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−104142234h[n]n−104132−234

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

x[n] = {4, 4, 2}, n = 0…2h[n] = {4, 3, −2}, n = 0…2
ymax =
2.Easy
x[n]n−102122232…h[n]n−101132134

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

h[n] = {1, 3, 1}x[n] = 2·u[n]
y[4] =
3.Medium
x(t)t1h(t)t1

The input x(t) = 1·e−4tu(t) is applied to a system with h(t) = e−5tu(t). Find the output y(0.5) at t = 0.5 s.

x(t) = 1·e−4tu(t)h(t) = e−5tu(t)t = 0.5 s
y(0.5) =
4.Medium
x(t)t4h(t)t21

A system has the impulse response h(t) = e−0.5tu(t). The input is a step x(t) = 4·u(t). Find the output y(2) at t = 2 s.

h(t) = e−0.5tu(t)x(t) = 4·u(t)t = 2 s
y(2) =
5.Hard
10cos(30t)h(t) = 5e−10t​u(t)y(t)

The input x(t) = 10·cos(30t) 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) = 10cos(30t)
Ym =
6.Hard
u[n]2+y[n]0.8·z⁻¹+

A system obeys y[n] = 0.8·y[n − 1] + 2·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[2].

y[n] = 0.8y[n − 1] + 2x[n]x[n] = u[n]
y[2] =

Answer key

  1. 1. ymax = 28
  2. 2. y[4] = 10
  3. 3. y(0.5) = 0.05325
  4. 4. y(2) = 5.057
  5. 5. Ym = 1.581
  6. 6. y[2] = 4.88

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