8.04 CTFT Pairs for Singularity & Common Functions | 8.06 Frequency Spectra, Phase Spectra & Modulation
8.05 Time-Domain Convolution & Multiplication Properties
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 [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]:
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 and , then [3.84]:
2.2 Mathematical Proof (12-Mark Exam Favorite) [PYQ 2022, 2018]
Let . By definition, we write [3.84]:
Applying the forward Continuous-Time Fourier Transform (CTFT) integral to yields [3.85]:
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.*