Signals and Systems: practice problems and exam questions

Continuous and discrete signals passing through linear time-invariant systems: convolution, impulse and step responses, the Laplace transform, frequency response and Fourier series. Exams mix sketching with short calculations.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Convolution & LTI systemsdiscrete convolution · continuous convolution · impulse response and Fourier seriescalculations
  2. Laplace transformstable pairs and value theorems · inverse transform by partial fractions · ODEs and transfer functionscalculations
  3. Step responsefirst-order systems · second-order parameters · physical systems and designcalculations
  4. Bode plots & marginsfilters and asymptotes · magnitude and phase · stability marginscalculations

Exam tips for Signals and Systems

  • Sketch the signals before you convolve: the overlap tells you the limits of the sum or integral.
  • The output of an LTI system starts at the sum of the start times and ends at the sum of the end times.
  • A sinusoid comes out with the same frequency, scaled by |H(jω)| and shifted by ∠H(jω).

Worked examples

Convolution & LTI systemsEasy
x[n]n−1041−32−234h[n]n−10311234

The sequences x[n] = {4, −3, −2} and h[n] = {3, 1} both start at n = 0. Find the sample y[1] of the convolution y[n] = x[n] ∗ h[n].

x[n] = {4, −3, −2}, n = 0…2h[n] = {3, 1}, n = 0…1

Find y[1]

Show hints
  1. y[n] = Σ x[k]·h[n − k].
  2. For n = 1 the terms are k = 0, 1.
  3. x[0]·h[1] + x[1]·h[0].
Show the answer and the step-by-step solution

Discrete convolution: the output of a linear time-invariant system is y[n] = Σk x[k]·h[n − k]: flip h, shift it by n, multiply sample by sample with x and add. If x has Lx samples starting at nx and h has Lh samples starting at nh, then y has Lx + Lh − 1 samples starting at nx + nh. The sum of all samples of y equals (Σx)·(Σh). An FIR filter y[n] = b₀x[n] + b₁x[n − 1] + … is a convolution with h[n] = bn.

Terms of the sum

Only the k for which both x[k] and h[n − k] exist contribute.

x[0]·h[1] = 4·1 = 4
x[1]·h[0] = −3·3 = −9

Add them

y[1] = −5

Answer

y[1] = −5
Laplace transformsMedium
ReIm−5−10−15−20−25

F(s) = (s + 22)/(s² + 14s + 48). Its inverse transform has the form f(t) = A·e−6t + B·e−8t. Find the coefficient A by partial fractions.

F(s) = (s + 22)/(s² + 14s + 48)

Find A

Show hints
  1. Factor the denominator: s² + 14s + 48 = (s + 6)(s + 8).
  2. Cover-up rule: A = [(s + 6)·F(s)] at s = −6.
  3. At s = −6: numerator 16, other factor 2.
Show the answer and the step-by-step solution

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Factor the denominator

s² + 14s + 48 = (s + 6)(s + 8)
F(s) = A/(s + 6) + B/(s + 8)

Cover-up rule

Multiply by (s + 6) and set s = −6: the B-term vanishes.

A = (s + 22)/(s + 8) |s = −6 = 16/2 = 8

Answer

A = 8
(B = −7, so f(t) = 8·e−6t − 7·e−8t)

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

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