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.

New worksheetPractise online with hints

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
x[n]n−101132234h[n]n−10−3142−334

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

x[n] = {1, 3, 2}, n = 0…2h[n] = {−3, 4, −3}, n = 0…2
y[2] =
2.Easy
x[n]n−105152535…h[n]n−1021−22−33−245

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

h[n] = {2, −2, −3, −2}x[n] = 5·u[n]
y[5] =
3.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−1tu(t). Find the output y(1) at t = 1 s, after the pulse has ended.

x(t) = 10, 0 ≤ t < 0.5 sh(t) = e−1tu(t)t = 1 s
y(1) =
4.Medium
x(t)t4h(t)t1

The input x(t) = 4·e−4tu(t) is applied to a system with h(t) = e−0.5tu(t). Find the output y(2) at t = 2 s.

x(t) = 4·e−4tu(t)h(t) = e−0.5tu(t)t = 2 s
y(2) =
5.Hard
2cos(20t)h(t) = 5e−10t​u(t)y(t)

The input x(t) = 2·cos(20t) is applied to a system with h(t) = 5·e−10tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 5e−10tu(t)x(t) = 2cos(20t)
Ym =
6.Hard
x(t)t (ms)510201−1

A square wave switches between +1 and −1 with period T = 10 ms (half a period at each level). Find the amplitude c3 of its Fourier component at 300 Hz.

±1T = 10 msf = 300 Hz
c3 =

Answer key

  1. 1. y[2] = 3
  2. 2. y[5] = −25
  3. 3. y(1) = 2.387
  4. 4. y(2) = 0.4201
  5. 5. Ym = 0.4472
  6. 6. c3 = 0.4244

Full step-by-step solutions: practise this topic on examproblems.com.