8.06 Frequency Spectra, Phase Spectra & Modulation | 9.00 Chapter Map - Laplace Transform & s-Domain Analysis

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8.07 CTFT System Analysis of Continuous Networks

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.

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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: $v_R(t)=R\cdot i(t)\leftrightarrow V_R(j\Omega )=R\cdot I(j\Omega )$

• Fourier Impedance: $Z_R(j\Omega )=R$ (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: $Z_L(j\Omega )=j\Omega L$ (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, $\Omega$)

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!

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2. Deriving the Continuous Frequency Response $H(j\Omega )$

Consider a passive series low-pass RC network with input voltage $x(t)=v_{extin}(t)$ and output voltage $y(t)=v_{extout}(t)$ 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:

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).*