8.04 CTFT Pairs for Singularity and Common Functions

8.03 Rayleigh’s Energy Theorem & Spectral Density | 8.05 Time-Domain Convolution & Multiplication Properties

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8.04 CTFT Pairs for Singularity & Common Functions

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.

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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.

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 ${\int }_{-\infty }^{\infty }x(t)\delta (t-t_0)dt=x(t_0)$:

ight|_{t=0} = e^{0} = 1$$

Therefore, we obtain the fundamental pair: $\delta (\mathbf{t})\leftrightarrow 1$

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.

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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.

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 ($\delta (-\Omega )=\delta (\Omega )$): $1\leftrightarrow 2\pi \delta (\mathbf{\Omega })$

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 $2\pi$}.

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1.3 The Signum Function, $extsgn(t)$

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.

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:

ight} = j\Omega \mathcal{F}{ ext{sgn}(t)}$$

3. Substitute the CTFT of the impulse ($2\mathcal{F}\{\delta (t)\}=2$): $j\Omega \mathcal{F}\{extsgn(t)\}=2$ 4. Solve for the transform: \mathbf{\mathcal{F}\{ ext{sgn}(t)\} = rac{2}{j\Omega}}

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1.4 The Heaviside Unit Step Function, $u(t)$

The Unit Step Function acts as a DC switch closing at $t=0$.

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.

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:

ight]$$

\mathbf{\mathcal{F}\{u(t)\} = rac{1}{j\Omega} + \pi \delta(\Omega)}

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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.

Gaussian Pulse Self-Transform Proof

Let the time-domain signal be defined as: $x(t)=e^{-b{t}^2}extwhereb>0$

1. Set up the forward CTFT integral: $X(j\Omega )={\int }_{-\infty }^{\infty }{e}^{-b{t}^2}{e}^{-j\Omega t}dt$
2. Group the exponents by completing the square:

ight)$$

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:

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 ${\int }_{-\infty }^{\infty }{e}^{-a{u}^2}dau=\sqrt{\pi }$: 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:

ight} = e^{-\pi f^2}} \quad ext{or} \quad \mathbf{e^{-\pi t^2} \leftrightarrow e^{-rac{\Omega^2}{4\pi}}}$$

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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.

Single-Sided Decay Derivation

Given $x(t)=e^{-at}u(t)$ with $a>0$: $X(j\Omega )={\int }_{-\infty }^{\infty }{e}^{-at}u(t)e^{-j\Omega t}dt={\int }_0^{\infty }{e}^{-(a+j\Omega )t}dt$

ight|_{0}^{\infty} = 0 - \left( -rac{1}{a+j\Omega} ight)$$

\mathbf{e^{-at}u(t) \leftrightarrow rac{1}{a + j\Omega}}

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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

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: $X(j\Omega )={\int }_{-\infty }^0{e}^{at}{e}^{-j\Omega t}dt+{\int }_0^{\infty }{e}^{-at}{e}^{-j\Omega t}dt$ $={\int }_{-\infty }^0{e}^{(a-j\Omega )t}dt+{\int }_0^{\infty }{e}^{-(a+j\Omega )t}dt$

ight]{-\infty}^{0} + \left[ rac{e^{-(a+j\Omega)t}}{-(a+j\Omega)} ight]{0}^{\infty}$$

Since $a>0$, as $to-\infty$, $e^{at}o0$. As $to\infty$, $e^{-at}o0$. 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}}

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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 $extsgn(t)$	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)$**	$T ext{sinc}\left(rac{\Omega T}{2\pi}
ight)$	$Textsinc(fT)$

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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:

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}:

ight] = rac{10\sin(3\Omega)}{\Omega}$$

ight] = rac{10\sin(2\Omega)}{\Omega}$4.CombinetheresultsusingtheLinearityProperty:$\mathbf{F(j\Omega) = 10 rac{\sin(3\Omega) + \sin(2\Omega)}{\Omega}}$$

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5.2 Symmetrical Discrete Impulse Train

Question: Calculate the Fourier transform of the discrete impulse train:

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.*