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−1013243−1435h[n]n−2−12041

The sequence x[n] is nonzero for n = 1…4 and h[n] for n = -1…0, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 1…4h[n]: n = -1…0
nlast =
2.Easy
x[n]n−10−21−22234h[n]n−10−31−3234

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

x[n] = {−2, −2, 2}, n = 0…2h[n] = {−3, −3}, n = 0…1
y[1] =
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)t22h(t)t11

The pulse x(t) = 2 for 0 ≤ t < 2 s is convolved with the pulse h(t) = 1 for 0 ≤ t < 1 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 2, 0 ≤ t < 2 sh(t) = 1, 0 ≤ t < 1 s
Ay =
5.Hard
δ[n]2+h[n]−0.8·z⁻¹+

A system obeys y[n] = −0.8·y[n − 1] + 2·x[n] and starts at rest. Find its impulse response sample h[2].

y[n] = −0.8y[n − 1] + 2x[n]x[n] = δ[n]
h[2] =
6.Hard
5cos(10t)h(t) = 2e−5t​u(t)y(t)

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

h(t) = 2e−5tu(t)x(t) = 5cos(10t)
Ym =

Answer key

  1. 1. nlast = 4
  2. 2. y[1] = 12
  3. 3. y(0.5) = 0.05325
  4. 4. Ay = 4
  5. 5. h[2] = 1.28
  6. 6. Ym = 0.8944

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