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−10−21−221314h[n]n−1031−12−334

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

x[n] = {−2, −2, 1, 1}, n = 0…3h[n] = {3, −1, −3}, n = 0…2
ymax =
2.Easy
x[n]n−10−31−32234h[n]n−102112−234

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

x[n] = {−3, −3, 2}, n = 0…2h[n] = {2, 1, −2}, n = 0…2
y[2] =
3.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(0.1) at t = 0.1 s.

h(t) = e−2tu(t)x(t) = 10·u(t)t = 0.1 s
y(0.1) =
4.Medium
x(t)t2h(t)t1

The input x(t) = 2·e−2tu(t) is applied to a system with h(t) = e−5tu(t). Find the output y(0.25) at t = 0.25 s.

x(t) = 2·e−2tu(t)h(t) = e−5tu(t)t = 0.25 s
y(0.25) =
5.Hard
x(t)t (ms)2484−4

A triangle wave rises and falls linearly between −4 and +4 with period T = 4 ms. Find the amplitude c5 of its Fourier component at 1250 Hz.

±4T = 4 msf = 1250 Hz
c5 =
6.Hard
x(t)t (ms)1483

A pulse train has height 3, pulse width 1 ms and period 4 ms. Find the amplitude c1 of harmonic number 1, at 250 Hz.

A = 3τ = 1 msT = 4 ms
c1 =

Answer key

  1. 1. ymax = 11
  2. 2. y[2] = 7
  3. 3. y(0.1) = 0.9063
  4. 4. y(0.25) = 0.2134
  5. 5. c5 = 0.1297
  6. 6. c1 = 1.35

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