Question 1 of 7: Fourier transform, spectra, energy and filtering of a delayed exponential (20 marks)

Question 1 — Fourier transform, spectra, energy and filtering of a delayed exponential (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. A causal, one-sided exponential switched on at $t = 2\ \text{s}$: $x(t) = u(t-2)\,e^{-2t}$, so $x(t) = e^{-2t}$ for $t \ge 2$ and zero before. Note carefully that the exponent is $-2t$ and not $-2(t-2)$: the signal starts at the reduced value $x(2^{+}) = e^{-4} = 1.832\times10^{-2}$, not at unity. The filter of part (d) is a unit-height rectangular window of duration $1\ \text{s}$, $h(t) = u(t) - u(t-1)$.

Find. (a) $X(f)$ with $f$ in hertz; (b) $|X(f)|$ and $\arg X(f)$; (c) the energy spectral density $\Psi_x(f)$ and the total energy $E_x$; (d) $y(t) = x(t) * h(t)$ evaluated entirely in the time domain.

Approach. Evaluate the defining Fourier integral over the support $[2,\infty)$, split the result into magnitude and angle, apply $\Psi_x = |X|^2$ together with Rayleigh’s energy theorem, and finish part (d) by recognising the rectangular $h(t)$ as a one-second running integral so that $y(t) = \int_{t-1}^{t} x(\tau)\,d\tau$.

1. Evaluate the Fourier integral over the signal’s support. Because $x(t)$ vanishes for $t \lt 2$, the transform integral runs from 2 to infinity: $$X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi f t}\,dt = \int_{2}^{\infty} e^{-2t}e^{-j2\pi f t}\,dt = \int_{2}^{\infty} e^{-(2 + j2\pi f)t}\,dt .$$ The integrand decays because $\operatorname{Re}\{2 + j2\pi f\} = 2 \gt 0$, so the upper limit contributes nothing and only the lower limit survives.
2. Carry out the integration and collect the delay factor. $$X(f) = \left[\frac{-e^{-(2+j2\pi f)t}}{2 + j2\pi f}\right]_{2}^{\infty} = \frac{e^{-2(2+j2\pi f)}}{2 + j2\pi f} = \boxed{\;X(f) = \frac{e^{-4}\,e^{-j4\pi f}}{2 + j2\pi f}\;}$$ The factor $e^{-j4\pi f} = e^{-j2\pi f(2)}$ is exactly the time-shift kernel for a 2 s delay, and the constant $e^{-4}$ is the amplitude the exponential has decayed to by the switch-on instant. Equivalently $x(t) = e^{-4}\,e^{-2(t-2)}u(t-2)$, whose transform is $e^{-4}e^{-j4\pi f}/(2+j2\pi f)$ by the shifting property — a useful independent check on the algebra.
3. Separate magnitude and phase. The delay factor has unit magnitude, so it affects only the angle: $$|X(f)| = \frac{e^{-4}}{\sqrt{4 + 4\pi^{2}f^{2}}} = \frac{e^{-4}}{2\sqrt{1 + \pi^{2}f^{2}}}, \qquad \arg X(f) = -4\pi f - \arctan\!\left(\pi f\right).$$ The amplitude spectrum is even and the phase spectrum odd, as required of a real signal. Its peak sits at d.c.: $$\boxed{\;|X(0)| = \tfrac{1}{2}e^{-4} = 9.158\times10^{-3}\ \text{V}\!\cdot\!\text{s}\;}$$ and $|X(f)|$ falls to half of that at $\pi f = \sqrt{3}$, i.e. $f = 0.551\ \text{Hz}$.

The phase is dominated by the straight line $-4\pi f$, which is the signature of the 2 s delay; the gentle $-\arctan(\pi f)$ term is the pole’s own contribution and saturates at $-\pi/2$. At $f = 1\ \text{Hz}$, for instance, $\arg X = -4\pi - \arctan\pi = -13.83\ \text{rad}$.

4. Form the energy spectral density. For an energy signal the ESD is the squared magnitude of the transform, and the delay drops out entirely: $$\Psi_x(f) = |X(f)|^{2} = \frac{e^{-8}}{4 + 4\pi^{2}f^{2}} \qquad\left[\text{V}^{2}\!\cdot\!\text{s}^{2}/\text{Hz}\right].$$ At d.c. $\Psi_x(0) = e^{-8}/4 = 8.387\times10^{-5}$, and the density is halved at the same $f = 0.551\ \text{Hz}$ found above.
5. Obtain the energy, and confirm it two ways. Directly in the time domain, $$E_x = \int_{-\infty}^{\infty}|x(t)|^{2}dt = \int_{2}^{\infty} e^{-4t}\,dt = \frac{e^{-8}}{4} \qquad\Longrightarrow\qquad \boxed{\;E_x = 8.387\times10^{-5}\ \text{J (into }1\ \Omega)\;}$$ Rayleigh’s theorem gives the same number from the ESD, since $\int_{-\infty}^{\infty} df/(1+\pi^{2}f^{2}) = 1$ and therefore $\int \Psi_x(f)\,df = (e^{-8}/4)\cdot 1$. The agreement confirms both the transform and the density.
6. Recognise the filter as a one-second running integral. Since $h(t)$ is unity on $[0,1]$ and zero elsewhere, $$y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau = \int_{t-1}^{t} x(\tau)\,d\tau .$$ This single statement replaces four separate flip-and-shift integrals: all that remains is to track how the moving one-second window overlaps the support $\tau \ge 2$.
7. Evaluate the three overlap regimes. For $t \lt 2$ the window lies entirely to the left of the support and $y(t) = 0$. For $2 \le t \le 3$ the window has entered the support but its trailing edge is still inside the dead zone, so the integral runs from 2 to $t$: $$y(t) = \int_{2}^{t} e^{-2\tau}d\tau = \frac{e^{-4} - e^{-2t}}{2}.$$ For $t \gt 3$ the window is fully inside the support and both limits are active: $$y(t) = \int_{t-1}^{t} e^{-2\tau}d\tau = \frac{e^{-2(t-1)} - e^{-2t}}{2} = \frac{e^{2}-1}{2}\,e^{-2t}.$$ Collecting the three pieces, $$\boxed{\;y(t) = \begin{cases} 0, & t \lt 2\\[4pt] \dfrac{e^{-4} - e^{-2t}}{2}, & 2 \le t \le 3\\[6pt] \dfrac{e^{2}-1}{2}\,e^{-2t}, & t \gt 3\end{cases}\;}$$ The two branches agree at $t = 3$, where both give $y(3) = (e^{-4}-e^{-6})/2 = 7.918\times10^{-3}$ — the peak of the response, and a check worth performing because a sign slip in either branch destroys the continuity.

Physically the output rises while the window fills with signal energy and then decays at the same rate $e^{-2t}$ as the input once the window is saturated, scaled by the constant $(e^{2}-1)/2 = 3.195$. A final consistency check: the area under $y$ must equal the product of the areas of $x$ and $h$, namely $(e^{-4}/2)(1) = 9.16\times10^{-3}$, which the piecewise expression reproduces.