8.01 Foundation of the Continuous-Time Fourier Transform (CTFT) | 8.03 Rayleigh’s Energy Theorem & Spectral Density
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].
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 and , then for any complex constants and [3.56]:
1.2 Mathematical Proof
Using the definition of the forward CTFT integral [3.45]:
Distribute the complex exponential term across the brackets:
Since integration is a linear operator, we factor out the constants and [3.56]:
\mathcal{F}\{a x_1(t) + b x_2(t)\} = a X_1(j\Omega) + b X_2(j\Omega) \quad lacksquare
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 , then for a real time delay [3.60]:
2.2 Mathematical Proof
By definition of the forward CTFT [3.45, 3.61]:
Perform a change of variable. Let and [3.61]. The limits of integration remain unchanged:
Factor out the constant exponent term since it does not depend on the variable of integration [3.61]:
Since is a dummy variable of integration [3.120], the integral is exactly [3.45]:
\mathcal{F}\{x(t - t_0)\} = X(j\Omega) e^{-j\Omega t_0} \quad lacksquare
2.3 Physical Intuition
- Magnitude Spectrum: [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 .
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 , then for any non-zero real constant [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.*