Question 1 of 7: Fourier Transform, Spectra and Filtering of a Delayed Exponential

Question 1: Fourier Transform, Spectra and Filtering of a Delayed Exponential (1/5 of paper)

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. $x(t)=u(t-2)e^{-2t}$, a right-shifted decaying exponential that switches on at $t=2$; a rectangular filter $h(t)=u(t)-u(t-1)$ of unit height on $0\le t\le1$.

Find. (a) $X(f)$; (b) $|X(f)|$ and $\angle X(f)$; (c) the energy spectral density $|X(f)|^2$ and total energy $E$; (d) $y(t)=x(t)*h(t)$ in the time domain.

Approach. Substitute $\tau=t-2$ to reduce $x(t)$ to a plain causal exponential and apply the time-shift property; get the energy two independent ways (Parseval and direct time-domain integration) as a cross-check; evaluate the convolution integral directly since $h(t)$ is a simple rectangular window.

1. Part (a) — Fourier transform. With $\tau=t-2$, $x(t)=e^{-2(\tau+2)}u(\tau)=e^{-4}e^{-2\tau}u(\tau)$, so $$X(f)=\int_{2}^{\infty}e^{-2t}e^{-j2\pi ft}\,dt=e^{-4}e^{-j4\pi f}\int_0^\infty e^{-(2+j2\pi f)\tau'}\,d\tau'$$ using the standard pair $e^{-at}u(t)\leftrightarrow 1/(a+j2\pi f)$ with $a=2$, $$X(f)=\boxed{\dfrac{e^{-4}\,e^{-j4\pi f}}{2+j2\pi f}}$$ the $e^{-j4\pi f}$ factor is exactly the time-shift phase for a 2-second delay.
2. Part (b) — amplitude and phase spectra. Writing $2+j2\pi f$ in polar form, $|2+j2\pi f|=2\sqrt{1+\pi^2f^2}$ and $\angle(2+j2\pi f)=\arctan(\pi f)$, so $$|X(f)|=\boxed{\dfrac{e^{-4}}{2\sqrt{1+\pi^2f^2}}}\qquad \angle X(f)=\boxed{-4\pi f-\arctan(\pi f)}$$ the amplitude spectrum is an even, low-pass-shaped function peaking at $f=0$ ($|X(0)|=e^{-4}/2\approx0.00916$); the phase is odd and dominated by the linear $-4\pi f$ delay term for the $f$ range where $x(t)$ has significant energy.
3. Part (c) — energy spectral density and total energy. Squaring the magnitude, $$\mathcal{E}(f)=|X(f)|^2=\boxed{\dfrac{e^{-8}}{4(1+\pi^2f^2)}}$$ Integrating $|x(t)|^2$ directly in time (Parseval's theorem says this must equal $\int|X(f)|^2df$): $$E=\int_2^\infty\big(e^{-2t}\big)^2dt=\int_2^\infty e^{-4t}\,dt=\Big[-\tfrac14e^{-4t}\Big]_2^\infty=\boxed{\tfrac14e^{-8}\approx8.387\times10^{-5}\ \text{J (per }\Omega\text{)}}$$ a numerical trapezoidal integration of $\mathcal{E}(f)$ over $f\in[-50,50]$ Hz reproduces this to within 0.4 %, confirming the transform pair.
4. Part (d) — time-domain convolution. $h(t-\tau)$ is a unit window that is 1 for $t-1\le\tau\le t$, so $$y(t)=\int_{\max(t-1,\,2)}^{t}e^{-2\tau}\,d\tau$$ and this integral is nonzero only once the sliding window $[t-1,t]$ overlaps $x(t)$'s support $\tau\ge2$, i.e. for $t>2$. Two sub-cases follow from where the window's left edge sits relative to $\tau=2$: while $t-1<2$ (i.e. $2\le t\le3$) the lower limit is pinned at 2; once $t-1\ge2$ (i.e. $t>3$) the window is entirely inside $x$'s support and both limits slide together. Evaluating $\int e^{-2\tau}d\tau=-\tfrac12e^{-2\tau}$ on each case, $$y(t)=\boxed{\begin{cases}0, & t<2\\[2pt] \tfrac12\big(e^{-4}-e^{-2t}\big), & 2\le t\le3\\[2pt] \tfrac12e^{-2t}\big(e^{2}-1\big), & t>3\end{cases}}$$ Both branches agree at $t=3$ ($y=0.00792$, confirmed by direct numerical convolution to 6 decimal places), so $y(t)$ rises smoothly from the switch-on at $t=2$ to a peak at $t=3$ (the instant the sliding window first fully covers $x$'s discontinuity), then decays as a pure exponential.