Fourier Transforms and Spectra of Two-Sided Exponentials: 22-Elec-A3 Signals and Communications, May 2013, Q4

Question 4: Fourier Transforms and Spectra of Two-Sided Exponentials (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two signals built from the two-sided exponential $g(t)=e^{-2|t|}$: (a) the same pulse delayed by 1 s; (b) the same pulse multiplied by $t$. The decay constant is $a=2\ \text{s}^{-1}$ in both cases.

Find. $X(j\omega)$ for each, plus sketches of $x(t)$, $|X(j\omega)|$ and $\arg X(j\omega)$.

Approach. Establish the parent pair $e^{-a|t|}\leftrightarrow\dfrac{2a}{a^{2}+\omega^{2}}$ once, then reach both answers with a single property each: the time-shift property for (a) and the frequency-differentiation property for (b). Neither requires a new integration.

Transform the parent pulse. Splitting the integral at the origin, $$G(j\omega)=\int_{-\infty}^{0}e^{at}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-at}e^{-j\omega t}dt=\frac{1}{a-j\omega}+\frac{1}{a+j\omega}=\frac{2a}{a^{2}+\omega^{2}}.$$ With $a=2$, $G(j\omega)=\dfrac{4}{4+\omega^{2}}$, a real, even, positive Lorentzian with $G(0)=1$ and half-power points at $\omega=\pm2$.
Part (a): apply the time-shift property. $x(t)=g(t-1)$ and $g(t-t_0)\leftrightarrow G(j\omega)e^{-j\omega t_0}$, so $$\boxed{\,X_a(j\omega)=\frac{4}{4+\omega^{2}}\,e^{-j\omega}\,}$$
Read the (a) spectra. The magnitude is unchanged by a delay: $|X_a|=\dfrac{4}{4+\omega^{2}}$, peaking at 1 at DC and falling to $\tfrac12$ at $\omega=\pm2\ \text{rad/s}$. The phase is purely linear, $\arg X_a=-\omega$ radians, whose constant slope $-1\ \text{s}$ is exactly the group delay — the 1 s shift shows up as slope, not as distortion.
Part (b): apply frequency differentiation. From $t\,g(t)\leftrightarrow j\dfrac{dG(j\omega)}{d\omega}$ and $\dfrac{d}{d\omega}\dfrac{4}{4+\omega^{2}}=\dfrac{-8\omega}{(4+\omega^{2})^{2}}$, $$\boxed{\,X_b(j\omega)=-\,j\,\frac{8\omega}{(4+\omega^{2})^{2}}\,}$$ which is purely imaginary and odd — exactly what the transform of a real odd signal must be.
Read the (b) spectra. $|X_b|=\dfrac{8|\omega|}{(4+\omega^{2})^{2}}$ vanishes at DC (the signal has zero area, being odd), rises to a maximum and decays as $\omega^{-3}$. Setting the derivative to zero gives $\omega_{\text{peak}}=\dfrac{a}{\sqrt3}=\dfrac{2}{\sqrt3}=1.1547$ rad/s, where $|X_b|=0.32476$. The phase is a step: $\arg X_b=-\tfrac{\pi}{2}$ for $\omega\gt 0$ and $+\tfrac{\pi}{2}$ for $\omega\lt 0$.

Part (a): x(t) = exp(-2|t-1|), the two-sided exponential delayed to t = 1.

Part (a) magnitude spectrum, unaffected by the delay; half-power at w = +/-2 rad/s.

Part (a) phase spectrum: the linear phase -w rad, slope -1 s = the group delay.

Part (b): x(t) = t exp(-2|t|), an odd signal with extrema at t = +/-1/2.

Part (b) magnitude spectrum: zero at DC, peak 0.32476 at w = 1.1547 rad/s, then w^-3 roll-off.

Part (b) phase spectrum: a constant -pi/2 for positive w and +pi/2 for negative w, the signature of a purely imaginary, odd transform.

Quantity	Result
Parent pair	$e^{-a\|t\|}\leftrightarrow \dfrac{2a}{a^{2}+\omega^{2}}$; $a=2\Rightarrow \dfrac{4}{4+\omega^{2}}$
(a) $X(j\omega)$	$\dfrac{4}{4+\omega^{2}}e^{-j\omega}$
(a) magnitude / phase	$\dfrac{4}{4+\omega^{2}}$ (peak 1 at DC, half-power $\omega=\pm2$) / $-\omega$ rad
(b) $X(j\omega)$	$-j\dfrac{8\omega}{(4+\omega^{2})^{2}}$
(b) magnitude peak	$0.32476$ at $\omega=2/\sqrt3=1.1547$ rad/s
(b) phase	$-\pi/2$ ($\omega\gt 0$), $+\pi/2$ ($\omega\lt 0$)

Topic. Fourier-transform properties: time shifting and frequency differentiation; symmetry of real signals.

Key relations. $e^{-a|t|}\leftrightarrow\frac{2a}{a^{2}+\omega^{2}}$; $x(t-t_0)\leftrightarrow X(j\omega)e^{-j\omega t_0}$; $t\,x(t)\leftrightarrow j\frac{dX}{d\omega}$; real even $\Rightarrow$ real even transform, real odd $\Rightarrow$ imaginary odd transform.

Why this works. A pure delay cannot change how much energy sits at each frequency — it only re-times the components — so it must act as a unit-magnitude, linear-phase factor. That is the physical content of the shift theorem, and the constant slope of the phase plot is the group delay of one second read directly off the graph. Multiplication by t, on the other hand, is the dual of differentiation: it differentiates the spectrum, which turns the smooth peak of the Lorentzian into an odd function that must pass through zero at DC. That DC null is also obvious in the time domain, since the odd signal has exactly zero area.

Common pitfall. Forgetting the factor j in the frequency-differentiation property, which produces a real (rather than imaginary) transform and contradicts the odd symmetry of x(t); and plotting the part-(a) phase modulo 2 pi as a sawtooth. Since the delay is a genuine linear phase, the unwrapped straight line is the physically meaningful plot. Also note that the magnitude of an imaginary transform is the absolute value, so the (b) magnitude sketch is a V-shaped null at the origin, not a signed curve passing through it.

Source. Lathi, Linear Systems and Signals 2nd ed., §7.3–7.4 (transform properties table); Oppenheim & Willsky, Signals and Systems 2nd ed., §4.3.

Question 4 of 6: Fourier Transforms and Spectra of Two-Sided Exponentials

Question 4: Fourier Transforms and Spectra of Two-Sided Exponentials (20 marks)