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−105152535…h[n]n−103122434

A system with the impulse response h[n] = {3, 2, 4} (n = 0…2) receives the step input x[n] = 5·u[n]. Find the output y[1].

h[n] = {3, 2, 4}x[n] = 5·u[n]
y[1] =
2.Easy
x[n]n−104132134h[n]n−10−2132434

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

x[n] = {4, 3, 1}, n = 0…2h[n] = {−2, 3, 4}, n = 0…2
ymax =
3.Medium
x(t)t10h(t)t1

The input x(t) = 10·e−2tu(t) is applied to a system with h(t) = e−5tu(t). Find the output y(1) at t = 1 s.

x(t) = 10·e−2tu(t)h(t) = e−5tu(t)t = 1 s
y(1) =
4.Medium
x(t)t12h(t)t02

The signal x(t) is nonzero only between t = 1 s and t = 2 s, and h(t) only between t = 0 s and t = 2 s. Find the time tend at which y(t) = x(t) ∗ h(t) ends.

x: 1 … 2 sh: 0 … 2 s
tend = s
5.Hard
2cos(15t)h(t) = 10e−5t​u(t)y(t)

The input x(t) = 2·cos(15t) is applied to a system with h(t) = 10·e−5tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 10e−5tu(t)x(t) = 2cos(15t)
Ym =
6.Hard
δ[n]0.5+h[n]0.4·z⁻¹+

A system obeys y[n] = 0.4·y[n − 1] + 0.5·x[n] and starts at rest. Find its impulse response sample h[5].

y[n] = 0.4y[n − 1] + 0.5x[n]x[n] = δ[n]
h[5] =

Answer key

  1. 1. y[1] = 25
  2. 2. ymax = 23
  3. 3. y(1) = 0.4287
  4. 4. tend = 4 s
  5. 5. Ym = 1.265
  6. 6. h[5] = 0.00512

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