Signals and Systems Practice Exam — Free

Because the system starts at rest, $Y(s) = H(s)U(s)$ with $U(s) = 1/s$. Factoring the denominator:

The cover-up rule gives each residue by deleting its factor and evaluating at that root:

Inverting term by term with $1/(s+a) \leftrightarrow e^{-at}u(t)$:

Two checks confirm the signs. At $t = 0$ the output of a system driven from rest must be zero: $1-5/3+2/3 = 0$. As $t \to \infty$ the final-value theorem gives $y(\infty) = H(0) = 1$, matching the constant term. Because the relative degree is 2, the initial slope must also vanish: $y'(0) = 10/3 - 10/3 = 0$.

Option B is the tempting sign slip: evaluating $C$ at $s = -5$ as $10/((-5)(3))$ instead of $10/((-5)(-3))$ gives $-2/3$, and the result has $y(0) = -4/3$, impossible for a system starting at rest. Option A is the impulse response $h(t)$, obtained by forgetting the $1/s$ of the step input. Option D swaps the residues; it satisfies $y(0) = 0$ but gives $y'(0) = 7$ rather than 0.

$y(t) = 1 - \tfrac{5}{3}e^{-2t} + \tfrac{2}{3}e^{-5t}$ for $t \ge 0$ (option C).

(a) The highest frequency present is 1300 Hz. The sampling theorem requires $f_s > 2f_{\max}$, so the Nyquist rate is $2\times 1300 = 2600\,\text{Hz}$; any rate above this avoids aliasing.

(b) At $f_s = 2000\,\text{Hz}$ the Nyquist frequency is $f_s/2 = 1000\,\text{Hz}$. The 400 Hz tone lies below it and is reconstructed unchanged. The 1300 Hz tone lies above it, so it aliases: a sampled sinusoid at frequency $f$ is indistinguishable from one at $|f - kf_s|$ for any integer $k$, and the alias that falls inside the band is $|1300 - 2000| = 700\,\text{Hz}$.

The sample values show this directly. With $t = n/2000$,

so the samples are exactly those of $3\cos(2\pi\cdot 400\,t) - 2\sin(2\pi\cdot 700\,t)$, and the ideal filter reconstructs that signal: the aliased tone appears at 700 Hz with its sign reversed.

Check: 700 Hz lies below the 1000 Hz cutoff, so it passes, while nothing in the sampled spectrum remains at 1300 Hz.

(a) 2600 Hz; (b) components at 400 Hz and 700 Hz.

Write the convolution as $y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$. Here $x(\tau) = 2$ only for $0\le\tau\le 3$, and $h(t-\tau) = e^{-2(t-\tau)}$ is non-zero only for $\tau \le t$. The integrand is therefore $2e^{-2(t-\tau)}$ on the overlap of $[0, 3]$ with $(-\infty, t]$.

For $t < 0$ there is no overlap, so $y(t) = 0$.

For $0\le t<3$ the overlap is $[0, t]$:

For $t \ge 3$ the whole pulse overlaps, so the limits are fixed at $[0, 3]$:

Both pieces equal $1-e^{-6}$ at $t = 3$, so $y$ is continuous there, as the convolution of bounded signals must be. On $[0, 3)$ the output rises monotonically; after $t = 3$ it decays, so the maximum occurs at $t = 3$, with $y(3) = 1-e^{-6} = 0.998$.

(b) $y(1) = 1-e^{-2} = 0.865$ and $y(4) = (1-e^{-6})e^{-2} = (0.9975)(0.1353) = 0.135$.

Laplace check: $Y(s) = \dfrac{2(1 - e^{-3s})}{s(s+2)} = (1 - e^{-3s})\left(\dfrac{1}{s} - \dfrac{1}{s+2}\right)$, which inverts to $(1 - e^{-2t})u(t) - (1 - e^{-2(t-3)})u(t-3)$; for $t \ge 3$ this is $e^{-2(t-3)} - e^{-2t} = e^{-2t}(e^{6} - 1)$, the same as above.

$y(t) = 0$ for $t < 0$, $y(t) = 1-e^{-2t}$ for $0\le t<3$, $y(t) = (1-e^{-6})e^{-2(t-3)}$ for $t \ge 3$; $y(1) \approx 0.865$, $y(4) \approx 0.135$, peak at $t = 3$.