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−10−31−2233445

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

h[n] = {−3, −2, 3, 4}x[n] = 1·u[n]
y[3] =
2.Easy
x[n]n−1012344−3536h[n]n−3−22−1101

The sequence x[n] is nonzero for n = 3…5 and h[n] for n = -2…-1, 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…-1
nlast =
3.Medium
x(t)t21h(t)t32

The pulse x(t) = 1 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with the impulse response h(t) = 2 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(3.5) at t = 3.5 s.

x(t) = 1, 0 ≤ t < 2 sh(t) = 2, 0 ≤ t < 3 st = 3.5 s
y(3.5) =
4.Medium
x(t)t24h(t)t1

The pulse x(t) = 4 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−2tu(t). Find the output y(3) at t = 3 s, after the pulse has ended.

x(t) = 4, 0 ≤ t < 2 sh(t) = e−2tu(t)t = 3 s
y(3) =
5.Hard
2cos(15t)h(t) = 5e−10t​u(t)y(t)

The input x(t) = 2·cos(15t) is applied to a system with h(t) = 5·e−10tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

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

The input x(t) = 2·cos(10t) 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) = 2cos(10t)
Ym =

Answer key

  1. 1. y[3] = 2
  2. 2. nlast = 4
  3. 3. y(3.5) = 3
  4. 4. y(3) = 0.2657
  5. 5. φ = −56.31 °
  6. 6. Ym = 0.2828

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