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−10−3112134h[n]n−1011−12−234

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

x[n] = {−3, 1, 1}, n = 0…2h[n] = {1, −1, −2}, n = 0…2
y[1] =
2.Easy
x[n]n−105152535…h[n]n−103112−134

A system with the impulse response h[n] = {3, 1, −1} (n = 0…2) receives the step input x[n] = 5·u[n]. Find the output y[2].

h[n] = {3, 1, −1}x[n] = 5·u[n]
y[2] =
3.Medium
x(t)t21h(t)t43

The pulse x(t) = 1 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with the impulse response h(t) = 3 for 0 ≤ t < 4 s (zero elsewhere). Find the output y(1) at t = 1 s.

x(t) = 1, 0 ≤ t < 2 sh(t) = 3, 0 ≤ t < 4 st = 1 s
y(1) =
4.Medium
x(t)t0.510h(t)t1

The pulse x(t) = 10 for 0 ≤ t < 0.5 s (zero elsewhere) is applied to a system with h(t) = e−4tu(t). Find the output y(2) at t = 2 s, after the pulse has ended.

x(t) = 10, 0 ≤ t < 0.5 sh(t) = e−4tu(t)t = 2 s
y(2) =
5.Hard
5cos(15t)h(t) = 2e−5t​u(t)y(t)

The input x(t) = 5·cos(15t) is applied to a system with h(t) = 2·e−5tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 2e−5tu(t)x(t) = 5cos(15t)
Ym =
6.Hard
x(t)t (ms)2.55103

A square wave switches between 3 and 0 with period T = 5 ms (half a period at each level). Find the amplitude c1 of its Fourier component at 200 Hz.

0 … 3T = 5 msf = 200 Hz
c1 =

Answer key

  1. 1. y[1] = 4
  2. 2. y[2] = 15
  3. 3. y(1) = 3
  4. 4. y(2) = 0.005358
  5. 5. Ym = 0.6325
  6. 6. c1 = 1.91

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