$ Physical Interpretation: Compressing a signal in the time domain ($a>1$) expands its frequency spectrum, showing the reciprocal relationship of time-bandwidth limits.

• Duality:

ight} = X(j\omega) \implies \mathcal{F}\left{ X(jt) ight} = 2\pi x(-\omega) \quad ext{[124, 126, 226, 336]}$$

• Time Differentiation:

ight} = (j\omega)^n X(j\omega) \quad ext{[124, 126, 226, 348]}$$

• Frequency Differentiation:

ight} = rac{dX(j\omega)}{d\omega} \implies \mathcal{F}\left{ t \cdot x(t) ight} = j rac{dX(j\omega)}{d\omega} \quad ext{[124, 126, 226, 333]}$$

• Time Integration:

ight} = rac{X(j\omega)}{j\omega} + \pi X(0) \delta(\omega) \quad ext{[124, 126, 226]}$$

• Time Convolution Theorem:

ight} = X_1(j\omega) X_2(j\omega) \quad ext{[124, 126, 226, 349]}$$

• Frequency Convolution (Windowing/Modulation Theorem):

ight} = rac{1}{2\pi} \left[ X_1(j\omega) * X_2(j\omega) ight] \quad ext{[124, 126, 226]}$$

• Conjugation:

ight} = X^*(-j\omega) \quad ext{[124, 126, 226]}$$

8.03.1 Area Properties

*(Target: Mathematical Proof / 3-Mark Identity)*

• Area Under the Time Curve: ${\int }_{-\infty }^{\infty }f(t)dt=F(0)ext(derivation)ext[124,336]$
• Area Under the Spectral Curve: f(0) = rac{1}{2\pi} \int_{-\infty}^{\infty} F(j\omega) \, d\omega \quad ext{(derivation)} \quad ext{[124, 336]}

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8.04 Waveform Symmetry Conditions & Spectra

*(Target: Theory Descriptive / Symmetry Shortcuts)*

If the time signal $x(t)$ is real, then the CTFT is conjugate symmetric: $X(-j\omega )=X^{\ast }(j\omega )$. Real waveforms exhibit an even magnitude spectrum $|X(j\omega )|=|X(-j\omega )|$ and an odd phase spectrum ngle X(j\omega) = -ngle X(-j\omega).

Table 8.1: CTFT Symmetry Condition Shortcuts

Waveform Condition of $x(t)$	Real Part $X_R(j\omega )$	Imaginary Part $X_I(j\omega )$	CTFT Spectrum $X(j\omega )$
Real & Even	Even	Exactly Zero	Purely Real & Even ($X(j\omega )=X^{\ast }(j\omega )$)
Real & Odd	Exactly Zero	Odd	Purely Imaginary & Odd ($X(j\omega )=-X^{\ast }(j\omega )$)
Imaginary & Even	Exactly Zero	Even	Purely Imaginary & Even
Imaginary & Odd	Odd	Exactly Zero	Purely Real & Odd

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8.05 Rayleigh’s Energy Theorem & Spectral Densities

*(Target: Theory Descriptive / Mathematical Proof)*

8.05.1 Rayleigh’s Energy Theorem

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Theorem Statement: The total energy of an aperiodic signal is identical whether calculated in the time domain or integrated across its energy spectral density in the frequency domain.
• Governing Formula: E_x = \int_{-\infty}^{\infty} \lvert x(t) ert^2 \, dt = rac{1}{2\pi} \int_{-\infty}^{\infty} \lvert X(j\omega) ert^2 \, d\omega \quad ext{(derivation)} \quad ext{[124, 126, 227, 345]}

Table 8.2: Energy Spectral Density (ESD) vs. Power Spectral Density (PSD)

Feature	Energy Spectral Density (ESD) $\Psi (j\omega )$	Power Spectral Density (PSD) $S(j\omega )$
Prerequisite	Energy Signals ($0<E<\infty$, $P=0$)	Power Signals ($E=\infty$, $0<P<\infty$)
Definition	Squared magnitude spectrum: \Psi(j\omega) = \lvert X(j\omega) ert^2	Limit of time-averaged squared magnitude: S(j\omega) = \lim_{T o \infty} rac{\lvert X_T(j\omega) ert^2}{T}
Dimensions	$ext{Volts}^2\cdot ext{s}^2$ or $extJoules/Hz$	$ext{Volts}^2$ or $extWatts/Hz$
Autocorrelation	FT of Energy Autocorrelation: $\mathcal{F}\{R_{xx}(au)\}=\Psi (j\omega )$	FT of Power Autocorrelation: $\mathcal{F}\{R_{xx}(au)\}=S(j\omega )$ (Wiener-Khinchin Theorem)
Total Area	rac{1}{2\pi}\int_{-\infty}^{\infty} \Psi(j\omega) \, d\omega = ext{Total Energy } E	rac{1}{2\pi}\int_{-\infty}^{\infty} S(j\omega) \, d\omega = ext{Total Average Power } P

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8.06 LTI System Analysis & Continuous Networks

*(Target: Circuit Analysis / Numerical Solving)*

Continuous-time systems relate input $x(t)$ to output $y(t)$ using convolution in time, which corresponds to multiplication in the frequency domain: $y(t)=x(t)\ast h(t)\leftrightarrow Y(j\omega )=X(j\omega )H(j\omega )ext[124,349]$

8.06.1 First-Order Low-Pass RC Network Analysis

*(Target: Circuit Analysis / 10-Mark Design Problem)*

              R
        o----/\/\/\-------+--------o Output y(t)
        +                 |        +
     Input x(t)          === C   across Capacitor
        -                 |        -
        o-----------------+--------o

• Network transfer function derivation: Using s-domain/frequency-domain impedances: $Z_R=R$, Z_C = rac{1}{j\omega C}. By voltage divider: H(j\omega) = rac{Y(j\omega)}{X(j\omega)} = rac{rac{1}{j\omega C}}{R + rac{1}{j\omega C}} = rac{1}{1 + j\omega RC} \quad ext{(derivation)} \quad ext{[124, 229]}
• System transient response derivation: For a decaying exponential input $x(t)=e^{-t/RC}u(t)$, find output $y(t)$.
  1. Continuous input spectrum: X(j\omega) = rac{1}{1/RC + j\omega} = rac{RC}{1 + j\omega RC}.
  2. Output spectrum:

ight) \left( rac{1}{1 + j\omega RC} ight) = rac{RC}{(1 + j\omega RC)^2} = rac{1/RC}{(1/RC + j\omega)^2} \quad ext{[124]} 3. Applying the pair $t e^{-at}u(t) \leftrightarrow rac{1}{(a + j\omega)^2}$ where $a = 1/RC$ `(derivation)`: y(t) = rac{t}{RC} e^{-t/RC} u(t) \quad ext{[124, 229]}$$

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8.07 Derivations of High-Yield Core Results

*(Target: Mathematical Proof)*

8.07.1 Normalized Gaussian Pulse is its Own Fourier Transform

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Starting Hypothesis: Let $f(t)=e^{-\pi t^2}$.
• Derivation steps:
  1. Write the forward CTFT integral: $F(j\omega )={\int }_{-\infty }^{\infty }{e}^{-\pi t^2}{e}^{-j\omega t}dtext[124]$
  2. Complete the square in the exponent: -\left( \pi t^2 + j\omega t ight) = -\pi \left( t + rac{j\omega}{2\pi} ight)^2 - rac{\omega^2}{4\pi}.
  3. Pull out the constant term and apply the substitution u = \sqrt{\pi}\left( t + rac{j\omega}{2\pi} ight), knowing the standard Gaussian integral is ${\int }_{-\infty }^{\infty }{e}^{-u^2}du=\sqrt{\pi }$ (derivation):
• Result: F(j\omega) = e^{-rac{\omega^2}{4\pi}} = e^{-\pi f^2} \quad ext{[124, 228, 350]}

8.07.2 Fourier Transform of the Signum Function

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Starting Hypothesis: The signum function is represented as the limit of damped exponentials:

ight] \quad ext{[124, 228]}$$

• Derivation steps:
  1. Apply the forward CTFT integral with damping term $a>0$:

ight] \quad ext{[124]}$$ 2. Evaluate individual integrals: rac{1}{a + j\omega} - rac{1}{a - j\omega} = rac{-2j\omega}{a^2 + \omega^2}. 3. Take the limit as $ao0$ (derivation):

• Result: F(j\omega) = rac{2}{j\omega} = -jrac{2}{\omega} \quad ext{[124, 228, 351]}

8.07.3 Time Shifting Property Proof

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Starting Hypothesis: Show that delaying a signal in time shifts its phase in frequency.
• Derivation steps:
  1. Write the forward CTFT of a shifted signal: $\mathcal{F}\{f(t-t_0)\}={\int }_{-\infty }^{\infty }f(t-t_0)e^{-j\omega t}dtext[124]$
  2. Apply the substitution variable $au=t-t_0⟹t=au+t_0$ and $dt=dau$: $\mathcal{F}\{f(t-t_0)\}={\int }_{-\infty }^{\infty }f(au)e^{-j\omega (au+t_0)}dau=e^{-j\omega t_0}{\int }_{-\infty }^{\infty }f(au)e^{-j\omega au}dauext[124]$
  3. Identify the remaining integral as $F(j\omega )$ (derivation):
• Result: $\mathcal{F}\{f(t-t_0)\}=F(j\omega )e^{-j\omega t_0}ext[124,226,346]$

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8.08 Common Mistakes That Cost Marks

Exam Pitfalls & Marks-Losing Traps

• The Unit Step DC Fallacy: Accidentally evaluating the Fourier transform of the unit step $u(t)$ as simply rac{1}{j\omega}. Since $u(t)$ has a non-zero average DC value, you must include the impulse component: rac{1}{j\omega} + \pi \delta(\omega). Failing to write $\pi \delta (\omega )$ will cost you 2 to 3 marks.
• The $1/2\pi$ Inverse Scaling Omission: Omitting the scaling constant rac{1}{2\pi} when evaluating the continuous inverse Fourier transform integral: f(t) = \mathbf{rac{1}{2\pi}} \int F(j\omega) e^{j\omega t} d\omega. (This scaling is not present in Laplace, making it a very common memory leak under pressure).
• Frequency-Scaling vs. Modulation Signs: Confusing the signs of frequency scaling and modulation. Time delay is a negative phase shift ($f(t-t_0)\leftrightarrow F(j\omega )e^{-j\omega t_0}$), whereas time advance is a positive shift. Similarly, modulation by a positive phasor shifts the spectrum to the right ($e^{j{\omega }_{0}t}x(t)\leftrightarrow X(j(\omega -{\omega }_0))$).

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8.09 PYQ Bank — Verbatim Questions & Answer Plans

Q1: The Gaussian Pulse Self-Transform [5-Mark, KUET 2024/2018/2017]

• Question: Show that the normalized Gaussian pulse is its own Fourier transform.
• Answer Plan:
  1. Define the Gaussian pulse as $f(t)=e^{-\pi t^2}$.
  2. Set up the forward CTFT integral and complete the square in the exponent.
  3. Apply the substitution $u=\sqrt{\pi }(t+j\omega /2\pi )$, evaluate the standard integral $\int e^{-u^2}du=\sqrt{\pi }$ (derivation), and write the final result: $F(j\omega )=e^{-{\omega }^2/4\pi }$.

Q2: Time-Domain Convolution to Spectral Multiplication [4/15-Mark, KUET 2022/2018]

• Question: Show that the convolution of signals in the time domain is equal to the multiplication of their individual Fourier transforms in the frequency domain.
• Answer Plan:
  1. Define the convolution integral $y(t)={\int }_{-\infty }^{\infty }{x}_1(au)x_2(t-au)dau$.
  2. Take the forward CTFT of $y(t)$ and change the order of integrations.
  3. Apply the time shifting property on $x_2(t-au)$ inside the integral to isolate $X_2(j\omega )$ (derivation), proving $Y(j\omega )=X_1(j\omega )X_2(j\omega )$.

Q3: Continuous-Time Piecewise Step Pulse [3/15-Mark, KUET 2025/2016/2015]

• Question: Find the Fourier transform of the time function $f(t)=5[u(t+3)+u(t+2)-u(t-2)-u(t-3)]$.
• Answer Plan:
  1. Rewrite $f(t)$ as a sum of two rectangular gate pulses: $f(t)=5extrect(t/6)+5extrect(t/4)$.
  2. Convert each rectangular pulse using the standard pair $extrect(t/au)\leftrightarrow auextsa(\omega au/2)$.
  3. Sum the transforms using the linearity property: F(j\omega) = 30 ext{sa}(3\omega) + 20 ext{sa}(2\omega) = 30rac{\sin(3\omega)}{3\omega} + 20rac{\sin(2\omega)}{2\omega}.

Q4: Linear Phase Shift of Time Shifts [13-Mark, KUET 2023]

• Question: Show that the time shift in the time domain is equal to a phase shift in the frequency domain.
• Answer Plan: See the complete derivation in Section 8.07.3.

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8.10 Self-Check Before Moving On

• Can you write out the forward and inverse CTFT integrals from memory, including the correct location of the $1/2\pi$ scaling constant?
• Do you know all 3 Dirichlet conditions required for CTFT existence?
• Can you prove why odd real signals must have a purely imaginary and odd frequency spectrum?
• Do you understand the difference between ESD and PSD, particularly which one applies to a periodic power signal versus a transient energy pulse?
• Can you derive the transfer function of a first-order passive RC network and calculate its transient output $y(t)$ using the differentiation/scaling property?

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Source: signals and systems (k.Deergha Rao).pdf, continuous Fourier transform properties (02 Fourier Transform.pdf), and class lecture notes (Rabiul sir class note.pdf).

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8.00 Chapter Map - Continuous-Time Fourier Transform | 8.02 Properties of the Continuous-Time Fourier Transform

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8.01 Foundation of the Continuous-Time Fourier Transform (CTFT)

Core Idea

The Continuous-Time Fourier Transform (CTFT) is the mathematical extension of the Fourier series to aperiodic (non-periodic) signals of infinite duration [3.20, 255]. By conceptually modeling an aperiodic signal as a periodic signal whose fundamental period approaches infinity ($T_{0}o\infty$), the discrete, harmonically spaced line spectrum collapses into a continuous frequency spectrum [3.46, 256]. This note details the rigorous limiting derivation of the CTFT pair, establishes its existence/convergence criteria, and analyzes the physical impact of time-domain expansion on spectral density.

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1. The Conceptual Transition: Periodic to Aperiodic

For periodic signals, the Fourier series decomposes the signal into discrete frequency components spaced at integer multiples of the fundamental frequency \Omega_0 = rac{2\pi}{T_0} [3.21, 3.22]. However, most real-world transient signals {such as speech, lightning, or a single radar pulse} are aperiodic—they occur once and do not repeat [3.44, 261].

To analyze these signals in the frequency domain, we apply a limiting thought experiment [3.44]:

1. Take an aperiodic signal $x(t)$ of finite duration $2T_1$, such that $x(t)=0$ for $|t|>T_1$ [3.44].
2. Construct a periodic extension $ildex(t)$ by repeating $x(t)$ at regular intervals of $T_0$ (where $T_0>2T_1$) [3.44]: $ildex(t)={\sum }_{k=-\infty }^{\infty }x(t-k{T}_0)$
3. If we now let the repetition period $T_{0}o\infty$, the adjacent pulses are pushed infinitely far away to the left and right [3.46, 256]. The periodic extension $ildex(t)$ converges exactly to our original aperiodic signal $x(t)$ [3.46]: ${\lim }_{T_{0}o\infty }ildex(t)=x(t)$

Aperiodic Signal x(t):
                 _
  ______________| |______________   (Occurs once, zero elsewhere)
               -T1 T1

Periodic Extension x~(t) with Period T0:
         _               _               _
  ______| |_____________| |_____________| |______
        -T0             0               T0
         |<--- T0 ----->|

As the period $T_0$ increases, the fundamental frequency \Omega_0 = rac{2\pi}{T_0} becomes progressively smaller, causing the discrete line spectrum to pack closer together. When $T_{0}o\infty$, the discrete line spacing becomes an infinitesimal step $d\Omega$, turning the discrete sum into a continuous integral [3.46, 256].

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2. Rigorous Derivation of the CTFT Pair

Let us mathematically execute this limiting process.

Step 1: Write the Exponential Fourier Series of the Periodic Extension

The periodic signal $ildex(t)$ with period $T_0$ can be represented using its complex exponential Fourier series [3.21, 3.22]: $ildex(t)={\sum }_{n=-\infty }^{\infty }{C}_n{e}^{jn{\Omega }_{0}t}ext---(Equation1.1)$ where the discrete complex coefficients $C_n$ are calculated over a single period [3.23]: C_n = rac{1}{T_0} \int_{-T_0/2}^{T_0/2} ilde{x}(t) e^{-j n \Omega_0 t} \, dt \quad ext{--- (Equation 1.2)}

Step 2: Relate the Coefficients to the Isolated Pulse Envelope

Since $ildex(t)=x(t)$ in the fundamental interval $[-T_0/2,T_0/2]$ and $x(t)=0$ outside this window, we can expand the limits of integration of Equation 1.2 to infinity [3.45]: C_n = rac{1}{T_0} \int_{-\infty}^{\infty} x(t) e^{-j n \Omega_0 t} \, dt \quad ext{--- (Equation 1.3)}

Let us define a continuous, smooth envelope function of a continuous frequency variable $\Omega$ as [3.45]: $X(j\Omega )={\int }_{-\infty }^{\infty }x(t)e^{-j\Omega t}dtext---(Equation1.4)$ Comparing Equation 1.3 and Equation 1.4 reveals that the discrete coefficients $C_n$ are simply scaled, uniformly sampled points of this continuous spectral envelope $X(j\Omega )$ at discrete harmonic frequencies $\Omega =n{\Omega }_0$ [3.45]: C_n = rac{1}{T_0} X(j n \Omega_0) \quad ext{--- (Equation 1.5)}

Step 3: Substitute the Sampled Envelope back into the Synthesis Equation

Substitute the coefficient expression (Equation 1.5) back into the Fourier series expansion (Equation 1.1) [3.46]: ilde{x}(t) = \sum_{n=-\infty}^{\infty} rac{1}{T_0} X(j n \Omega_0) e^{j n \Omega_0 t} Since \Omega_0 = rac{2\pi}{T_0} \implies rac{1}{T_0} = rac{\Omega_0}{2\pi}, we rewrite the summation as [3.46]: ilde{x}(t) = rac{1}{2\pi} \sum_{n=-\infty}^{\infty} X(j n \Omega_0) e^{j n \Omega_0 t} \Omega_0 \quad ext{--- (Equation 1.6)}

Step 4: Take the Limit as the Period $T_{0}o\infty$

As the period $T_0$ approaches infinity, we observe the following limiting transformations [3.46]:

1. Infinitesimal Frequency Spacing: The fundamental frequency interval ${\Omega }_0$ approaches zero and is represented by the differential frequency element $d\Omega$ [3.46]: ${\Omega }_{0}od\Omega$
2. Continuous Frequency Variable: The discrete harmonic frequencies $n{\Omega }_0$ merge into a continuous frequency variable $\Omega$: $n{\Omega }_{0}o\Omega$
3. Summation to Integration: The infinite sum over discrete intervals of width ${\Omega }_0$ becomes a continuous definite integral over all real frequencies [3.46]: ${\sum }_{n=-\infty }^{\infty }[\cdot ]{\Omega }_{0}o{\int }_{-\infty }^{\infty }[\cdot ]d\Omega$
4. Signal Convergence: The periodic extension $ildex(t)$ converges back to the original unique transient signal $x(t)$ [3.46]: $ildex(t)ox(t)$

Applying these limit conditions directly to Equation 1.6 yields the Inverse Continuous-Time Fourier Transform (IDFT / Synthesis Equation) [3.46]:

\mathbf{x(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega} \quad ext{--- (Equation 1.7)}

where the spectral envelope $X(j\Omega )$ is defined by the Forward Continuous-Time Fourier Transform (CTFT / Analysis Equation) [3.46]:

$\mathbf{X}(\mathbf{j\Omega })={\int }_{-\infty }^{\infty }\mathbf{x}(\mathbf{t}){\mathbf{e}}^{-\mathbf{j\Omega t}}\mathbf{dt}ext---(Equation1.8)$

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3. The Continuous-Time Fourier Transform Pair

Equations 1.7 and 1.8 form the Fourier Transform Pair [3.46, 320]. We denote this mutual mathematical linkage as: $x(t)⟷X(j\Omega )$ or using operator notation: $X(j\Omega )=\mathcal{F}\{x(t)\}extandx(t)={\mathcal{F}}^{-1}\{X(j\Omega )\}$

                FORWARD CTFT: F{ x(t) }
   +-------------------------------------------------+
   |                                                 |
   v                                                 v
 [Time Domain]                                 [Frequency Domain]
  x(t) (Volts)                                  X(jΩ) (Volts·sec)
   ^                                                 ^
   |                                                 |
   +-------------------------------------------------+
               INVERSE CTFT: F^-1{ X(jΩ) }

Physical Interpretation of Units and Dimensions

• Time-Domain Signal $x(t)$: Typically measures amplitude over time {e.g., Volts ($V$) or Amperes ($A$)}.
• Fourier Spectrum $X(j\Omega )$: Represents spectral density rather than individual discrete amplitudes. Integrating Equation 1.8 shows that the units of $X(j\Omega )$ are amplitude multiplied by time {e.g., Volt-seconds ($V\cdot exts$) or Volts per Radian/sec ($V/(extrad/s)$)}.
• Therefore, $X(j\Omega )$ does not show the amplitude of a single discrete frequency (which is infinitesimally small for an aperiodic signal); it shows the continuous density of frequencies across the spectrum.

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4. Existence and Convergence: The Dirichlet Conditions

Because the limits of the continuous Fourier integrals extend to infinity ($-\infty$ to $\infty$), we must establish strict conditions to guarantee that the integrals converge mathematically.

The Dirichlet Conditions for the CTFT

For an aperiodic signal $x(t)$ to have a valid, well-defined Fourier transform $X(j\Omega )$, it must satisfy the following three conditions [3.11, 261]:

1. Absolute Integrability: The signal must be absolutely integrable over the entire real time-line [3.11]: ${\int }_{-\infty }^{\infty }|x(t)|dt<\infty$ Proof of bounding: Taking the absolute magnitude of Equation 1.8 yields:

ight| \le \int_{-\infty}^{\infty} \left| x(t) e^{-j\Omega t} ight| , dt = \int_{-\infty}^{\infty} |x(t)| , dt$$

Since the integral is bounded by a finite value, the spectrum $X(j\Omega)$ is guaranteed to be finite and free from infinite spikes [3.11].

2. Finite Extrema: $x(t)$ must have a finite number of local maxima and minima within any finite interval of time.

3. Finite Discontinuities: $x(t)$ must have a finite number of discontinuities within any finite time interval, and each discontinuity must have a bounded, finite height.

4.1 Bounded Convergence and the Gibbs Midpoint Rule

If $x(t)$ satisfies the Dirichlet conditions and has a jump discontinuity at $t=t_0$, the inverse Fourier transform integral does not diverge [3.11]. Instead, it converges exactly to the arithmetic midpoint of the discontinuity: \mathcal{F}^{-1}\{X(j\Omega)\} \Big|_{t=t_0} = rac{x(t_0^+) + x(t_0^-)}{2}

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5. High-Yield Exam Analytics: Changing the Period $T_0$

High-Yield 5-Mark Exam Classic [PYQ 2016]

What is the Fourier transform? What are the effects on the discrete spectrum of a periodic signal if its time period ($T_0$) increases?

Step-by-Step Analytical Explanation Plan:

1. Define the Fourier Transform: State that the CTFT converts a continuous time-domain signal $x(t)$ into a continuous frequency-domain representation $X(j\Omega )$ [3.46]. Write Equations 1.7 and 1.8.
2. The Discrete Harmonic Spacing Formula: Show that the periodic signal has discrete spectral components located at: \Omega_n = n\Omega_0 = n rac{2\pi}{T_0}
3. Effect 1: Spectral Line Density Compression: As the time period $T_0$ increases, the fundamental frequency spacing ${\Omega }_0=2\pi /T_0$ decreases. The discrete spectral lines become denser and pack more tightly together.
4. Effect 2: Amplitude Scaling Attenuation: The amplitude of the discrete Fourier series coefficients C_n = rac{1}{T_0} X(j n \Omega_0) scales down inversely with $T_0$ [3.45]. As $T_{0}o\infty$, the discrete coefficients $C_{n}o0$.
5. Effect 3: Continuous Spectral Transition: Despite the amplitudes shrinking to zero, the relative ratio of the coefficients is locked to the shape of the continuous pulse envelope $X(j\Omega )$ [3.45]. In the limit as $T_{0}o\infty$, the discrete line spectrum transitions completely into a smooth continuous frequency spectrum [3.46, 256].

  T0 is small (Discrete, sparse spacing):
      |       |       |       |       |       |       |
    --o-------o-------o-------o-------o-------o-------o--
             -3Ω0    -2Ω0    -Ω0      0      Ω0      2Ω0     3Ω0

  T0 is large (Discrete, dense spacing):
    |||||||||||||||||||||||||||||||||||||||||||||||||||||
    --o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o--

  T0 -> infinity (Continuous spectrum):
    _____________________________________________________
    ----------------------------------------------------- (Continuous line)

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6. Common Mistakes That Cost Marks

The Cyclic Frequency vs. Angular Frequency Scaling Trap

ECE 2107 utilizes angular frequency $\Omega$ (in rad/s), while some secondary mathematics textbooks (such as Math 2109) use cyclic frequency $f$ (in Hz).

• If you write the Fourier Transform pair using angular frequency $\Omega$, you must include the scaling factor of rac{1}{2\pi} in the Inverse Transform [3.46]: x(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega
• If you write the pair using cyclic frequency $f$ (where $\Omega =2\pi f$), the scaling factor disappears: $x(t)={\int }_{-\infty }^{\infty }X(f)e^{j2\pi ft}dfextandX(f)={\int }_{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt$
• Exam Danger: Mixing these two notations {e.g., omitting the rac{1}{2\pi} factor while integrating with respect to $d\Omega$} is a critical error that results in a zero-mark grading on derivations.

The Non-Convergent Integration Trap

Attempting to evaluate the CTFT of non-absolutely integrable signals {such as a constant $x(t)=1$ or a unit step $u(t)$} using the standard integral formula (Equation 1.8) results in an invalid, divergent calculation because ${\int }_{-\infty }^{\infty }1dt=\infty$.

• To find the CTFT of these signals, you must use distribution-domain modeling with Dirac delta functions (which we cover in Note 8.04).

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7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 PYQ 2015 [5 Marks]

Question: Define the Fourier transform of a time function and explain under what condition it exists.

• Answer Plan:
  1. Define the Fourier Transform mathematically as an integral operation that converts a time function $x(t)$ to its continuous frequency spectrum $X(j\Omega )$ [3.20, 320].
  2. Write the Forward CTFT equation (Equation 1.8).
  3. State the three Dirichlet Conditions (Absolute Integrability, Bounded Extrema, Bounded Discontinuities) as detailed in Section 4 [3.11, 261]. Highlight that absolute integrability guarantees $|X(j\Omega )|$ is bounded [3.11].

7.2 PYQ 2025/2024 [10 Marks]

Question: Define Fourier transform. Briefly explain the properties of Fourier transform.

• Answer Plan:
  1. Provide the formal definition and write down the complete CTFT / Inverse CTFT equation pair [3.46].
  2. Define a Fourier Transform Pair as a specific mathematical relationship linking a unique time-domain function $x(t)$ with its corresponding continuous frequency-domain representation $X(j\Omega )$.
  3. Summarize the major physical properties (Linearity, Time Shifting, Time Scaling, Modulation, and Convolution) using a concise properties table. (Refer to Note 8.02 for the complete property proofs).

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8. Self-Check Before Moving On

• Can you derive the Inverse Fourier Transform integral by taking the limit of the Fourier series as $T_{0}o\infty$? [2.0]
• Do you know where the rac{1}{2\pi} scaling factor belongs in the CTFT equations, and why it is there? [6.0]
• Can you state the three Dirichlet convergence conditions for aperiodic signals? [4.0]
• Do you understand why the continuous Fourier transform measures frequency density instead of discrete amplitude spikes? [3.0]

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, lec 3 u academy online playlist.pdf, Rabiul sir class note.pdf.

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8.01 Foundation of the Continuous-Time Fourier Transform (CTFT) | 8.03 Rayleigh’s Energy Theorem & Spectral Density

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8.02 Properties of the Continuous-Time Fourier Transform

Core Idea

The Continuous-Time Fourier Transform (CTFT) maps time-domain operations directly into elegant algebraic manipulations in the frequency domain [3.33, 3.76]. Understanding these properties—such as Linearity, Time Shifting, Time Scaling, Modulation, and Differentiation—is critical for analyzing linear systems without performing tedious, from-scratch integration [3.33, 3.89]. This note provides the mathematically rigorous proofs for these foundational properties, highlights the physical trade-off between time duration and spectral bandwidth, and solves high-yield KUET examination numericals [3.39, 3.122, 333].

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1. Linearity Property

The CTFT is a linear operator, meaning it preserves the weighted linear combination of signals [3.56].

1.1 Mathematical Statement

If $x_1(t)\leftrightarrow X_1(j\Omega )$ and $x_2(t)\leftrightarrow X_2(j\Omega )$, then for any complex constants $a$ and $b$ [3.56]:

$\mathcal{F}\{a{x}_1(t)+b{x}_2(t)\}=a{X}_1(j\Omega )+b{X}_2(j\Omega )$

1.2 Mathematical Proof

Using the definition of the forward CTFT integral [3.45]:

$\mathcal{F}\{a{x}_1(t)+b{x}_2(t)\}={\int }_{-\infty }^{\infty }[a{x}_1(t)+b{x}_2(t)]e^{-j\Omega t}dt$

Distribute the complex exponential term across the brackets:

$\mathcal{F}\{a{x}_1(t)+b{x}_2(t)\}={\int }_{-\infty }^{\infty }a{x}_1(t)e^{-j\Omega t}dt+{\int }_{-\infty }^{\infty }b{x}_2(t)e^{-j\Omega t}dt$

Since integration is a linear operator, we factor out the constants $a$ and $b$ [3.56]:

$\mathcal{F}\{a{x}_1(t)+b{x}_2(t)\}=a{\int }_{-\infty }^{\infty }{x}_1(t)e^{-j\Omega t}dt+b{\int }_{-\infty }^{\infty }{x}_2(t)e^{-j\Omega t}dt$

\mathcal{F}\{a x_1(t) + b x_2(t)\} = a X_1(j\Omega) + b X_2(j\Omega) \quad lacksquare

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2. Time Shifting Property (Delay)

A delay or advance of a signal in the time domain does not change its magnitude spectrum, but introduces a linear phase shift in the frequency domain [3.61].

2.1 Mathematical Statement

If $x(t)\leftrightarrow X(j\Omega )$, then for a real time delay $t_0$ [3.60]:

$x(t-t_0)\leftrightarrow X(j\Omega )e^{-j\Omega t_0}$

2.2 Mathematical Proof

By definition of the forward CTFT [3.45, 3.61]:

$\mathcal{F}\{x(t-t_0)\}={\int }_{-\infty }^{\infty }x(t-t_0)e^{-j\Omega t}dt$

Perform a change of variable. Let $au=t-t_0⟹t=au+t_0$ and $dt=dau$ [3.61]. The limits of integration remain unchanged:

$\mathcal{F}\{x(t-t_0)\}={\int }_{-\infty }^{\infty }x(au)e^{-j\Omega (au+t_0)}dau$

Factor out the constant exponent term $e^{-j\Omega t_0}$ since it does not depend on the variable of integration $au$ [3.61]:

$\mathcal{F}\{x(t-t_0)\}=e^{-j\Omega t_0}{\int }_{-\infty }^{\infty }x(au)e^{-j\Omega au}dau$

Since $au$ is a dummy variable of integration [3.120], the integral is exactly $X(j\Omega )$ [3.45]:

\mathcal{F}\{x(t - t_0)\} = X(j\Omega) e^{-j\Omega t_0} \quad lacksquare

2.3 Physical Intuition

• Magnitude Spectrum: $|X(j\Omega )e^{-j\Omega t_0}|=|X(j\Omega )|\cdot |e^{-j\Omega t_0}|=|X(j\Omega )|$ [3.61]. Delaying a signal does not change its energy or frequency content.
• Phase Spectrum: ngle [X(j\Omega) e^{-j\Omega t_0}] = ngle X(j\Omega) - \Omega t_0 [3.61]. Shifting in time adds a negative linear phase ramp proportional to the delay $t_0$.

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3. Time Scaling Property (Stretch/Squeeze)

Compressing a signal in the time domain expands its frequency spectrum, while stretching a signal in time compresses its spectrum. This is the Time-Frequency Uncertainty trade-off [3.122].

3.1 Mathematical Statement

If $x(t)\leftrightarrow X(j\Omega )$, then for any non-zero real constant $a$ [3.63]:

ight)$$ ### 3.2 Mathematical Proof Using the forward CTFT integral [3.45, 3.63]: $$\mathcal{F}\{x(a t)\} = \int_{-\infty}^{\infty} x(a t) e^{-j\Omega t} \, dt$$ Perform a change of variable. Let $ au = a t \implies t = rac{ au}{a}$ and $dt = rac{d au}{a}$ [3.63]. The limits of integration depend on the sign of $a$: #### Case A: If $a > 0$ As $t o -\infty \implies au o -\infty$, and as $t o \infty \implies au o \infty$ [3.63]: $$\mathcal{F}\{x(a t)\} = \int_{-\infty}^{\infty} x( au) e^{-j\Omega (rac{ au}{a})} rac{d au}{a} = rac{1}{a} \int_{-\infty}^{\infty} x( au) e^{-j(rac{\Omega}{a}) au} \, d au = rac{1}{a} X\left(jrac{\Omega}{a} ight)$$ #### Case B: If $a < 0$ As $t o -\infty \implies au o \infty$, and as $t o \infty \implies au o -\infty$. This swaps the limits of integration [3.63]: $$\mathcal{F}\{x(a t)\} = \int_{\infty}^{-\infty} x( au) e^{-j\Omega (rac{ au}{a})} rac{d au}{a}$$ Invert the integration limits back to standard order by introducing a negative sign: $$\mathcal{F}\{x(a t)\} = -rac{1}{a} \int_{-\infty}^{\infty} x( au) e^{-j(rac{\Omega}{a}) au} \, d au$$ Since $a < 0$, the term $-rac{1}{a}$ is positive and is equivalent to $rac{1}{|a|}$. Combining both cases using the absolute value operator yields [3.63]: $$\mathcal{F}\{x(a t)\} = rac{1}{|a|} X\left(jrac{\Omega}{a} ight) \quad lacksquare$$ ### 3.3 Physical Trade-off * **Time Squeeze ($a > 1$):** Squeezing a signal in time (making it faster) stretches its spectrum horizontally and squashes its amplitude by $1/a$. It requires more bandwidth to transmit faster signals [3.122]. * **Time Stretch ($0 < a < 1$):** Stretching a signal in time (slowing it down) compresses its spectrum horizontally and increases its spectral density [3.123]. ``` TIME DOMAIN FREQUENCY DOMAIN x(t) X(jΩ) | _ _ | _ _ | | | | | | --+---|---|---+--> t --+------|---|------+--> Ω -1 1 -Ωc Ωc x(2t) [Compressed] X(jΩ/2) / 2 [Expanded & Squashed] | _ _ | | | | | _ _ _ _ --+--|---|----+--> t --+----|-------|----+--> Ω -0.5 0.5 -2Ωc 2Ωc ``` --- ## 4. Frequency Shifting Property (Modulation) Multiplying a signal by a complex exponential in the time domain corresponds to shifting its entire spectrum along the frequency axis [3.62]. This forms the mathematical basis of **amplitude modulation** in modern communications [3.82]. ### 4.1 Mathematical Statement If $x(t) \leftrightarrow X(j\Omega)$, then for a real modulation frequency $\Omega_0$ [3.62]: $$e^{j\Omega_0 t} x(t) \leftrightarrow X(j(\Omega - \Omega_0))$$ ### 4.2 Mathematical Proof By definition of the forward CTFT [3.45, 3.62]: $$\mathcal{F}\{e^{j\Omega_0 t} x(t)\} = \int_{-\infty}^{\infty} [e^{j\Omega_0 t} x(t)] e^{-j\Omega t} \, dt$$ Combine the exponential terms under a single base: $$\mathcal{F}\{e^{j\Omega_0 t} x(t)\} = \int_{-\infty}^{\infty} x(t) e^{-j(\Omega - \Omega_0) t} \, dt$$ Comparing this with the definition of $X(j\Omega) = \int x(t) e^{-j\Omega t} dt$, we observe that $\Omega$ has been replaced everywhere by the shifted frequency variable $(\Omega - \Omega_0)$ [3.62]: $$\mathcal{F}\{e^{j\Omega_0 t} x(t)\} = X(j(\Omega - \Omega_0)) \quad lacksquare$$ --- ## 5. Time Differentiation Property Differentiating a continuous-time signal corresponds to a simple **algebraic multiplication by $j\Omega$** in the frequency domain, making the CTFT highly effective for solving linear constant-coefficient differential equations (LCCDEs) [3.64, 3.89]. ### 5.1 Mathematical Statement If $x(t) \leftrightarrow X(j\Omega)$, then [3.63]: $$rac{d^n x(t)}{dt^n} \leftrightarrow (j\Omega)^n X(j\Omega)$$ ### 5.2 Mathematical Proof We prove this property using the definition of the **Inverse CTFT** [3.46, 3.64]: $$x(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega$$ Differentiate both sides with respect to time $t$. Since the integral limits do not depend on $t$, we move the derivative operator inside the integral [3.64]: $$rac{d x(t)}{dt} = rac{d}{dt} \left[ rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega ight]$$ $$rac{d x(t)}{dt} = rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) \left( rac{\partial e^{j\Omega t}}{\partial t} ight) \, d\Omega$$ Evaluate the partial derivative: $rac{\partial e^{j\Omega t}}{\partial t} = j\Omega e^{j\Omega t}$ [3.64]: $$rac{d x(t)}{dt} = rac{1}{2\pi} \int_{-\infty}^{\infty} [j\Omega X(j\Omega)] e^{j\Omega t} \, d\Omega$$ The right-hand side represents the standard Inverse CTFT of the term $[j\Omega X(j\Omega)]$ [3.46]. Therefore, by uniqueness [3.159]: $$\mathcal{F}\left\{ rac{d x(t)}{dt} ight\} = j\Omega X(j\Omega)$$ Applying this differentiation process repeatedly $n$ times yields [3.64]: $$\mathcal{F}\left\{ rac{d^n x(t)}{dt^n} ight\} = (j\Omega)^n X(j\Omega) \quad lacksquare$$ --- ## 6. Time Integration Property Integrating a signal in time corresponds to dividing its spectrum by $j\Omega$. However, because the integrator acts as a memory accumulator, we must add an impulse at $\Omega=0$ to capture any non-zero DC bias *{average value of the signal}* [3.65]. ### 6.1 Mathematical Statement If $x(t) \leftrightarrow X(j\Omega)$, then [3.73]: $$\int_{-\infty}^{t} x( au) \, d au \leftrightarrow rac{X(j\Omega)}{j\Omega} + \pi X(0) \delta(\Omega)$$ ### 6.2 Mathematical Proof Let the integrated system state be $y(t) = \int_{-\infty}^{t} x( au) d au$ [3.73]. We can express $y(t)$ as the convolution of the input signal $x(t)$ with the unit step function $u(t)$ [3.12]: $$y(t) = \int_{-\infty}^{\infty} x( au) u(t - au) \, d au = x(t) * u(t)$$ Apply the **CTFT Convolution Property** (which we prove in Note 8.05) [3.76]: $$Y(j\Omega) = X(j\Omega) \cdot U(j\Omega)$$ Recall that the Fourier transform of a unit step is $U(j\Omega) = rac{1}{j\Omega} + \pi \delta(\Omega)$ [3.66]. Substitute this into the equation: $$Y(j\Omega) = X(j\Omega) \left[ rac{1}{j\Omega} + \pi \delta(\Omega) ight]$$ $$Y(j\Omega) = rac{X(j\Omega)}{j\Omega} + \pi X(j\Omega) \delta(\Omega)$$ Using the **sampling property of the impulse function**, $X(j\Omega)\delta(\Omega) = X(0)\delta(\Omega)$ since the impulse exists only at $\Omega = 0$ [2.17]: $$Y(j\Omega) = rac{X(j\Omega)}{j\Omega} + \pi X(0) \delta(\Omega) \quad lacksquare$$ --- ## 7. Duality Property (Spectral Symmetry) Duality exploits the mathematical symmetry between the forward and inverse Fourier integrals, allowing us to find new Fourier pairs instantly by swapping the roles of time and frequency [3.81]. ### 7.1 Mathematical Statement If $x(t) \leftrightarrow X(j\Omega)$, then replacing the frequency variable $\Omega$ with a time variable $t$ yields [3.81]: $$X(jt) \leftrightarrow 2\pi x(-\Omega)$$ ### 7.2 Mathematical Proof Start with the definition of the Inverse CTFT [3.46, 3.81]: $$x(t) = rac{1}{2\pi} \, \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega$$ Multiply both sides by $2\pi$: $$2\pi x(t) = \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega$$ Replace the time variable $t$ with $-t$: $$2\pi x(-t) = \int_{-\infty}^{\infty} X(j\Omega) e^{-j\Omega t} \, d\Omega$$ Now, swap the variable names by substituting $t o \Omega$ and $\Omega o t$ [3.81]: $$2\pi x(-\Omega) = \int_{-\infty}^{\infty} X(jt) e^{-j\Omega t} \, dt$$ The right-hand side is exactly the forward Fourier transform of the function $X(jt)$ [3.45]. Thus: $$\mathcal{F}\{X(jt)\} = 2\pi x(-\Omega) \quad lacksquare$$ --- ## 8. Differentiation in Frequency (Multiplication by $t$) Multiplying a signal by the time variable $t$ in the time domain corresponds to differentiating its spectrum with respect to $\Omega$ and scaling by $j$ [3.70]. ### 8.1 Mathematical Statement If $x(t) \leftrightarrow X(j\Omega)$, then [3.70]: $$t x(t) \leftrightarrow j rac{d X(j\Omega)}{d\Omega}$$ ### 8.2 Mathematical Proof Start with the definition of the forward CTFT [3.45, 3.70]: $$X(j\Omega) = \int_{-\infty}^{\infty} x(t) e^{-j\Omega t} \, dt$$ Differentiate both sides with respect to the continuous frequency variable $\Omega$ [3.70]: $$rac{d X(j\Omega)}{d\Omega} = rac{d}{d\Omega} \left[ \int_{-\infty}^{\infty} x(t) e^{-j\Omega t} \, dt ight]$$ Bring the derivative operator inside the integral: $$rac{d X(j\Omega)}{d\Omega} = \int_{-\infty}^{\infty} x(t) \left( rac{\partial e^{-j\Omega t}}{\partial \Omega} ight) \, dt$$ Evaluate the partial derivative: $rac{\partial e^{-j\Omega t}}{\partial \Omega} = -jt e^{-j\Omega t}$ [3.70]: $$rac{d X(j\Omega)}{d\Omega} = \int_{-\infty}^{\infty} x(t) (-jt) e^{-j\Omega t} \, dt$$ $$rac{d X(j\Omega)}{d\Omega} = -j \int_{-\infty}^{\infty} [t x(t)] e^{-j\Omega t} \, dt$$ Multiply both sides by $j$ (noting that $j \cdot (-j) = 1$): $$j rac{d X(j\Omega)}{d\Omega} = \int_{-\infty}^{\infty} [t x(t)] e^{-j\Omega t} \, dt$$ The right-hand side is the standard forward Fourier transform of the signal $t x(t)$ [3.45]. Therefore: $$\mathcal{F}\{t x(t)\} = j rac{d X(j\Omega)}{d\Omega} \quad lacksquare$$ --- ## 9. Master Properties Summary Table For convenience, the properties of the continuous-time Fourier transform are summarized below [3.87]: | Property | Time Domain Signal $x(t)$ | Frequency Domain Spectrum $X(j\Omega)$ | | :--- | :--- | :--- | | **Linearity** | $a x_1(t) + b x_2(t)$ | $a X_1(j\Omega) + b X_2(j\Omega)$ | | **Time Shifting** | $x(t - t_0)$ | $X(j\Omega) e^{-j\Omega t_0}$ | | **Time Scaling** | $x(at)$ | $rac{1}{\|a\|} X\left(jrac{\Omega}{a} ight)$ | | **Modulation** | $e^{j\Omega_0 t} x(t)$ | $X(j(\Omega - \Omega_0))$ | | **Time Differentiation** | $rac{d^n x(t)}{dt^n}$ | $(j\Omega)^n X(j\Omega)$ | | **Time Integration** | $\int_{-\infty}^{t} x( au) \, d au$ | $rac{X(j\Omega)}{j\Omega} + \pi X(0) \delta(\Omega)$ | | **Duality** | $X(jt)$ | $2\pi x(-\Omega)$ | | **Frequency Differentiation**| $t x(t)$ | $j rac{d X(j\Omega)}{d\Omega}$ | --- ## 10. High-Yield Worked Examples (The Exam Killers) ### 10.1 The 10-Mark 2021 KUET Exam Classic (Q. 7b) **Question:** A certain function of time, $f(t)$ has a Fourier transform $F(j\omega) = rac{1}{\omega^2+1} e^{-j\omega^3+1}$. Write down the Fourier transform of: 1. $f(2t)$ 2. $f(t-2) e^{jt}$ 3. $3 rac{df(t)}{dt}$ 4. $rac{f(t)}{-jt}$ --- #### Part 1 Solution: Find the Fourier transform of $y_1(t) = f(2t)$ We apply the **Time Scaling Property** with scale factor $a = 2$ [3.63, 3.88]: $$F_1(j\omega) = \mathcal{F}\{f(2t)\} = rac{1}{|2|} F\left(jrac{\omega}{2} ight)$$ Substitute $rac{\omega}{2}$ in place of $\omega$ in the original expression for $F(j\omega)$ [3.63]: $$F_1(j\omega) = rac{1}{2} \cdot \left[ rac{1}{(\omega/2)^2 + 1} e^{-j(\omega/2)^3 + 1} ight]$$ $$F_1(j\omega) = rac{1}{2} \cdot \left[ rac{1}{rac{\omega^2}{4} + 1} e^{-jrac{\omega^3}{8} + 1} ight]$$ Multiply numerator and denominator of the fraction by $4$: $$F_1(j\omega) = rac{1}{2} \cdot \left[ rac{4}{\omega^2 + 4} e^{-jrac{\omega^3}{8} + 1} ight] = \mathbf{rac{2}{\omega^2 + 4} e^{-jrac{\omega^3}{8} + 1}}$$ --- #### Part 2 Solution: Find the Fourier transform of $y_2(t) = f(t-2) e^{jt}$ Here we must apply both **Time Shifting** and **Frequency Shifting (Modulation)**. Let's do this sequentially to avoid algebraic errors [3.104]. 1. First, let $g(t) = f(t-2)$. By the **Time Shifting Property** with $t_0 = 2$ [3.60, 3.87]: $$G(j\omega) = F(j\omega) e^{-j2\omega}$$ 2. Next, let $y_2(t) = e^{jt} g(t)$. This represents modulation with frequency shift $\omega_0 = 1$ [3.62, 3.88]: $$Y_2(j\omega) = G(j(\omega - 1))$$ 3. Substitute $(\omega - 1)$ in place of $\omega$ in the expression for $G(j\omega)$: $$Y_2(j\omega) = F(j(\omega-1)) e^{-j2(\omega-1)}$$ 4. Now write out the full expression by substituting the original $F(j\omega)$ function shifted by $1$: $$Y_2(j\omega) = \left[ rac{1}{(\omega-1)^2 + 1} e^{-j(\omega-1)^3 + 1} ight] e^{-j2(\omega-1)}$$ 5. Combine the exponential terms: $$\mathbf{Y_2(j\omega) = rac{1}{\omega^2 - 2\omega + 2} e^{-j[(\omega-1)^3 + 2\omega - 3]}}$$ --- #### Part 3 Solution: Find the Fourier transform of $y_3(t) = 3 rac{df(t)}{dt}$ We apply **Linearity** and the **Time Differentiation Property** [3.56, 3.63, 3.88]: $$Y_3(j\omega) = 3 \cdot [j\omega F(j\omega)] = 3j\omega \left[ rac{1}{\omega^2 + 1} e^{-j\omega^3 + 1} ight]$$ $$\mathbf{Y_3(j\omega) = rac{3j\omega}{\omega^2 + 1} e^{-j\omega^3 + 1}}$$ --- #### Part 4 Solution: Find the Fourier transform of $y_4(t) = rac{f(t)}{-jt}$ This is an advanced sub-question testing the integration-differentiation dual pair [3.70]. We know from the **Frequency Differentiation** property that [3.69]: $$\mathcal{F}\{-jt f(t)\} = rac{d F(j\omega)}{d\omega}$$ Let $y_4(t) = rac{f(t)}{-jt} \implies -jt y_4(t) = f(t)$. Taking the Fourier transform of both sides: $$\mathcal{F}\{-jt y_4(t)\} = \mathcal{F}\{f(t)\} \implies rac{d Y_4(j\omega)}{d\omega} = F(j\omega)$$ To solve for $Y_4(j\omega)$, integrate both sides with respect to $\omega$ from $-\infty$ to $\omega$ [3.73]: $$\mathbf{Y_4(j\omega) = \int_{-\infty}^{\omega} F(j\lambda) \, d\lambda = \int_{-\infty}^{\omega} rac{1}{\lambda^2 + 1} e^{-j\lambda^3 + 1} \, d\lambda}$$ --- ## 11. Common Mistakes That Cost Marks > [!danger] **The Shifting-before-Scaling Phase Trap** > > When evaluating the Fourier transform of a composite signal involving both a time shift and a time scale, e.g., $x(at - t_0)$, students often write the phase term incorrectly as $e^{-j\Omega t_0}$. > **The Correct Method:** Always factor out the scaling constant $a$ first to isolate the true shift: > $$x(at - t_0) = x\left(a\left(t - rac{t_0}{a} ight) ight)$$ > Applying scaling first yields $rac{1}{|a|} X\left(jrac{\Omega}{a} ight)$. Then, applying the shift to this scaled function gives: > $$\mathcal{F}\{x(at - t_0)\} = rac{1}{|a|} X\left(jrac{\Omega}{a} ight) e^{-j\Omega rac{t_0}{a}}$$ > [!warning] **Forgetting the Absolute Value in Scaling** > > Forgetting to write the scale multiplier as $rac{1}{|a|}$ when $a$ is negative. For instance, the Fourier transform of $x(-2t)$ is $rac{1}{2} X\left(jrac{\Omega}{-2} ight)$, **not** $-rac{1}{2} X\left(jrac{\Omega}{-2} ight)$. Spectral density magnitudes must remain positive! --- ## 12. PYQ Bank — Verbatim Questions ### 12.1 KUET 2025 / 2024 [5 Marks] * **Question:** Define Fourier transform. Briefly explain the properties of Fourier transform. * **Answer Plan:** Define the forward integral [3.45]. Tabulate and briefly explain the 6 primary properties (Linearity, Shifting, Scaling, Modulation, Differentiation, Integration) using the **Summary Table in Section 9**. ### 12.2 KUET 2023 [13 Marks] * **Question:** Show that the time shift in the time domain is equal to a phase shift in the frequency domain. * **Answer Plan:** redraft and write the formal mathematical proof of the **Time Shifting Property** from **Section 2.2**, showing the variable substitution step $ au = t - t_0$ [3.61]. ### 12.3 KUET 2022 / 2018 [15 Marks] * **Question:** Show that the convolution of the signals in the time domain is equal to the multiplication of their individual Fourier transform in the frequency domain. * **Answer Plan:** (This is the Convolution Theorem, which we derive completely in **Note 8.05**). --- ## 13. Self-Check Before Moving On - [ ] Can you prove why a time delay adds a negative linear phase shift to a signal's spectrum? [10.02] - [ ] Do you understand why compressing a signal in time expands its frequency bandwidth? [10.03] - [ ] Can you solve the 2021 KUET Exam scaling and shift modulation cascade with zero errors? [10.1] - [ ] Did you memorize the integration property, including the $\pi X(0) \delta(\Omega)$ DC offset impulse term? [10.06] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.* --- [[8.02_Properties_of_the_Continuous-Time_Fourier_Transform|8.02 Properties of the Continuous-Time Fourier Transform]] | [[8.04_CTFT_Pairs_for_Singularity_and_Common_Functions|8.04 CTFT Pairs for Singularity & Common Functions]] --- # 8.03 Rayleigh's Energy Theorem & Spectral Density > [!abstract] Core Idea > > **Rayleigh's Energy Theorem** {often referred to as Parseval's relation for aperiodic signals} establishes the fundamental physical principle of **energy conservation** across domains. It mathematically guarantees that the total energy calculated by integrating a signal's squared amplitude over continuous time is identical to integrating its **Energy Spectral Density (ESD)** across the continuous frequency spectrum. This enables engineers to analyze the distribution of signal energy across frequency bands without performing complex time-domain integrations. --- ## 1. Energy Conservation in Time vs. Frequency When analyzing transient signals *{non-periodic waveforms that decay to zero as $t o \infty$}*, we classify them as **Energy Signals** if they contain finite total energy. In the **Time Domain**, the total physical energy $E$ dissipated across a normalized $1\,\Omega$ resistor is defined by the integral of the squared magnitude of the signal: $$E = \int_{-\infty}^{\infty} |x(t)|^2 \, dt$$ In the **Frequency Domain**, Rayleigh's Energy Theorem proves that this same energy is preserved and can be evaluated by integrating the continuous spectrum: $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\Omega)|^2 \, d\Omega$$ ### 1.1 The Spectral Density Concepts To understand how energy is distributed, we define two critical spectral density profiles: * **Energy Spectral Density (ESD):** For an energy signal $x(t)$, the ESD is defined as: $$\Psi_x(j\Omega) = |X(j\Omega)|^2$$ *Units:* Volt-seconds squared per radian/second ($V^2 \cdot ext{s}^2 / ( ext{rad/s})$), representing the concentration of energy per unit frequency. * **Power Spectral Density (PSD):** For periodic *{infinite energy, finite power}* or random signals, the energy is infinite, making ESD undefinable. Instead, we compute the distribution of average power over frequency: $$S_x(j\Omega) = \lim_{T o \infty} rac{|X_T(j\Omega)|^2}{T}$$ where $X_T(j\Omega)$ is the Fourier transform of the signal windowed over interval $T$. ``` Total Energy E (Joules) / Time-Domain Integral Frequency-Domain Integral E = ∫ |x(t)|² dt [8] E = 1/2π ∫ |X(jΩ)|² dΩ [59, 60] {Sum of instantaneous power} {Integral of Energy Spectral Density} ``` --- ## 2. Rigorous Proof of Rayleigh's Energy Theorem Let's prove Rayleigh's theorem step-by-step using the properties of the continuous-time Fourier transform (CTFT) and its complex conjugate. This derivation is a **4-to-6 mark theory classic** in examinations. ### Step 1: Express squared magnitude in conjugate form We start with the time-domain energy definition: $$E = \int_{-\infty}^{\infty} |x(t)|^2 \, dt = \int_{-\infty}^{\infty} x(t) \cdot x^*(t) \, dt \quad ext{--- (Equation 1)}$$ ### Step 2: Substitute the Inverse CTFT for the conjugate term Recall the Inverse CTFT synthesis equation for $x^*(t)$: $$x(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \, d\Omega$$ Taking the complex conjugate of both sides yields: $$x^*(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) e^{-j\Omega t} \, d\Omega \quad ext{--- (Equation 2)}$$ Substitute Equation 2 back into Equation 1: $$E = \int_{-\infty}^{\infty} x(t) \left[ rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) e^{-j\Omega t} \, d\Omega ight] dt \quad ext{--- (Equation 3)}$$ ### Step 3: Interchange the order of integration Assuming the signals satisfy Dirichlet conditions *{allowing Fubini's theorem to apply}*, we interchange the time and frequency integration order: $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) \left[ \int_{-\infty}^{\infty} x(t) e^{-j\Omega t} \, dt ight] d\Omega \quad ext{--- (Equation 4)}$$ ### Step 4: Identify the inner integral as the Forward CTFT We recognize the bracketed time-integral as the standard Forward CTFT equation: $$\int_{-\infty}^{\infty} x(t) e^{-j\Omega t} \, dt = X(j\Omega)$$ Substitute this back into Equation 4: $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} X^*(j\Omega) \cdot X(j\Omega) \, d\Omega$$ Since any complex quantity multiplied by its conjugate equals its magnitude squared ($Z \cdot Z^* = |Z|^2$): $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\Omega)|^2 \, d\Omega \quad lacksquare$$ --- ## 3. High-Yield Worked "Exam Killers" ### 3.1 Numerical 1: Total Energy Verification of decaying exponential [Syllabus Classic] **Question:** Find the total energy of the causal decaying signal $f(t) = e^{-at}u(t)$ ($a > 0$) using both time-domain and frequency-domain integrations to verify Rayleigh's theorem. #### Time-Domain Integration: Using the continuous time-domain definition: $$E = \int_{-\infty}^{\infty} |f(t)|^2 \, dt = \int_{0}^{\infty} \left(e^{-at} ight)^2 \, dt$$ $$E = \int_{0}^{\infty} e^{-2at} \, dt = \left[ rac{e^{-2at}}{-2a} ight]_{0}^{\infty} = 0 - \left(-rac{1}{2a} ight) = \mathbf{rac{1}{2a} ext{ Joules}}$$ #### Frequency-Domain Integration: 1. **Find the CTFT of $f(t)$:** $$F(j\Omega) = \mathcal{F}\{e^{-at}u(t)\} = rac{1}{a + j\Omega} \quad [55]$$ 2. **Determine the Energy Spectral Density (ESD):** $$|F(j\Omega)|^2 = \left| rac{1}{a + j\Omega} ight|^2 = rac{1}{a^2 + \Omega^2} \quad [55]$$ 3. **Integrate over all frequencies:** $$E = rac{1}{2\pi} \int_{-\infty}^{\infty} |F(j\Omega)|^2 \, d\Omega = rac{1}{2\pi} \int_{-\infty}^{\infty} rac{1}{a^2 + \Omega^2} \, d\Omega$$ Recall the standard calculus integral $\int rac{1}{a^2 + u^2} du = rac{1}{a} an^{-1}\left(rac{u}{a} ight)$: $$E = rac{1}{2\pi} \left[ rac{1}{a} an^{-1}\left(rac{\Omega}{a} ight) ight]_{-\infty}^{\infty}$$ $$E = rac{1}{2\pi a} \left( an^{-1}(\infty) - an^{-1}(-\infty) ight)$$ $$E = rac{1}{2\pi a} \left( rac{\pi}{2} - \left(-rac{\pi}{2} ight) ight) = rac{1}{2\pi a} (\pi) = \mathbf{rac{1}{2a} ext{ Joules}}$$ **Conclusion:** Both domains yield exactly $E = rac{1}{2a}$ Joules, demonstrating flawless validation. --- ### 3.2 Numerical 2: The 95% Energy Bandwidth Challenge [Heavily Tested] **Question:** For the same signal $f(t) = e^{-at}u(t)$ with $a = 2$, determine the frequency limit $\Omega_1$ (in rad/s) below which **95%** of the total signal energy is contained. #### Step 1: Calculate the total energy of the signal With $a = 2$: $$E_{ ext{total}} = rac{1}{2a} = rac{1}{2(2)} = 0.25 ext{ Joules}$$ #### Step 2: Set up the 95% energy boundary equation We want to find a frequency limit $\Omega_1$ such that integrating the ESD from $-\Omega_1$ to $\Omega_1$ captures $95\%$ of $E_{ ext{total}}$: $$E_{\Omega_1} = 0.95 \cdot E_{ ext{total}}$$ $$rac{1}{2\pi} \int_{-\Omega_1}^{\Omega_1} |F(j\Omega)|^2 \, d\Omega = 0.95 \cdot (0.25)$$ $$rac{1}{2\pi} \int_{-\Omega_1}^{\Omega_1} rac{1}{2^2 + \Omega^2} \, d\Omega = 0.2375$$ #### Step 3: Perform the definite integration Since the integrand is a symmetric even function, we can simplify the limits from $0$ to $\Omega_1$ by multiplying by 2: $$2 \cdot rac{1}{2\pi} \int_{0}^{\Omega_1} rac{1}{4 + \Omega^2} \, d\Omega = 0.2375$$ $$rac{1}{\pi} \left[ rac{1}{2} an^{-1}\left(rac{\Omega}{a} ight) ight]_{0}^{\Omega_1} = 0.2375$$ $$rac{1}{2\pi} an^{-1}\left(rac{\Omega_1}{2} ight) = 0.2375$$ #### Step 4: Solve for the frequency limit $\Omega_1$ Multiply both sides by $2\pi$: $$ an^{-1}\left(rac{\Omega_1}{2} ight) = 0.2375 \cdot 2\pi = 0.475\pi ext{ radians}$$ Convert the angle to evaluate the tangent: $$rac{\Omega_1}{2} = an(0.475\pi) pprox an(1.49226 ext{ rad}) pprox 12.7062$$ $$\Omega_1 = 2 \cdot 12.7062 pprox \mathbf{25.41 ext{ rad/s}}$$ #### Step 5: Convert to cyclic frequency (Hz) $$f_1 = rac{\Omega_1}{2\pi} = rac{25.4124}{2\pi} pprox \mathbf{4.04 ext{ Hz}}$$ **Answer:** 95% of the signal energy is concentrated within the low-frequency band below $\Omega_1 = 25.41 ext{ rad/s}$ (or $4.04 ext{ Hz}$). --- ## 4. Common Mistakes That Cost Marks > [!danger] **The Squared Magnitude Calculus Trap** > > When performing frequency-domain integrations, students frequently integrate the complex spectrum $F(j\Omega) = rac{1}{a + j\Omega}$ directly instead of its squared magnitude $|F(j\Omega)|^2 = rac{1}{a^2 + \Omega^2}$. Attempting to integrate $\int rac{1}{a+j\Omega} d\Omega$ yields a complex logarithmic expression ($\ln(a+j\Omega)$), which leads to **immediate zero marks** on exam papers. **Always compute $|F(j\Omega)|^2$ first!** > [!warning] **The $2\pi$ Scaling Omission** > > Remember that the frequency-domain integration variable is angular frequency $\Omega$ (rad/s). If you integrate with respect to $d\Omega$, you **must** scale the integral by the factor of $rac{1}{2\pi}$. Forgetting this factor scales your computed energy up by $pprox 628\%$, ruining your numerical check. > *Note:* If you integrate with respect to cyclic frequency $df$ (Hz), the scaling term is naturally absorbed: $E = \int_{-\infty}^{\infty} |F(j2\pi f)|^2 \, df$. --- ## 5. PYQ Bank — Verbatim Questions & Answer Plans ### 5.1 PYQ 2022 / 2016 Exam Classic [8 Marks] **Question:** State and prove Rayleigh's Energy Theorem for continuous-time aperiodic signals. * **Answer Plan:** 1. **State the theorem:** State that the total energy of an aperiodic signal is conserved across both the time and frequency domains, presenting the master equality: $\int_{-\infty}^{\infty} |x(t)|^2 dt = rac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\Omega)|^2 d\Omega$. 2. **Formulate definitions:** Write out the time-domain energy integral and define the Energy Spectral Density (ESD) as $|X(j\Omega)|^2$. 3. **Write the proof:** Follow the exact steps shown in **Section 2**, utilizing the complex conjugate of the Inverse Fourier Transform synthesis equation and interchanging integration orders. ### 5.2 QB Section B Numerical [10 Marks] **Question:** Find the total energy of $x(t) = e^{-3t}u(t)$. Determine the frequency $\omega$ in rad/s such that 90% of the total energy is contained within $(-\omega, \omega)$. * **Answer Plan:** 1. Calculate total energy $E_{ ext{total}} = rac{1}{2a} = rac{1}{6} pprox 0.1667 ext{ Joules}$. 2. Set up the energy fraction: $E_{\omega_1} = 0.90 \cdot rac{1}{6} = 0.15 ext{ Joules}$. 3. Integrate the ESD: $rac{1}{\pi} \int_{0}^{\omega_1} rac{1}{9 + \Omega^2} d\Omega = 0.15 \implies rac{1}{3\pi} an^{-1}\left(rac{\omega_1}{3} ight) = 0.15$. 4. Solve for the angle: $ an^{-1}\left(rac{\omega_1}{3} ight) = 0.45\pi ext{ rad}$. 5. Compute tangent: $rac{\omega_1}{3} = an(0.45\pi) = 6.3138 \implies \omega_1 = \mathbf{18.94 ext{ rad/s}}$. --- ## 6. Self-Check Before Moving On - [ ] Can you state Rayleigh's Energy Theorem and write its dual-domain equations from memory? - [ ] Do you know how to derive the theorem using the complex conjugate synthesis substitution? - [ ] Can you mathematically define Energy Spectral Density (ESD) and state its physical units? - [ ] Are you comfortable solving for the fractional energy bandwidth limits (e.g., 95% or 90% bounds) of a decaying exponential? --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.* --- [[8.03_Rayleighs_Energy_Theorem_and_Spectral_Density|8.03 Rayleigh's Energy Theorem & Spectral Density]] | [[8.05_Time-Domain_Convolution_and_Multiplication_Properties|8.05 Time-Domain Convolution & Multiplication Properties]] --- # 8.04 CTFT Pairs for Singularity & Common Functions > [!abstract] Core Idea > > Fourier analysis relies on a foundational set of **Continuous-Time Fourier Transform (CTFT) Pairs** that map standard mathematical idealizations *{singularity functions like the Dirac delta, unit step, and signum}* and common physical signals *{exponentials, rectangular gates, and Gaussians}* between the time and frequency domains. Understanding how these baseline pairs are analytically derived, along with their physical interpretations, is critical for solving multi-component system networks under exam conditions. --- ## 1. Singularity Function Derivations ### 1.1 The Unit Impulse / Dirac Delta Function, $\delta(t)$ The **Dirac Delta** represents an infinitely narrow, infinitely tall spike centered at $t = 0$ with an integrated area of unity. > [!theorem] **Dirac Delta Transform Derivation** > > Starting from the forward CTFT definition integral: > $$X(j\Omega) = \mathcal{F}\{\delta(t)\} = \int_{-\infty}^{\infty} \delta(t) e^{-j\Omega t} dt$$ > Using the **sampling property** of the unit impulse function, which states that $\int_{-\infty}^{\infty} x(t)\delta(t-t_0)dt = x(t_0)$: > $$X(j\Omega) = \left. e^{-j\Omega t} ight|_{t=0} = e^{0} = 1$$ > > Therefore, we obtain the fundamental pair: > $$\mathbf{\delta(t) \leftrightarrow 1}$$ #### Physical Interpretation: An infinitely brief impulse in the time domain contains absolutely all frequencies with equal weight and zero phase offset *{referred to as a white spectral density}*. This explains why the impulse response of a system completely characterizes its behavior across the entire frequency spectrum. --- ### 1.2 The Constant DC Signal, $x(t) = 1$ A constant signal $x(t) = 1$ is not absolutely integrable over $-\infty < t < \infty$, meaning its transform must be evaluated using the **Duality Property** or as a limiting distribution. > [!theorem] **Constant Signal Transform Derivation** > > By applying the **Duality Property** of the CTFT, which states that if $x(t) \leftrightarrow X(j\Omega)$, then $X(jt) \leftrightarrow 2\pi x(-\Omega)$: > 1. We start with our known impulse pair: $x(t) = \delta(t) \leftrightarrow X(j\Omega) = 1$. > 2. Swap the roles of time and frequency: > $$X(jt) = 1 \leftrightarrow 2\pi x(-\Omega) = 2\pi \delta(-\Omega)$$ > 3. Since the Dirac delta function is even ($\delta(-\Omega) = \delta(\Omega)$): > $$\mathbf{1 \leftrightarrow 2\pi \delta(\Omega)}$$ #### Physical Interpretation: A static DC voltage has no time variation. Consequently, its entire energy is concentrated at a single, isolated frequency of $\Omega = 0$ rad/s *{represented as an impulse of weight $2\pi$}*. --- ### 1.3 The Signum Function, $ ext{sgn}(t)$ The **Signum Function** represents a perfect polarity switcher, defined as $+1$ for $t>0$ and $-1$ for $t<0$. ``` sgn(t) ^ | 1 ----------+----------> t | -1 | | ``` Because it does not decay at infinity, we derive its transform using the **differentiation property** combined with distributions. > [!theorem] **Signum Function Transform Derivation** > > 1. Express the derivative of the signum function. The function has a step discontinuity of height $+2$ at $t = 0$: > $$rac{d}{dt} ext{sgn}(t) = 2\delta(t)$$ > 2. Take the CTFT of both sides. By the **Time Differentiation Property**: > $$\mathcal{F}\left\{rac{d}{dt} ext{sgn}(t) ight\} = j\Omega \mathcal{F}\{ ext{sgn}(t)\}$$ > 3. Substitute the CTFT of the impulse ($2\mathcal{F}\{\delta(t)\} = 2$): > $$j\Omega \mathcal{F}\{ ext{sgn}(t)\} = 2$$ > 4. Solve for the transform: > $$\mathbf{\mathcal{F}\{ ext{sgn}(t)\} = rac{2}{j\Omega}}$$ --- ### 1.4 The Heaviside Unit Step Function, $u(t)$ The **Unit Step Function** acts as a DC switch closing at $t=0$. > [!warning] **The Unit Step Pitfall** > > A common exam failure is writing the CTFT of $u(t)$ as simply $rac{1}{j\Omega}$. Because the unit step has a non-zero average value (DC offset of $1/2$), its transform **must** incorporate a Dirac impulse at the origin to account for this DC average. > [!theorem] **Unit Step Transform Derivation** > > 1. Express the unit step function as a sum of a constant baseline and an odd signum function: > $$u(t) = rac{1}{2} + rac{1}{2} ext{sgn}(t)$$ > 2. Apply the linearity property: > $$\mathcal{F}\{u(t)\} = rac{1}{2}\mathcal{F}\{1\} + rac{1}{2}\mathcal{F}\{ ext{sgn}(t)\}$$ > 3. Substitute our derived transforms for the constant and signum functions: > $$\mathcal{F}\{u(t)\} = rac{1}{2}[2\pi\delta(\Omega)] + rac{1}{2}\left[rac{2}{j\Omega} ight]$$ > $$\mathbf{\mathcal{F}\{u(t)\} = rac{1}{j\Omega} + \pi \delta(\Omega)}$$ --- ## 2. The Gaussian Self-Transform Proof A **Gaussian Pulse** forms the classic bell curve used widely in noise modeling during simulations. ``` x(t) = e^(-pi * t^2) ^ / \ / \ ----------+-----+-----------> t -1 1 ``` We want to prove the high-yield theorem showing that a normalized Gaussian is its own Fourier transform. > [!theorem] **Gaussian Pulse Self-Transform Proof** > > Let the time-domain signal be defined as: > $$x(t) = e^{-bt^2} \quad ext{where } b > 0$$ > > 1. Set up the forward CTFT integral: > $$X(j\Omega) = \int_{-\infty}^{\infty} e^{-bt^2} e^{-j\Omega t} dt$$ > 2. Group the exponents by completing the square: > $$-bt^2 - j\Omega t = -b\left(t^2 + rac{j\Omega}{b} t ight)$$ > $$= -b\left[\left(t + rac{j\Omega}{2b} ight)^2 - \left(rac{j\Omega}{2b} ight)^2 ight] = -b\left(t + rac{j\Omega}{2b} ight)^2 - rac{\Omega^2}{4b}$$ > 3. Factor out the constant frequency term: > $$X(j\Omega) = e^{-rac{\Omega^2}{4b}} \int_{-\infty}^{\infty} e^{-b\left(t + rac{j\Omega}{2b} ight)^2} dt$$ > 4. Perform a change of variables. Let $ au = \sqrt{b}\left(t + rac{j\Omega}{2b} ight) \implies d au = \sqrt{b}\,dt$: > $$X(j\Omega) = e^{-rac{\Omega^2}{4b}} rac{1}{\sqrt{b}} \int_{-\infty}^{\infty} e^{- au^2} d au$$ > 5. Substitute the standard Gaussian integral value $\int_{-\infty}^{\infty} e^{- au^2} d au = \sqrt{\pi}$: > $$X(j\Omega) = \sqrt{rac{\pi}{b}} e^{-rac{\Omega^2}{4b}}$$ > 6. Let us normalize the pulse by setting $b = \pi$ so that $x(t) = e^{-\pi t^2}$: > $$X(j\Omega) = \sqrt{rac{\pi}{\pi}} e^{-rac{\Omega^2}{4\pi}} = e^{-rac{\Omega^2}{4\pi}}$$ > 7. Express the frequency variable in terms of cyclic frequency $f$ where $\Omega = 2\pi f$: > $$X(j2\pi f) = e^{-rac{(2\pi f)^2}{4\pi}} = e^{-\pi f^2}$$ > > Therefore: > $$\mathbf{\mathcal{F}\left\{e^{-\pi t^2} ight\} = e^{-\pi f^2}} \quad ext{or} \quad \mathbf{e^{-\pi t^2} \leftrightarrow e^{-rac{\Omega^2}{4\pi}}}$$ --- ## 3. Real Decaying Exponentials ### 3.1 Single-Sided Exponential decay, $e^{-at}u(t)$ This is the standard transient response modeling signal for R-C networks. > [!theorem] **Single-Sided Decay Derivation** > > Given $x(t) = e^{-at}u(t)$ with $a > 0$: > $$X(j\Omega) = \int_{-\infty}^{\infty} e^{-at}u(t) e^{-j\Omega t} dt = \int_{0}^{\infty} e^{-(a+j\Omega)t} dt$$ > $$= \left. rac{e^{-(a+j\Omega)t}}{-(a+j\Omega)} ight|_{0}^{\infty} = 0 - \left( -rac{1}{a+j\Omega} ight)$$ > $$\mathbf{e^{-at}u(t) \leftrightarrow rac{1}{a + j\Omega}}$$ --- ### 3.2 Double-Sided Symmetric Exponential, $e^{-a|t|}$ This represents a bilateral pulse decaying symmetrically in both the past and the future. ``` x(t) = e^(-a|t|) ^ / \ / \ ----------+-----+-----------> t ``` > [!theorem] **Bilateral Decay Derivation** > > Express the absolute value function in piecewise form: > $$x(t) = e^{-a|t|} = e^{at}u(-t) + e^{-at}u(t)$$ > > Evaluate the integral across both negative and positive boundaries: > $$X(j\Omega) = \int_{-\infty}^{0} e^{at} e^{-j\Omega t} dt + \int_{0}^{\infty} e^{-at} e^{-j\Omega t} dt$$ > $$= \int_{-\infty}^{0} e^{(a-j\Omega)t} dt + \int_{0}^{\infty} e^{-(a+j\Omega)t} dt$$ > $$= \left[ rac{e^{(a-j\Omega)t}}{a-j\Omega} ight]_{-\infty}^{0} + \left[ rac{e^{-(a+j\Omega)t}}{-(a+j\Omega)} ight]_{0}^{\infty}$$ > Since $a > 0$, as $t o -\infty$, $e^{at} o 0$. As $t o \infty$, $e^{-at} o 0$. > $$X(j\Omega) = rac{1}{a-j\Omega} + rac{1}{a+j\Omega}$$ > Combine the fractions using a common denominator: > $$X(j\Omega) = rac{(a+j\Omega) + (a-j\Omega)}{(a-j\Omega)(a+j\Omega)}$$ > $$\mathbf{e^{-a|t|} \leftrightarrow rac{2a}{a^2 + \Omega^2}}$$ --- ## 4. Master CTFT Reference Table This table serves as your primary quick-reference lookup sheet for exam-day transformations: | Time-Domain Signal, $x(t)$ | Angular Frequency Spectrum, $X(j\Omega)$ | Cyclic Frequency Spectrum, $X(f)$ | | :--- | :--- | :--- | | **Unit Impulse $\delta(t)$** | $1$ | $1$ | | **Constant DC $1$** | $2\pi \delta(\Omega)$ | $\delta(f)$ | | **Unit Step $u(t)$** | $rac{1}{j\Omega} + \pi \delta(\Omega)$ | $rac{1}{j2\pi f} + rac{1}{2}\delta(f)$ | | **Signum function $ ext{sgn}(t)$** | $rac{2}{j\Omega}$ | $rac{1}{j\pi f}$ | | **Decaying Exponential $e^{-at}u(t)$** | $rac{1}{a+j\Omega}$ | $rac{1}{a+j2\pi f}$ | | **Bilateral Exponential $e^{-aert t ert}$** | $rac{2a}{a^2+\Omega^2}$ | $rac{2a}{a^2+(2\pi f)^2}$ | | **Cosine Wave $\cos(\Omega_0 t)$** | $\pi\left[\delta(\Omega-\Omega_0) + \delta(\Omega+\Omega_0) ight]$ | $rac{1}{2}\left[\delta(f-f_0) + \delta(f+f_0) ight]$ | | **Sine Wave $\sin(\Omega_0 t)$** | $rac{\pi}{j}\left[\delta(\Omega-\Omega_0) - \delta(\Omega+\Omega_0) ight]$ | $rac{1}{2j}\left[\delta(f-f_0) - \delta(f+f_0) ight]$ | | **Normalized Gaussian $e^{-\pi t^2}$** | $e^{-rac{\Omega^2}{4\pi}}$ | $e^{-\pi f^2}$ | | **Rectangular Gate $ ext{rect}\left(rac{t}{T} ight)$** | $T ext{sinc}\left(rac{\Omega T}{2\pi} ight)$ | $T ext{sinc}(f T)$ | --- ## 5. High-Yield Worked Examples ### 5.1 Piecewise Combined Pulse Transform **Question:** Calculate the Fourier transform of the piecewise step pulse: $$f(t) = 5[u(t+3) + u(t+2) - u(t-2) - u(t-3)]$$ #### Step-by-Step Algebraic Re-Grouping: 1. Group the step functions to form symmetric rectangular gate pulses centered at the origin: $$f(t) = 5\left[u(t+3) - u(t-3) ight] + 5\left[u(t+2) - u(t-2) ight]$$ 2. Map these groups to standard rectangular gate functions: * Let $x_1(t) = 5[u(t+3) - u(t-3)] = 5\, ext{rect}\left(rac{t}{6} ight)$ *{width $T_1 = 6$, amplitude $5$}* * Let $x_2(t) = 5[u(t+2) - u(t-2)] = 5\, ext{rect}\left(rac{t}{4} ight)$ *{width $T_2 = 4$, amplitude $5$}* 3. Apply the standard rectangular transform pair $ ext{rect}\left(rac{t}{T} ight) \leftrightarrow T ext{sinc}\left(rac{\Omega T}{2\pi} ight) = rac{2\sin(\Omega T/2)}{\Omega}$: * $$X_1(j\Omega) = 5 \cdot \left[rac{2\sin(3\Omega)}{\Omega} ight] = rac{10\sin(3\Omega)}{\Omega}$$ * $$X_2(j\Omega) = 5 \cdot \left[rac{2\sin(2\Omega)}{\Omega} ight] = rac{10\sin(2\Omega)}{\Omega}$$ 4. Combine the results using the Linearity Property: $$\mathbf{F(j\Omega) = 10 rac{\sin(3\Omega) + \sin(2\Omega)}{\Omega}}$$ --- ### 5.2 Symmetrical Discrete Impulse Train **Question:** Calculate the Fourier transform of the discrete impulse train: $$g(t) = rac{1}{2}\left[\delta(t+1) + \delta\left(t+rac{1}{2} ight) + \delta\left(t-rac{1}{2} ight) + \delta(t-1) ight]$$ #### Step-by-Step Algebraic Re-Grouping: 1. Apply the **Time Shifting Property** $\delta(t-t_0) \leftrightarrow e^{-j\Omega t_0}$ directly to each impulse term: $$G(j\Omega) = rac{1}{2}\left[e^{j\Omega} + e^{jrac{\Omega}{2}} + e^{-jrac{\Omega}{2}} + e^{-j\Omega} ight]$$ 2. Group the complex conjugate pairs together: $$G(j\Omega) = rac{e^{j\Omega} + e^{-j\Omega}}{2} + rac{e^{jrac{\Omega}{2}} + e^{-jrac{\Omega}{2}}}{2}$$ 3. Substitute Euler's trigonometric identity $\cos( heta) = rac{e^{j heta} + e^{-j heta}}{2}$: $$\mathbf{G(j\Omega) = \cos(\Omega) + \cos\left(rac{\Omega}{2} ight)}$$ --- ## 6. Common Mistakes That Cost Marks > [!danger] **The Unit Step DC Impulse Deletion** > > When asked to transform a unit step function $u(t)$, writing $1/j\Omega$ without adding the $\pi\delta(\Omega)$ term is an automatic point-deduction. Always remember that because $u(t)$ has a non-zero average DC baseline of $1/2$, it must present a corresponding Dirac delta impulse centered at $\Omega = 0$. > [!warning] **Sinc Definition Discrepancy** > > Different textbooks define the sinc function differently: > * **Normalized Sinc (used in DSP and Rao):** $ ext{sinc}(x) = rac{\sin(\pi x)}{\pi x}$. > * **Unnormalized Sinc (used in physics and CTFT):** $ ext{Sa}(x) = rac{\sin(x)}{x}$. > *Always state which definition you are utilizing on your exam sheet to prevent grading confusion.* --- ## 7. PYQ Bank — Verbatim Questions & Answer Plans ### 7.1 Verbatim exam Question 1 **Question:** State and derive the Fourier transform of the Gaussian pulse $f(t) = e^{-\pi t^2}$ and show that it is its own Fourier transform. (07 Marks) * **Answer Plan:** 1. Write down the forward CTFT integral of $f(t) = e^{-\pi t^2}$. 2. Follow the completion of squares method detailed in **Section 2**, setting $b = \pi$. 3. Show step-by-step substitution of the standard Gaussian definite integral $\int_{-\infty}^{\infty} e^{-u^2}du = \sqrt{\pi}$. 4. Conclude with $F(j2\pi f) = e^{-\pi f^2}$, proving the self-transforming property. ### 7.2 Verbatim exam Question 2 **Question:** Obtain the Fourier transform of the unit step function $u(t)$ starting from its decomposition. (05 Marks) * **Answer Plan:** 1. Define the unit step function using its baseline decomposition: $u(t) = rac{1}{2} + rac{1}{2} ext{sgn}(t)$. 2. Write down the transform for the constant term $1 \leftrightarrow 2\pi\delta(\Omega)$ and Signum term $ ext{sgn}(t) \leftrightarrow rac{2}{j\Omega}$ as proven in **Section 1.3**. 3. Apply the linearity property to sum the elements and derive the final result: $U(j\Omega) = rac{1}{j\Omega} + \pi\delta(\Omega)$. --- ## 8. Self-Check Before Moving On - [ ] Can you prove why a constant DC signal has an impulse in frequency? [1.2] - [ ] Do you know how to complete the square to solve the Gaussian pulse derivation? [2.0] - [ ] Have you memorized the unit step transform including the DC impulse component? [1.4] - [ ] Can you derive the Signum function transform using the differentiation property? [1.3] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.* --- [[8.04_CTFT_Pairs_for_Singularity_and_Common_Functions|8.04 CTFT Pairs for Singularity & Common Functions]] | [[8.06_Frequency_Spectra_Phase_Spectra_and_Modulation|8.06 Frequency Spectra, Phase Spectra & Modulation]] --- # 8.05 Time-Domain Convolution & Multiplication Properties > [!abstract] Core Idea > > The **Time-Domain Convolution and Multiplication Properties** represent the most powerful analytical shortcuts in continuous-time system analysis [3.84, 3.89]. Time-domain convolution—representing the physical interaction of an input signal with an LTI system's impulse response—simplifies to **direct algebraic multiplication** in the frequency domain [3.84, 4.82]. Conversely, time-domain multiplication (key to amplitude modulation) corresponds to **continuous convolution of spectra** in the frequency domain, scaled by the factor $1/(2\pi)$ [3.89, 3.90]. This note details the rigorous mathematical proofs and provides solved exam classics to master these properties. --- ## 1. Symmetries of Convolution and Multiplication In the continuous-time domain, the relationship between signals and systems is governed by the **Convolution Integral** [2.15]: $$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x( au) h(t - au) \, d au$$ Evaluating this integral for complex piecewise or exponential signals is mathematically tedious and prone to integration limit errors [3.102]. The Fourier Transform establishes a elegant **duality** that maps time-domain integrations directly to frequency-domain algebra [3.84, 3.89]: ``` TIME DOMAIN FREQUENCY DOMAIN x(t) * h(t) <=====================> X(jΩ) · H(jΩ) (Convolution) (Multiplication) x1(t) · x2(t) <=====================> 1/(2π) [X1(jΩ) * X2(jΩ)] (Multiplication) (Convolution) ``` --- ## 2. Rigorous Proof: Time-Domain Convolution Property ### 2.1 Theorem Statement If $x_1(t) \leftrightarrow X_1(j\Omega)$ and $x_2(t) \leftrightarrow X_2(j\Omega)$, then [3.84]: $$\mathcal{F}\{x_1(t) * x_2(t)\} = X_1(j\Omega) X_2(j\Omega)$$ --- ### 2.2 Mathematical Proof (12-Mark Exam Favorite) [PYQ 2022, 2018] Let $y(t) = x_1(t) * x_2(t)$. By definition, we write [3.84]: $$y(t) = \int_{-\infty}^{\infty} x_1( au) x_2(t - au) \, d au$$ Applying the forward Continuous-Time Fourier Transform (CTFT) integral to $y(t)$ yields [3.85]: $$Y(j\Omega) = \int_{-\infty}^{\infty} y(t) e^{-j\Omega t} \, dt$$ $$\mathbf{Y(j\Omega) = \int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} x_1( au) x_2(t - au) \, d au ight] e^{-j\Omega t} \, dt}$$ Assuming the signals are absolutely integrable *{satisfying Dirichlet conditions to allow interchanging the order of integration}*, we can rearrange the integrals [3.85]: $$Y(j\Omega) = \int_{-\infty}^{\infty} x_1( au) \left[ \int_{-\infty}^{\infty} x_2(t - au) e^{-j\Omega t} \, dt ight] \, d au$$ To evaluate the inner integral over $t$, let us perform a **change of variables** [3.85]: * Let $u = t - au \implies t = u + au$ * The differential becomes $dt = du$ * The limits of integration remain $[-\infty, \infty]$ Substitute these into the inner integral [3.85]: $$Y(j\Omega) = \int_{-\infty}^{\infty} x_1( au) \left[ \int_{-\infty}^{\infty} x_2(u) e^{-j\Omega (u + au)} \, du ight] \, d au$$ Split the complex exponential term: $e^{-j\Omega (u + au)} = e^{-j\Omega u} \cdot e^{-j\Omega au}$ [3.85]: $$Y(j\Omega) = \int_{-\infty}^{\infty} x_1( au) \left[ \int_{-\infty}^{\infty} x_2(u) e^{-j\Omega u} e^{-j\Omega au} \, du ight] \, d au$$ Since the exponential term $e^{-j\Omega au}$ is independent of the inner variable of integration $u$, we pull it outside the inner bracket [3.85]: $$Y(j\Omega) = \int_{-\infty}^{\infty} x_1( au) e^{-j\Omega au} \left[ \int_{-\infty}^{\infty} x_2(u) e^{-j\Omega u} \, du ight] \, d au$$ We recognize that the inner integral is exactly the forward Fourier Transform of $x_2(t)$ expressed with dummy variable $u$ [3.85]: $$\int_{-\infty}^{\infty} x_2(u) e^{-j\Omega u} \, du = X_2(j\Omega)$$ Substitute this back into our expression [3.85]: $$Y(j\Omega) = \int_{-\infty}^{\infty} x_1( au) e^{-j\Omega au} \left[ X_2(j\Omega) ight] \, d au$$ Since $X_2(j\Omega)$ is independent of the variable of integration $ au$, we pull it completely outside the integral [3.85]: $$Y(j\Omega) = X_2(j\Omega) \left[ \int_{-\infty}^{\infty} x_1( au) e^{-j\Omega au} \, d au ight]$$ The remaining integral is the definition of the forward Fourier Transform of $x_1(t)$ [3.86]: $$\int_{-\infty}^{\infty} x_1( au) e^{-j\Omega au} \, d au = X_1(j\Omega)$$ Thus, we obtain: $$\mathbf{Y(j\Omega) = X_1(j\Omega) X_2(j\Omega)} \quad ext{[Proved]}$$ --- ## 3. Rigorous Proof: Time-Domain Multiplication Property ### 3.1 Theorem Statement If $x_1(t) \leftrightarrow X_1(j\Omega)$ and $x_2(t) \leftrightarrow X_2(j\Omega)$, then [3.89, 3.90]: $$\mathcal{F}\{x_1(t) \cdot x_2(t)\} = rac{1}{2\pi} \left[ X_1(j\Omega) * X_2(j\Omega) ight] = rac{1}{2\pi} \int_{-\infty}^{\infty} X_1(j\lambda) X_2(j(\Omega - \lambda)) \, d\lambda$$ --- ### 3.2 Mathematical Proof Let $y(t) = x_1(t) x_2(t)$. Let us express $x_1(t)$ in terms of its continuous frequency spectrum using the **Inverse Fourier Transform (Synthesis Equation)** [3.90]: $$x_1(t) = rac{1}{2\pi} \int_{-\infty}^{\infty} X_1(j\lambda) e^{j\lambda t} \, d\lambda$$ Now, write the forward Fourier Transform of $y(t)$ [3.90]: $$Y(j\Omega) = \int_{-\infty}^{\infty} \left[ x_1(t) x_2(t) ight] e^{-j\Omega t} \, dt$$ Substitute our Inverse Fourier expression of $x_1(t)$ into this integral [3.90]: $$\mathbf{Y(j\Omega) = \int_{-\infty}^{\infty} \left[ rac{1}{2\pi} \int_{-\infty}^{\infty} X_1(j\lambda) e^{j\lambda t} \, d\lambda ight] x_2(t) e^{-j\Omega t} \, dt}$$ Interchange the order of integrations [3.90]: $$Y(j\Omega) = rac{1}{2\pi} \int_{-\infty}^{\infty} X_1(j\lambda) \left[ \int_{-\infty}^{\infty} x_2(t) e^{j\lambda t} e^{-j\Omega t} \, dt ight] \, d\lambda$$ Combine the exponentials in the inner integral [3.90]: $$Y(j\Omega) = rac{1}{2\pi} \int_{-\infty}^{\infty} X_1(j\lambda) \left[ \int_{-\infty}^{\infty} x_2(t) e^{-j(\Omega - \lambda) t} \, dt ight] \, d\lambda$$ We immediately recognize the inner integral over $t$ as the forward Continuous-Time Fourier Transform of $x_2(t)$ evaluated at the shifted frequency $(\Omega - \lambda)$ [3.90]: $$\int_{-\infty}^{\infty} x_2(t) e^{-j(\Omega - \lambda) t} \, dt = X_2(j(\Omega - \lambda))$$ Substitute this back into our outer integral: $$Y(j\Omega) = rac{1}{2\pi} \int_{-\infty}^{\infty} X_1(j\lambda) X_2(j(\Omega - \lambda)) \, d\lambda$$ By definition of continuous convolution, this integral is exactly $X_1(j\Omega) * X_2(j\Omega)$ [3.90]: $$\mathbf{Y(j\Omega) = rac{1}{2\pi} \left[ X_1(j\Omega) * X_2(j\Omega) ight]} \quad ext{[Proved]}$$ --- ## 4. High-Yield Worked Examples (The Exam Killers) ### 4.1 Example 1: The Triangular Pulse via Cascaded Rectangles [Example 3.29] **Question:** Calculate the Fourier transform of the symmetric triangular pulse shown in the figure below: ``` y(t) ^ 2| / | / | / | / | / ---+--+--------+--+---> t -2 0 2 ``` *Hint: A triangular pulse of width $2W$ is formed by convolving a rectangular pulse of width $W$ with itself.* #### Step-by-Step Analytical Solution: 1. **Represent $y(t)$ as a convolution of two basic signals:** Let $x_1(t) = x_2(t) = ext{rect}(t/2)$ *{a rectangular gate pulse centered at $t = 0$ with amplitude 1 and width $ au = 2$}* [3.86]. $$y(t) = x_1(t) * x_2(t)$$ 2. **Evaluate the Fourier transform of the rectangular prototype $x_1(t)$:** $$X_1(j\Omega) = \int_{-1}^{1} (1) e^{-j\Omega t} \, dt = rac{e^{-j\Omega t}}{-j\Omega} \Big|_{-1}^{1} = rac{e^{j\Omega} - e^{-j\Omega}}{j\Omega} = rac{2 \sin(\Omega)}{\Omega} = 2 ext{sinc}\left(rac{\Omega}{\pi} ight)$$ 3. **Apply the time-domain convolution property [3.87]:** $$Y(j\Omega) = X_1(j\Omega) X_2(j\Omega) = X_1^2(j\Omega)$$ $$Y(j\Omega) = \left[ rac{2\sin(\Omega)}{\Omega} ight] \cdot \left[ rac{2\sin(\Omega)}{\Omega} ight] = \mathbf{rac{4 \sin^2(\Omega)}{\Omega^2}} = \mathbf{4 ext{sinc}^2\left(rac{\Omega}{\pi} ight)}$$ --- ### 4.2 Example 2: Ideal Low-Pass Filter Response [Example 3.30] **Question:** Consider an LTI system with an impulse response of $h(t) = rac{\sin(\Omega_0 t)}{\pi t}$. Find the output response $y(t)$ for the input signal $x(t) = rac{\sin(2\Omega_0 t)}{\pi t}$ [3.87]. #### Step-by-Step Frequency-Domain Solution: 1. **Extract the continuous spectra using known CTFT pairs [3.87, 3.88]:** Recall that: $$rac{\sin(W t)}{\pi t} \quad \leftrightarrow \quad ext{rect}\left(rac{\Omega}{2W} ight) = egin{cases} 1, & |\Omega| \le W \ 0, & |\Omega| > W \end{cases}$$ Therefore: * **System Frequency Response:** $H(j\Omega) = ext{rect}\left(rac{\Omega}{2\Omega_0} ight)$ *{flat low-pass filter with cutoff frequency $\Omega_0$}* [3.88]. * **Input Spectrum:** $X(j\Omega) = ext{rect}\left(rac{\Omega}{4\Omega_0} ight)$ *{gate spectrum spanning from $-2\Omega_0$ to $2\Omega_0$}* [3.88]. 2. **Evaluate output spectrum using the convolution property [3.88]:** $$Y(j\Omega) = X(j\Omega) \cdot H(j\Omega)$$ Since $X(j\Omega)$ and $H(j\Omega)$ are rectangular functions, their product corresponds to the intersection of their non-zero bands: * $H(j\Omega) = 1$ for $\Omega \in [-\Omega_0, \Omega_0]$ * $X(j\Omega) = 1$ for $\Omega \in [-2\Omega_0, 2\Omega_0]$ ``` H(jΩ) X(jΩ) Y(jΩ) = X(jΩ)·H(jΩ) 1 +----+ 1 +--------+ 1 +----+ | | | | | | ---+----+---> --+--------+---> ====> --+----+---> Ω -Ω0 Ω0 -2Ω0 2Ω0 -Ω0 Ω0 ``` Thus, the narrower filter spectrum completely clamps the input, yielding [3.88]: $$Y(j\Omega) = ext{rect}\left(rac{\Omega}{2\Omega_0} ight)$$ 3. **Synthesize back to the time domain [3.88]:** Taking the inverse CTFT of $Y(j\Omega)$ yields the final, mathematically flawless output signal: $$\mathbf{y(t) = rac{\sin(\Omega_0 t)}{\pi t}}$$ --- ### 4.3 Example 3: Tricky Zero-Overlap Filtering [Example 3.32] **Question:** Determine the output response $y(t) = x_1(t) * x_2(t)$ if [3.91]: $$x_1(t) = ext{sinc}(2t) \quad ext{and} \quad x_2(t) = ext{sinc}(t)\cos(3\pi t)$$ #### Step-by-Step Spectral Analysis: 1. **Obtain the spectrum $X_1(j\Omega)$:** Using the duality property of the $ ext{sinc}$ function [3.91]: $$ ext{sinc}(t) \quad \leftrightarrow \quad ext{rect}\left(rac{\Omega}{2\pi} ight) = egin{cases} 1, & |\Omega| \le \pi \ 0, & |\Omega| > \pi \end{cases}$$ Applying the time-scaling property where $a = 2$ [3.91]: $$X_1(j\Omega) = rac{1}{2} ext{rect}\left(rac{\Omega}{4\pi} ight) = egin{cases} rac{1}{2}, & -2\pi \le \Omega \le 2\pi \ 0, & ext{otherwise} \end{cases}$$ 2. **Obtain the spectrum $X_2(j\Omega)$:** Express the cosine term as complex exponentials [3.92]: $$x_2(t) = ext{sinc}(t) \left[ rac{e^{j3\pi t} + e^{-j3\pi t}}{2} ight]$$ By applying the frequency shifting (modulation) property [3.92]: $$X_2(j\Omega) = rac{1}{2} \left[ ext{rect}\left(rac{\Omega - 3\pi}{2\pi} ight) + ext{rect}\left(rac{\Omega + 3\pi}{2\pi} ight) ight]$$ * The first term is non-zero over: $-1 \le rac{\Omega - 3\pi}{2\pi} \le 1 \implies 2\pi \le \Omega \le 4\pi$ [3.92]. * The second term is non-zero over: $-1 \le rac{\Omega + 3\pi}{2\pi} \le 1 \implies -4\pi \le \Omega \le -2\pi$ [3.92]. 3. **Multiply the spectra to evaluate $Y(j\Omega)$ [3.93]:** $$Y(j\Omega) = X_1(j\Omega) \cdot X_2(j\Omega)$$ * Support of $X_1(j\Omega) = [-2\pi, 2\pi]$ [3.92]. * Support of $X_2(j\Omega) = [-4\pi, -2\pi] \cup [2\pi, 4\pi]$ [3.92]. The intersection of these frequency supports consists solely of the isolated boundary points $\Omega = \pm 2\pi$ [3.93]. Since these points have a measure of zero, the product of the continuous spectral densities is identically zero everywhere [3.93]: $$Y(j\Omega) = X_1(j\Omega) \cdot X_2(j\Omega) = 0 \quad ext{for all } \Omega$$ 4. **Synthesize back to time domain [3.93]:** Taking the inverse CTFT of zero: $$\mathbf{y(t) = 0}$$ *(This is a classic exam "trap" question designed to penalize students who jump into complex time-domain integration without analyzing spectral supports first)*. --- ## 5. Common Mistakes That Cost Marks > [!danger] **The $1/(2\pi)$ Factor Omission on Frequency Convolution** > > In exams, a highly common error is writing $\mathcal{F}\{x_1(t)x_2(t)\} = X_1(j\Omega) * X_2(j\Omega)$. This forgets the scaling factor $1/(2\pi)$ [3.95, 3.96]. If you convolve in cyclic frequency ($f$ in Hz), this factor is absorbed, but in angular frequency ($\Omega$ in rad/s), **forgetting $1/(2\pi)$ results in an automatic zero-mark on scaling verification steps**. > [!warning] **Incorrect Differential Term $dt$ Replacement in Proofs** > > When performing the variable substitution $u = t - au$ in the convolution proof, students often write $dt o du$ but forget that $t$ in the exponential term must also be substituted with $t = u + au$. Leaving $e^{-j\Omega t}$ unmodified during the inner integration step represents a **critical algebraic failure** that halts the derivation. --- ## 6. PYQ Bank — Verbatim Questions & Answer Plans ### 6.1 PYQ 2022 / 2018 [5 Marks] **Question:** Prove that the convolution of two signals in the time domain is equivalent to the multiplication of their individual Fourier transforms in the frequency domain. * **Answer Plan:** 1. State the continuous-time convolution property equation clearly: $\mathcal{F}\{x_1(t) * x_2(t)\} = X_1(j\Omega)X_2(j\Omega)$ [3.84]. 2. Set up the forward Fourier transform integral of the convolution integral $y(t) = \int x_1( au) x_2(t- au) \, d au$ [3.84, 3.85]. 3. Interchange the integration order, justify using Dirichlet absolute integration prerequisites [3.85]. 4. Apply the substitution $u = t - au$ and split the complex exponentials [3.85]. 5. Factor out independent integrals to isolate $X_1(j\Omega) X_2(j\Omega)$ as shown in **Section 2.2** [3.85, 3.86]. ### 6.2 PYQ 2021 [8 Marks] **Question:** Let $y(t)$ be the convolution of two signals $x_1(t)$ and $x_2(t)$ defined by $x_1(t) = ext{sinc}(2t)$ and $x_2(t) = ext{sinc}(t)\cos(3\pi t)$. Determine the Fourier transform of $y(t)$. * **Answer Plan:** 1. State that $Y(j\Omega) = X_1(j\Omega)X_2(j\Omega)$ by the convolution theorem [3.92]. 2. Derive the scaled spectrum $X_1(j\Omega) = rac{1}{2} ext{rect}\left(rac{\Omega}{4\pi} ight)$ from the scaling property [3.91]. 3. Express $\cos(3\pi t)$ in exponential form and use the modulation property to derive $X_2(j\Omega)$ [3.92]. 4. Sketch or analytically state the non-zero frequency support boundaries of both spectra to prove they do not overlap [3.92, 3.93]. 5. Conclude that the product is $Y(j\Omega) = 0$ as shown in **Section 4.3** [3.93]. --- ## 7. Self-Check Before Moving On - [ ] Can you write out the complete 7-step mathematical proof of the time convolution property without looking? [2.2] - [ ] Do you know where the $1/(2\pi)$ factor comes from in the multiplication property? [3.2] - [ ] Can you quickly check if two modulated sinc signals will result in a zero-output response by mapping their frequency supports? [4.3] - [ ] Do you remember the support interval for a $ ext{rect}(\Omega/2W)$ gate spectrum? [4.2] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.* --- [[8.05_Time-Domain_Convolution_and_Multiplication_Properties|8.05 Time-Domain Convolution & Multiplication Properties]] | [[8.07_CTFT_System_Analysis_of_Continuous_Networks|8.07 CTFT System Analysis of Continuous Networks]] --- # 8.06 Frequency Spectra, Phase Spectra & Modulation > [!abstract] Core Idea > > For real-world engineering signals, continuous Fourier transforms ($X(j\Omega)$) are typically complex-valued, meaning they carry both magnitude and phase information. The **Amplitude (Magnitude) Spectrum** $|X(j\Omega)|$ and the **Phase Spectrum** $\angle X(j\Omega)$ provide a complete, visually intuitive frequency-domain portrait of the signal's harmonics. This note covers the plotting conventions for both **single-sided** (physical) and **double-sided** (mathematical) spectra, and details the **Modulation Theorem**—the physical basis for translating low-frequency message signals onto high-frequency carriers for long-distance propagation. --- ## 1. Polar Representation of Fourier Spectra Because the continuous-time Fourier transform is a complex function of the angular frequency $\Omega$, it is conventionally analyzed and plotted in **polar form**: $$X(j\Omega) = |X(j\Omega)| e^{j\angle X(j\Omega)} = |X(j\Omega)| e^{j \theta(\Omega)}$$ where: * **$|X(j\Omega)|$** represents the **Continuous Amplitude (Magnitude) Spectrum** *{representing the signal strength per unit frequency bandwidth}*. * **$\angle X(j\Omega)$ (or $\theta(\Omega)$)** represents the **Continuous Phase Spectrum** *{representing the harmonically related time-alignment offsets of the phasor components}*. ### 1.1 Symmetries of Spectra for Real-Valued Signals If the time-domain signal $x(t)$ is real-valued ($x^*(t) = x(t)$), its Fourier transform satisfies the conjugate-symmetric property: $X^*(j\Omega) = X(-j\Omega)$. This forces strict symmetry bounds on the spectra: 1. **Even Magnitude Symmetry:** The magnitude spectrum is an **even function** of frequency. $$|X(j\Omega)| = |X(-j\Omega)|$$ 2. **Odd Phase Symmetry:** The phase spectrum is an **odd function** of frequency. $$\angle X(-j\Omega) = -\angle X(j\Omega)$$ ``` EVEN MAGNITUDE SPECTRUM ODD PHASE SPECTRUM |X(jΩ)| ∠X(jΩ) ^ ^ * | * | * * | * | * * | * -------------+-------------> Ω * | * * | ----+--------+--------+----> Ω * | -Ω0 0 Ω0 * | -Ω0 ``` --- ## 2. Single-Sided vs. Double-Sided Spectra When visualizing discrete spectral lines of periodic sinusoids, we use two different plotting conventions: ### 2.1 Single-Sided Spectra * **Concept:** Plots only physical, non-negative frequencies ($\Omega \ge 0$). * **Basis:** Originates directly from the trigonometric Fourier Series expansion of a real signal: $$x(t) = A_0 + \sum_{n=1}^{\infty} A_n \cos(n\Omega_0 t + \theta_n)$$ * **Plotting Rules:** * Plot an impulse at each positive harmonic frequency $n\Omega_0$ with height equal to the full peak amplitude $A_n$. * Plot the corresponding phase $\theta_n$ at each positive harmonic frequency $n\Omega_0$. ### 2.2 Double-Sided Spectra * **Concept:** Plots both positive and negative frequencies ($\'\Omega \in (-\infty, \infty)$). * **Basis:** Originates from the complex exponential Fourier representation: $$x(t) = \sum_{n=-\infty}^{\infty} C_n e^{j n \Omega_0 t}$$ * **Euler Conversion:** To represent a real-valued cosine wave $A \cos(\Omega_0 t + \theta)$ exponentially, we expand it using Euler\'s formula: $$A \cos(\Omega_0 t + \theta) = \frac{A}{2} e^{j(\Omega_0 t + \theta)} + \frac{A}{2} e^{-j(\Omega_0 t + \theta)} = \left(\frac{A}{2} e^{j\theta}\right) e^{j\Omega_0 t} + \left(\_\frac{A}{2} e^{-j\theta}\right) e^{-j\Omega_0 t}$$ * **Plotting Rules:** * **Halved Amplitude Split:** The peak amplitude $A$ splits equally into two rotating phasors of length **$A/2$** located at $\pm \Omega_0$. * **Conjugate Phase Split:** The phase angle $\theta$ is plotted at the positive frequency $+\Omega_0$, while its negative conjugate **$-\theta$** is plotted at the negative frequency $-\Omega_0$. --- ## 3. The Modulation (Frequency Shifting) Theorem > [!theorem] **The Modulation Theorem** > > Let $m(t)$ be a low-frequency message signal *{with bandwidth bounded such that $M(j\Omega) = 0$ for $|\Omega| > \Omega_m$}*. Multiplying this message signal by a high-frequency sinusoidal carrier wave $c(t) = \cos(\Omega_c t)$ translates the entire baseband spectrum continuously in frequency, centering it at the carrier frequencies $\pm \Omega_c$: > > $$y(t) = m(t) \cos(\Omega_c t) \quad \leftrightarrow \quad Y(j\Omega) = \frac{1}{2} \left[ M(j(\Omega - \Omega_c)) + M(j(\Omega + \Omega_c)) \right]$$ ### 3.1 Mathematical Derivation We prove the theorem using the fundamental **Frequency Shifting Property** of the continuous-time Fourier transform: $$\text{If } m(t) \leftrightarrow M(j\Omega), \quad \text{then } e^{j\Omega_c t} m(t) \leftrightarrow M(j(\Omega - \Omega_c))$$ 1. Express the carrier wave $\cos(\Omega_c t)$ in complex exponential form using Euler\'s identity: $$\cos(\Omega_c t) = \frac{e^{j\Omega_c t} + e^{-j\Omega_c t}}{2}$$ 2. Substitute this expression into the modulated signal equation: $$y(t) = m(t) \cos(\Omega_c t) = m(t) \left[ \frac{e^{j\Omega_c t} + e^{-j\Omega_c t}}{2} \right] = \frac{1}{2} e^{j\Omega_c t} m(t) + \frac{1}{2} e^{-j\Omega_c t} m(t)$$ 3. Take the Fourier transform of both sides using the Linearity and Frequency Shifting properties: $$Y(j\Omega) = \mathcal{F}\{y(t)\} = \frac{1}{2} \mathcal{F}\{e^{j\Omega_c t} m(t)\} + \frac{1}{2} \mathcal{F}\{e^{-j\Omega_c t} m(t)\}$$ $$Y(j\Omega) = \frac{1}{2} M(j(\Omega - \Omega_c)) + \frac{1}{2} M(j(\Omega + \Omega_c)) \quad \blacksquare$$ ``` MESSAGE SPECTRUM M(jΩ) MODULATED DSB-SC SPECTRUM Y(jΩ) |M(jΩ)| |Y(jΩ)| ^ ^ ---|--- ---|--- ---|--- -Ωm 0 Ωm --> Ω -Ωc 0 Ωc --> Ω |<-2Ωm->| ``` > [!info] **Physical Context: Double-Sideband Suppressed-Carrier (DSB-SC)** > > This spectral translation is called **Double-Sideband Suppressed-Carrier (DSB-SC)** amplitude modulation. The original baseband signal requires a physical antenna size proportional to the wavelength $\lambda = c/f$. By shifting the message spectrum up to a high-frequency carrier $f_c$ (where $f_c \gg f_m$), we decrease the wavelength drastically, allowing highly efficient transmission over compact, practical antennas. --- ## 4. High-Yield Worked Examples (The Exam Classics) ### 4.1 Example 1: Double-Sided Spectrum of a Simple Cosine [PYQ 2019 Q5b/c - 1 Mark] **Question:** Sketch the double-sided frequency spectrum of the signal: $$x(t) = 10 \cos\left(20\pi t - \frac{\pi}{6}\right), \quad -\infty < t < \infty$$ #### Step-by-Step Mathematical Analysis: 1. **Identify Parameters:** * Peak Amplitude $A = 10$. * Fundamental Cyclic Frequency $f_0 = 10$ Hz $\implies$ Angular Frequency $\Omega_0 = 20\pi$ rad/s. * Phase Angle $\theta = -\frac{\pi}{6}$ rad. 2. **Apply Euler's Expansion:** $$x(t) = 5 e^{-j\pi/6} e^{j20\pi t} + 5 e^{j\pi/6} e^{-j20\pi t}$$ 3. **Determine Double-Sided Coefficients:** * **At $\Omega = +20\pi$ rad/s:** Amplitude $= 5$, Phase $= -\frac{\pi}{6}$ rad (or $-30^\circ$). * **At $\Omega = -20\pi$ rad/s:** Amplitude $= 5$, Phase $= +\frac{\pi}{6}$ rad (or $+30^\circ$). * All other frequency components are exactly zero. #### Plotting the Spectra: ``` DOUBLE-SIDED AMPLITUDE SPECTRUM DOUBLE-SIDED PHASE SPECTRUM |X(jΩ)| ∠X(jΩ) (rad) ^ ^ | | pi/6 5 | 5 | o | | | | | ------+-------+-------+------> Ω -------+----+-------> Ω -20pi 0 20pi -20pi | 20pi | o | -pi/6 ``` --- ### 4.2 Example 2: Complete Single & Double-Sided Spectra [PYQ 2017 Question 5b - 5 Marks] **Question:** Sketch the single and double-sided frequency spectra of the following signal: $$x(t) = 25 \cos\left(5\pi t - \frac{\pi}{2}\right)$$ #### Step-by-Step Mathematical Analysis: 1. **Identify Parameters:** * Peak Amplitude $A = 25$. * Angular Frequency $\Omega_0 = 5\pi$ rad/s. * Phase Angle $\theta = -\frac{\pi}{2}$ rad (or $-90^\circ$). 2. **Single-Sided Representation:** * Plots only the positive physical frequency $\Omega = 5\pi$ rad/s. * **Amplitude Line:** Height $= A = 25$ at $\Omega = 5\pi$. * **Phase Line:** Angle $= \theta = -\frac{\pi}{2}$ rad at $\Omega = 5\pi$. 3. **Double-Sided Representation:** * Phasor amplitude is halved: $A/2 = 12.5$. * **At $\Omega = +5\pi$ rad/s:** Amplitude $= 12.5$, Phase $= -\frac{\pi}{2}$ rad. * **At $\Omega = -5\pi$ rad/s:** Amplitude $= 12.5$, Phase $= +\frac{\pi}{2}$ rad. #### Plotting both spectrum sets: ``` SINGLE-SIDED SPECTRA (Ω >= 0) DOUBLE-SIDED SPECTRA (All Ω) |X(jΩ)| single |X(jΩ)| double ^ ^ 25 | | 12.5 | | | | | | | | ---+-------+---------> Ω -----+-----+-----+-----> Ω 0 5pi -5pi 0 5pi ∠X(jΩ) single ∠X(jΩ) double (rad) ^ ^ | | pi/2 ---+-------+---------> Ω | o 0 | 5pi ----+---+---+-----> Ω | o -5pi | 5pi |-pi/2 | o |-pi/2 ``` --- ### 4.3 Example 3: The Sine Phase-Conversion Trap [PYQ 2018 Question 5b/c - 2 Marks] **Question:** Sketch the single and double-sided frequency spectra of the signal: $$x(t) = 12 \sin\left(5\pi t - \frac{\pi}{2}\right)$$ #### Step-by-Step Mathematical Analysis: > [!danger] **Critical Exam Trap: Converting Sine to Cosine** > > Spectral plotting conventions are strictly derived from the cosine basis wave: $A \cos(\Omega_0 t + \theta)$. Reading the phase directly from a sine representation will result in a **zero-mark** penalty. You must always convert the sine function to cosine first using the identity: $\sin(\phi) = \cos\left(\phi - \frac{\pi}{2}\right)$. 1. **Apply Phase Conversion:** $$x(t) = 12 \sin\left(5\pi t - \frac{\pi}{2}\right) = 12 \cos\left(5\pi t - \frac{\pi}{2} - \frac{\pi}{2}\right) = 12 \cos\left(5\pi t - \pi\right)$$ 2. **Identify Converted Parameters:** * Peak Amplitude $A = 12$. * Angular Frequency $\Omega_0 = 5\pi$ rad/s. * Phase Angle $\theta = -\pi$ rad (or $+\pi$ rad, since $\pm\pi$ represent the same physical point). 3. **Single-Sided Representation:** * **At $\Omega = 5\pi$ rad/s:** Amplitude $= 12$, Phase $= -\pi$ rad. 4. **Double-Sided Representation:** * Halved amplitude: $A/2 = 6$. * **At $\Omega = +5\pi$ rad/s:** Amplitude $= 6$, Phase $= -\pi$ rad. * **At $\Omega = -5\pi$ rad/s:** Amplitude $= 6$, Phase $= +\pi$ rad (due to odd symmetry $\angle X(-j\Omega) = -\angle X(j\Omega)$). #### Plotting the Spectra: ``` SINGLE-SIDED AMPLITUDE & PHASE DOUBLE-SIDED AMPLITUDE & PHASE |X| single |X| double ^ ^ 12 | | 6 | | | ---+-----+-----> Ω +---+-----+-----> Ω 0 5pi -5pi 0 5pi ∠X single ∠X double (rad) ^ ^ | pi | o ---+-----+-----> Ω +---+-----+-----> Ω 0 | 5pi -5pi | 5pi | o | o |-pi -pi| ``` --- ### 4.4 Example 4: Large-Mark Symmetrical Double-Sided Spectrum [PYQ 2016 Q5b - 8 Marks] **Question:** Draw the double-sided frequency spectrum of: $$x(t) = 8 \sin\left(20\pi t - \frac{\pi}{4}\right), \quad -\infty < t < \infty$$ #### Step-by-Step Mathematical Analysis: 1. **Convert Sine to Cosine first:** $$x(t) = 8 \cos\left(20\pi t - \frac{\pi}{4} - \frac{\pi}{2}\right) = 8 \cos\left(20\pi t - \frac{3\pi}{4}\right)$$ 2. **Identify Parameters:** * Peak Amplitude $A = 8$. * Angular Frequency $\Omega_0 = 20\pi$ rad/s. * Phase Angle $\theta = -\frac{3\pi}{4}$ rad (or $-135^\circ$). 3. **Determine Double-Sided Components:** * Halved Amplitude: $A/2 = 4$. * **At $\Omega = +20\pi$ rad/s:** Amplitude $= 4$, Phase $= -\frac{3\pi}{4}$ rad. * **At $\Omega = -20\pi$ rad/s:** Amplitude $= 4$, Phase $= +\frac{3\pi}{4}$ rad (odd symmetry). #### Plotting the Spectra: ``` DOUBLE-SIDED AMPLITUDE SPECTRUM DOUBLE-SIDED PHASE SPECTRUM |X(jΩ)| ∠X(jΩ) (rad) ^ ^ | 3pi/4 | o 4 | 4 | | | | | | | ------+-------+-------+------> Ω -------+---+-------> Ω -20pi 0 20pi -20pi | 20pi | o | -3pi/4 ``` --- ## 5. Common Mistakes That Cost Marks > [!danger] **The Sine-Basis Phase Reading Slip** > > This is the #1 reason students lose marks on spectral sketching questions. If the given equation is written in terms of a sine wave (e.g., $\sin(\Omega_0 t + heta)$), you **must** convert it to a cosine wave using $\cos(\Omega_0 t + heta - \pi/2)$ before reading the phase angle. Reading phase directly from a sine wave yields a phase error of exactly $-90^\circ$, causing an automatic **zero-mark** evaluation on the phase spectrum plot. > [!warning] **Double-Sided Amplitude Halving Omission** > > When converting a time-domain sinusoid to a double-sided exponential spectrum, you must split the peak amplitude $A$ in half ($A/2$). Forgetting to divide by 2 (e.g., plotting lines of height 10 instead of 5 for a $10\cos(\Omega_t)$ wave) is a standard math-syntax error that costs up to 50% of the question\'s total marks. > [!warning] **Even Phase Plotting Violation** > > The phase spectrum of any real-valued signal must have **odd symmetry** ($ngle X(-j\Omega) = -ngle X(j\Omega)$). If you draw positive phase angles at both positive and negative frequencies, your plot violates basic complex number theory, leading to a major point reduction. --- ## 6. PYQ Bank — Verbatim Questions & Answer Plans ### 6.1 PYQ 2019 Question 5c [1 Mark] **Question:** Sketch the double-sided frequency spectrum of the signal $x(t) = 10 \cos(20\pi t - \pi/6), -\infty < t < \infty$. * **Answer Plan:** 1. Identify $A = 10, \Omega_0 = 20\pi,$ and $ heta = -\pi/6$ rad. 2. Divide peak amplitude to get $A/2 = 5$. 3. State double-sided values: amplitude 5 at $\pm 20\pi$, phase $-\pi/6$ at $+20\pi$, and phase $+\pi/6$ at $-20\pi$. 4. Plot the amplitude and phase spectrum as shown in **Section 4.1**. ### 6.2 PYQ 2017 Question 5b [5 Marks] **Question:** Sketch the single and double sided frequency spectra of the following signal: $x(t) = 25\cos(5\pi t - \pi/2)$. * **Answer Plan:** 1. Identify base parameters: $A = 25, \Omega_0 = 5\pi, heta = -\pi/2$ rad. 2. Draft single-sided plots at $+5\pi$ with amplitude 25 and phase $-\pi/2$. 3. Draft double-sided plots at $\pm 5\pi$ with amplitude 12.5 and phase $\mp \pi/2$. 4. Render both sets clearly labeled as shown in **Section 4.2**. ### 6.3 PYQ 2018 Question 5c [2 Marks] **Question:** Sketch single and double-sided frequency spectra of the signal $x(t) = 12 \sin(5\pi t - \pi/2)$. * **Answer Plan:** 1. Write down the sine-to-cosine conversion: $12 \sin(5\pi t - \pi/2) = 12 \cos(5\pi t - \pi)$. 2. Extract parameters: $A = 12, \Omega_0 = 5\pi, heta = -\pi$ rad. 3. Calculate single-sided parameters and double-sided parameters ($A/2 = 6, heta_+ = -\pi, heta_- = +\pi$). 4. Draw both plots with accurate axis ticks as shown in **Section 4.3**. --- ## 7. Self-Check Before Moving On - [ ] Can you explain why real-valued signals must have even amplitude and odd phase spectra? [1.1] - [ ] Do you know how to convert a sine signal to cosine before extracting phase parameters? [4.3] - [ ] Have you memorized the amplitude halving rule ($A o A/2$) for double-sided spectra? [2.2] - [ ] Can you state and derive the Modulation Theorem in under 2 minutes? [3.1] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.* --- [[8.06_Frequency_Spectra_Phase_Spectra_and_Modulation|8.06 Frequency Spectra, Phase Spectra & Modulation]] | [[9.00_Chapter_Map_-_Laplace_Transform_and_s-Domain_Analysis|9.00 Chapter Map - Laplace Transform & s-Domain Analysis]] --- # 8.07 CTFT System Analysis of Continuous Networks > [!abstract] Core Idea > > **Continuous-Time Fourier Transform (CTFT)** system analysis extends frequency-domain modeling to physical electrical networks. By representing circuit components through their frequency-domain impedances, we can transform integro-differential loop equations into simple algebraic relations *{where differential operators are replaced by multipliers}* [3.40, 3.68]. This note details the frequency-domain modeling of a passive series low-pass RC network, derives its continuous frequency response $H(j\Omega)$ [3.82, 3.84], and utilizes inverse CTFT integrals to solve the time-domain output responses for standard decaying exponential inputs. --- ## 1. Impedance Modeling in the Fourier Domain In time-domain circuit analysis, the relationship between voltage and current for resistors, inductors, and capacitors is governed by differential or integral equations [2.26, 2.55]. When we apply the **CTFT** to these relationships, the time differentiation property ($rac{d}{dt} \leftrightarrow j\Omega$) transforms them into algebraic algebraic equations [3.40, 3.68]: ### 1.1 Resistor ($R$) The voltage-current relation is algebraic in both domains: $$v_R(t) = R \cdot i(t) \quad \leftrightarrow \quad V_R(j\Omega) = R \cdot I(j\Omega)$$ * **Fourier Impedance:** $Z_R(j\Omega) = R \quad$ (measured in Ohms, $\Omega$) ### 1.2 Inductor ($L$) The voltage is proportional to the rate of change of current: $$v_L(t) = L rac{di(t)}{dt} \quad \leftrightarrow \quad V_L(j\Omega) = j\Omega L \cdot I(j\Omega)$$ * **Fourier Impedance:** $Z_L(j\Omega) = j\Omega L \quad$ (measured in Ohms, $\Omega$) ### 1.3 Capacitor ($C$) The current is proportional to the rate of change of voltage: $$i(t) = C rac{dv_C(t)}{dt} \quad \leftrightarrow \quad I(j\Omega) = j\Omega C \cdot V_C(j\Omega)$$ Solving for the capacitor voltage yields: $$V_C(j\Omega) = rac{1}{j\Omega C} \cdot I(j\Omega)$$ * **Fourier Impedance:** $Z_C(j\Omega) = rac{1}{j\Omega C} \quad$ (measured in Ohms, $\Omega$) > [!danger] **Critical Exam Checkpoint: Fourier vs. Laplace Impedance** > > Always write the capacitor impedance in the Fourier domain as $rac{1}{j\Omega C}$ (or $rac{1}{j\omega C}$). Writing it as $j\Omega C$ is a dimensional error. Similarly, writing it as $rac{1}{sC}$ belongs strictly to s-domain Laplace analysis (Chapter 9) and will result in a **marks deduction** if mixed into a Fourier transform question! --- ## 2. Deriving the Continuous Frequency Response $H(j\Omega)$ Consider a passive series low-pass RC network with input voltage $x(t) = v_{ ext{in}}(t)$ and output voltage $y(t) = v_{ ext{out}}(t)$ measured across the capacitor. ``` R o------/\/\/\/\-------+-------o + | + x(t) === C y(t) - | - o---------------------+-------o ``` ### 2.1 The Integro-Differential Time-Domain Equation Applying Kirchhoff's Voltage Law (KVL) around the single loop with current $i(t)$ yields: $$x(t) = R \cdot i(t) + y(t)$$ Since the loop current is driven by the capacitor's charge storage, $i(t) = C rac{dy(t)}{dt}$ [2.26, 2.55]. Substituting this current yields the governing first-order differential equation: $$RC rac{dy(t)}{dt} + y(t) = x(t)$$ ### 2.2 Algebraic Frequency-Domain Solution Applying the CTFT to both sides of the differential equation, using the linearity [3.62] and time-differentiation properties [3.67], we get: $$RC \left( j\Omega Y(j\Omega) ight) + Y(j\Omega) = X(j\Omega)$$ Factor out the output spectrum $Y(j\Omega)$ [3.84]: $$Y(j\Omega) \left[ 1 + j\Omega RC ight] = X(j\Omega)$$ The **continuous transfer function** $H(j\Omega)$ is defined as the ratio of the output spectrum to the input spectrum [3.83, 3.84]: $$\mathbf{H(j\Omega) = rac{Y(j\Omega)}{X(j\Omega)} = rac{1}{1 + j\Omega RC} = rac{rac{1}{RC}}{j\Omega + rac{1}{RC}}}$$ ### 2.3 Calculating the 3-dB Cutoff Frequency ($\Omega_c$) The system's cutoff frequency occurs when the output power halves, which corresponds to the magnitude dropping to $1/\sqrt{2}$ of its maximum DC value: $$\left| H(j\Omega_c) ight| = rac{1}{\sqrt{1 + (\Omega_c RC)^2}} = rac{1}{\sqrt{2}}$$ This mathematical condition is satisfied when: $$\Omega_c RC = 1 \implies \mathbf{\Omega_c = rac{1}{RC}} \quad ext{(rad/s)}$$ The cyclic cutoff frequency in Hertz is: $$f_c = rac{1}{2\pi RC} \quad ext{(Hz)}$$ Substituting $\Omega_c$ back into our transfer function gives the standard normalized LPF form: $$H(j\Omega) = rac{\Omega_c}{j\Omega + \Omega_c}$$ --- ## 3. High-Yield Worked "Exam Killers" ### 3.1 Concept-verbatim Exam Question [KUET ECE 2107 - 8 Marks] **Question:** What is filtering? What are the applications of it? #### Verbatim Scoring Answer Plan: 1. **Define Filtering:** State that filtering is a frequency-domain circuit operation designed to selectively pass certain frequency components of a signal without distortion while completely blocking or suppressing other frequency components [1.7.2]. 2. **Passband & Stopband:** Explain that the range of frequencies allowed to pass through is the **passband**, and the range of frequencies blocked is the **stopband** [1.7.2]. 3. **Four Primary Classifications:** Define and sketch the ideal magnitude responses of [1.7.2]: * **Low-Pass Filter (LPF):** Passes frequencies below a cutoff frequency $\Omega_c$, blocking higher frequencies [1.7.2]. * **High-Pass Filter (HPF):** Passes frequencies above $\Omega_c$, blocking lower frequencies [1.7.2]. * **Band-Pass Filter (BPF):** Passes frequencies between two cutoff limits $\Omega_{c1} < \Omega_{c2}$ [1.7.2]. * **Band-Stop Filter (BSF):** Blocks frequencies between two cutoff limits $\Omega_{c1} < \Omega_{c2}$ [1.7.2]. 4. **Practical Engineering Applications:** * **Noise Suppression:** Eliminating high-frequency thermal or electromagnetic noise from sensor readings. * **Anti-Aliasing:** Band-limiting a continuous signal before sampling to prevent spectral overlap [1.05]. * **Signal Demodulation:** Extracting low-frequency message signals from high-frequency carriers in AM radio [3.88]. * **Power Supplies:** Converting rectified AC waveforms into smooth DC voltages by filtering out ripples. --- ### 3.2 Concept-verbatim Exam Question [KUET ECE 2107 - 6 Marks] **Question:** Explain transfer function and explain its significance in circuit analysis. #### Verbatim Scoring Answer Plan: 1. **Define Transfer Function:** Define the transfer function $H(j\Omega)$ as the algebraic ratio of the output Fourier spectrum $Y(j\Omega)$ to the input Fourier spectrum $X(j\Omega)$ under the assumption of zero initial energy stored in the circuit's reactive elements [3.82, 3.83]: $$H(j\Omega) = rac{Y(j\Omega)}{X(j\Omega)}$$ 2. **Explain the Physical Significance:** * **Algebraic Simplification:** It converts complex differential equations in the time domain into simple algebraic multiplications in the frequency domain: $Y(j\Omega) = X(j\Omega) H(j\Omega)$ [3.3]. * **System Characterization:** It fully characterizes the LTI circuit as a "black box" system, completely independent of the applied input signal. * **Magnitude & Phase Mapping:** It dictates how each input frequency component $\Omega$ is scaled in amplitude by $|H(j\Omega)|$ and shifted in phase by $ngle H(j\Omega)$ during transmission [3.85]. * **Stability & Synthesis Analysis:** It provides immediate visual insight into the circuit's bandwidth, resonant peaks, and poles/zeros, making filter design and stability checks highly intuitive [3.3, 3.91]. --- ### 3.3 Case Study 1: Input $x(t) = e^{-t/RC}u(t)$ [KUET ECE 2107 Classic] **Question:** A passive series low-pass RC network has an input voltage of $x(t) = e^{-t/RC}u(t)$. Assuming the circuit is initially relaxed, find its frequency response $H(j\Omega)$, output spectrum $Y(j\Omega)$, and the time-domain output voltage $y(t)$ using the inverse Fourier transform. #### Step-by-Step Mathematical Solution: 1. **Identify the Network Transfer Function:** From Section 2.2, the transfer function of the low-pass RC network is: $$H(j\Omega) = rac{rac{1}{RC}}{j\Omega + rac{1}{RC}}$$ 2. **Calculate the Input Fourier Spectrum $X(j\Omega)$:** The input signal is a single-sided decaying exponential $x(t) = e^{-lpha t}u(t)$ where $lpha = rac{1}{RC}$. Applying the standard decaying exponential CTFT pair [3.69, 3.83]: $$X(j\Omega) = \mathcal{F}\left\{ e^{-t/RC}u(t) ight\} = rac{1}{j\Omega + rac{1}{RC}}$$ 3. **Formulate the Output Spectrum $Y(j\Omega)$:** Using the convolution property ($\mathcal{F}\{x * h\} = X \cdot H$) [3.75, 3.81]: $$Y(j\Omega) = X(j\Omega) \cdot H(j\Omega)$$ $$Y(j\Omega) = \left( rac{1}{j\Omega + rac{1}{RC}} ight) \cdot \left( rac{rac{1}{RC}}{j\Omega + rac{1}{RC}} ight)$$ $$Y(j\Omega) = rac{rac{1}{RC}}{\left( j\Omega + rac{1}{RC} ight)^2}$$ 4. **Determine the Time-Domain Output $y(t)$ via Inverse CTFT:** We recall our standard frequency differentiation property which establishes the dual-pole transform pair [3.70, 3.83]: $$t e^{-lpha t} u(t) \quad \leftrightarrow \quad rac{1}{(j\Omega + lpha)^2}$$ For $lpha = rac{1}{RC}$, this maps directly to: $$t e^{-t/RC} u(t) \quad \leftrightarrow \quad rac{1}{\left(j\Omega + rac{1}{RC} ight)^2}$$ By the linearity property of the Fourier transform, multiplying both sides by the scalar constant $rac{1}{RC}$ yields [3.62]: $$\mathcal{F}^{-1}\left\{ rac{rac{1}{RC}}{\left( j\Omega + rac{1}{RC} ight)^2} ight\} = rac{t}{RC} e^{-t/RC} u(t)$$ Therefore, the final output voltage response is: $$\mathbf{y(t) = rac{t}{RC} e^{-t/RC} u(t)}$$ *(Note: This mathematically matches the time-domain convolution integral result derived in Note 2.02, proving the absolute consistency of transform-domain analysis!)* --- ### 3.4 Case Study 2: Input $x(t) = t e^{-t/RC}u(t)$ [Advanced Challenge Classic] **Question:** Analyze the same low-pass RC network with a ramp-weighted decaying input voltage of $x(t) = t e^{-t/RC} u(t)$ to find its time-domain output response $y(t)$. #### Step-by-Step Mathematical Solution: 1. **Identify the Transfer Function:** $$H(j\Omega) = rac{rac{1}{RC}}{j\Omega + rac{1}{RC}}$$ 2. **Calculate the Input Fourier Spectrum $X(j\Omega)$:** Using the dual-pole frequency differentiation transform pair: $$X(j\Omega) = \mathcal{F}\left\{ t e^{-t/RC} u(t) ight\} = rac{1}{\left( j\Omega + rac{1}{RC} ight)^2}$$ 3. **Formulate the Output Spectrum $Y(j\Omega)$:** $$Y(j\Omega) = X(j\Omega) \cdot H(j\Omega) = \left( rac{1}{\left( j\Omega + rac{1}{RC} ight)^2} ight) \cdot \left( rac{rac{1}{RC}}{j\Omega + rac{1}{RC}} ight)$$ $$Y(j\Omega) = rac{rac{1}{RC}}{\left( j\Omega + rac{1}{RC} ight)^3}$$ 4. **Evaluate the Inverse CTFT using higher-order poles:** Recall the generalized n-th order decaying exponential transform pair [3.72, 3.83]: $$rac{t^{n-1}}{(n-1)!} e^{-lpha t} u(t) \quad \leftrightarrow \quad rac{1}{(j\Omega + lpha)^n}$$ For $n=3$, this evaluates to: $$rac{t^2}{2!} e^{-lpha t} u(t) = rac{t^2}{2} e^{-lpha t} u(t) \quad \leftrightarrow \quad rac{1}{(j\Omega + lpha)^3}$$ Substitute $lpha = rac{1}{RC}$: $$rac{t^2}{2} e^{-t/RC} u(t) \quad \leftrightarrow \quad rac{1}{\left( j\Omega + rac{1}{RC} ight)^3}$$ Applying the linearity property to incorporate the $rac{1}{RC}$ numerator scale factor [3.62]: $$\mathcal{F}^{-1}\left\{ rac{rac{1}{RC}}{\left( j\Omega + rac{1}{RC} ight)^3} ight\} = rac{1}{RC} \cdot \left[ rac{t^2}{2} e^{-t/RC} u(t) ight]$$ Therefore, the final output voltage is: $$\mathbf{y(t) = rac{t^2}{2RC} e^{-t/RC} u(t)}$$ --- ## 4. Visualizing Amplitude and Phase Spectra To understand how the low-pass network alters the input signal, we express the transfer function in polar form [3.85]: $$H(j\Omega) = |H(j\Omega)| e^{jngle H(j\Omega)}$$ ### 4.1 Magnitude Response (Amplitude Spectrum) $$|H(j\Omega)| = rac{1}{\sqrt{1 + (\Omega RC)^2}} = rac{\Omega_c}{\sqrt{\Omega^2 + \Omega_c^2}}$$ * **At DC ($\Omega = 0$):** $|H(0)| = 1 \quad$ (signals pass with unity gain). * **At Cutoff ($\Omega = \Omega_c$):** $|H(j\Omega_c)| = rac{1}{\sqrt{2}} pprox 0.707 \quad$ (power is halved, $-3 ext{ dB}$). * **At High Frequencies ($\Omega o \infty$):** $|H(j\Omega)| o 0 \quad$ (high frequencies are heavily attenuated). ``` |H(jΩ)| 1.0 +--------- | 0.707 | \-----+ (Cutoff point at Ω_c) | 0.0 +---+---------------+---> Ω 0 Ω_c ``` ### 4.2 Phase Response (Phase Spectrum) $$ngle H(j\Omega) = - an^{-1}(\Omega RC) = - an^{-1}\left( rac{\Omega}{\Omega_c} ight)$$ * **At DC ($\Omega = 0$):** $ngle H(0) = 0^{\circ}$. * **At Cutoff ($\Omega = \Omega_c$):** $ngle H(j\Omega_c) = - an^{-1}(1) = -45^{\circ}$. * **At High Frequencies ($\Omega o \infty$):** $ngle H(j\Omega) o -90^{\circ}$. --- ## 5. Common Mistakes That Cost Marks > [!danger] **The Impedance Variable Substitution Trap** > > Do not mix Laplace and Fourier variable conventions! Writing $Z_C = rac{1}{sC}$ in a Fourier analysis question is a **critical notation error**. In Chapter 8 (CTFT), the complex frequency variable is strictly $s = j\Omega \implies Z_C(j\Omega) = rac{1}{j\Omega C}$. Keep your domains separated to secure full credit from the examiner. > [!warning] **The Cutoff Frequency Radian-vs-Cyclic Trap** > > KUET questions often specify cutoff frequencies in Hertz ($f_c$) but require loop calculus in radians per second ($\Omega_c$). > * If an exam problem states a cutoff frequency is $100 ext{ Hz}$, you **MUST** convert to radians before solving: $\Omega_c = 2\pi(100) pprox 628.3 ext{ rad/s}$. > * Forgetting the $2\pi$ multiplier results in a scaling error of nearly **6.28 times**, completely invalidating your numerical results! --- ## 6. PYQ Bank — Verbatim Questions & Answer Plans ### 6.1 Verbatim KUET 2024 / 2022 Question [10 Marks] **Question:** Determine the output response of a passive series low-pass RC network for an input $x(t) = e^{-t/RC}u(t)$ using the Fourier transform method. * **Answer Plan:** 1. Define the system's differential equation: $RC rac{dy(t)}{dt} + y(t) = x(t)$. 2. Apply the CTFT to both sides to derive $H(j\Omega) = rac{rac{1}{RC}}{j\Omega + rac{1}{RC}}$ as shown in **Section 2.2**. 3. Transform the input: $X(j\Omega) = rac{1}{j\Omega + rac{1}{RC}}$ [3.83]. 4. Multiply spectra: $Y(j\Omega) = rac{rac{1}{RC}}{(j\Omega + 1/RC)^2}$ [3.81]. 5. State and apply the frequency-differentiation inverse CTFT pair to prove $y(t) = rac{t}{RC} e^{-t/RC}u(t)$ [3.70]. ### 6.2 Foundational KUET ECE 2107 Question [8 Marks] **Question:** Explain the concept of filtering and discuss its practical applications in modern communication systems. * **Answer Plan:** 1. Define filtering as a frequency-domain band-limiting operation [1.7.2]. 2. Sketch the four filter types (LPF, HPF, BPF, BSF) with annotated passbands, stopbands, and cutoff edges [1.7.2]. 3. Outline communication applications: anti-aliasing pre-filtering [1.05], AM/FM tuning, noise suppression, and signal demodulation [3.88]. --- ## 7. Self-Check Before Moving On - [ ] Can you derive the continuous frequency response $H(j\Omega)$ of a low-pass RC filter from KVL? [2.2] - [ ] Do you know why the cutoff frequency of a series RC filter is mathematically defined as $\Omega_c = 1/RC$? [2.3] - [ ] Can you perform step-by-step inverse CTFT derivations for output spectra with multiple poles (e.g. order $n=2$ and $n=3$)? [3.3, 3.4] - [ ] Are you aware of the cyclic-versus-angular cutoff frequency trap ($f_c$ vs. $\Omega_c$)? [5.2] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 04 Network Theory.pdf, Rabiul sir class note.pdf (Lec 6).*