Convolution & LTI systems worksheet with answers

Free printable Convolution & LTI systems worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
x[n]n−103132333…h[n]n−101122−234

A system with the impulse response h[n] = {1, 2, −2} (n = 0…2) receives the step input x[n] = 3·u[n]. Find the output y[4].

h[n] = {1, 2, −2}x[n] = 3·u[n]
y[4] =
2.Easy
x[n]n−10−1142−234h[n]n−103122334

The sequences x[n] = {−1, 4, −2} and h[n] = {3, 2, 3} are convolved. Find the sum Σy[n] of all samples of y[n] = x[n] ∗ h[n].

x[n] = {−1, 4, −2}, n = 0…2h[n] = {3, 2, 3}
Σy[n] =
3.Medium
x(t)t12h(t)t41

The pulse x(t) = 2 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 1 for 0 ≤ t < 4 s (zero elsewhere). Find the output y(4.5) at t = 4.5 s.

x(t) = 2, 0 ≤ t < 1 sh(t) = 1, 0 ≤ t < 4 st = 4.5 s
y(4.5) =
4.Medium
x(t)t10h(t)t0.51

A system has the impulse response h(t) = e−2tu(t). The input is a step x(t) = 10·u(t). Find the output y(1.5) at t = 1.5 s.

h(t) = e−2tu(t)x(t) = 10·u(t)t = 1.5 s
y(1.5) =
5.Hard
δ(t)y′ + 2y = 3xh(t)

A system obeys y′(t) + 2·y(t) = 3·x(t). Find its impulse response at t = 0.25 s, h(0.25).

y′ + 2y = 3xt = 0.25 s
h(0.25) =
6.Hard
x(t)t (ms)2484−4

A square wave switches between +4 and −4 with period T = 4 ms (half a period at each level). Find the amplitude c5 of its Fourier component at 1250 Hz.

±4T = 4 msf = 1250 Hz
c5 =

Answer key

  1. 1. y[4] = 3
  2. 2. Σy[n] = 8
  3. 3. y(4.5) = 1
  4. 4. y(1.5) = 4.751
  5. 5. h(0.25) = 1.82
  6. 6. c5 = 1.019

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