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−2−14021−12h[n]n−1012243

The sequence x[n] is nonzero for n = -1…1 and h[n] for n = 1…2, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -1…1h[n]: n = 1…2
nlast =
2.Easy
x[n]n−105152535…h[n]n−1021−1213−145

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

h[n] = {2, −1, 1, −1}x[n] = 5·u[n]
y[2] =
3.Medium
x(t)t15h(t)t1

The pulse x(t) = 5 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with h(t) = e−4tu(t). Find the output y(2.5) at t = 2.5 s, after the pulse has ended.

x(t) = 5, 0 ≤ t < 1 sh(t) = e−4tu(t)t = 2.5 s
y(2.5) =
4.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(1) at t = 1 s.

h(t) = e−2tu(t)x(t) = 10·u(t)t = 1 s
y(1) =
5.Hard
5cos(15t)h(t) = 10e−20t​u(t)y(t)

The input x(t) = 5·cos(15t) is applied to a system with h(t) = 10·e−20tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 10e−20tu(t)x(t) = 5cos(15t)
Ym =
6.Hard
5cos(15t)h(t) = 5e−5t​u(t)y(t)

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

h(t) = 5e−5tu(t)x(t) = 5cos(15t)
Ym =

Answer key

  1. 1. nlast = 3
  2. 2. y[2] = 10
  3. 3. y(2.5) = 0.003042
  4. 4. y(1) = 4.323
  5. 5. Ym = 2
  6. 6. Ym = 1.581

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