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−101112131…h[n]n−103112334

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

h[n] = {3, 1, 3}x[n] = 1·u[n]
y[3] =
2.Easy
x[n]n−10123−14−15−36h[n]n−101213−14

The sequence x[n] is nonzero for n = 3…5 and h[n] for n = 2…3, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 3…5h[n]: n = 2…3
nlast =
3.Medium
x(t)t45h(t)t22

The pulse x(t) = 5 for 0 ≤ t < 4 s is convolved with the pulse h(t) = 2 for 0 ≤ t < 2 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 5, 0 ≤ t < 4 sh(t) = 2, 0 ≤ t < 2 s
Ay =
4.Medium
x(t)t1h(t)t21

A system has the impulse response h(t) = e−0.5tu(t). The input is a step x(t) = 1·u(t). Find the output y(4) at t = 4 s.

h(t) = e−0.5tu(t)x(t) = 1·u(t)t = 4 s
y(4) =
5.Hard
10cos(15t)h(t) = 5e−5t​u(t)y(t)

The input x(t) = 10·cos(15t) is applied to a system with h(t) = 5·e−5tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

h(t) = 5e−5tu(t)x(t) = 10cos(15t)
φ = °
6.Hard
10cos(30t)h(t) = 2e−10t​u(t)y(t)

The input x(t) = 10·cos(30t) is applied to a system with h(t) = 2·e−10tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 2e−10tu(t)x(t) = 10cos(30t)
Ym =

Answer key

  1. 1. y[3] = 7
  2. 2. nlast = 8
  3. 3. Ay = 80
  4. 4. y(4) = 1.729
  5. 5. φ = −71.57 °
  6. 6. Ym = 0.6325

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