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−1041−12234h[n]n−10−214234

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

x[n] = {4, −1, 2}, n = 0…2h[n] = {−2, 4}, n = 0…1
y[2] =
2.Easy
x[n]n−103132333…h[n]n−1041−12434

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

h[n] = {4, −1, 4}x[n] = 3·u[n]
y[2] =
3.Medium
x(t)t12h(t)t21

The pulse x(t) = 2 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 1 for 0 ≤ t < 2 s (zero elsewhere). Find the output y(2.5) at t = 2.5 s.

x(t) = 2, 0 ≤ t < 1 sh(t) = 1, 0 ≤ t < 2 st = 2.5 s
y(2.5) =
4.Medium
x(t)t−12h(t)t13

The signal x(t) is nonzero only between t = −1 s and t = 2 s, and h(t) only between t = 1 s and t = 3 s. Find the duration D of y(t) = x(t) ∗ h(t).

x: −1 … 2 sh: 1 … 3 s
D = s
5.Hard
10cos(10t)h(t) = 5e−5t​u(t)y(t)

The input x(t) = 10·cos(10t) 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(10t)
φ = °
6.Hard
δ(t)y′ + 4y = 5xh(t)

A system obeys y′(t) + 4·y(t) = 5·x(t). Find its impulse response at t = 2 s, h(2).

y′ + 4y = 5xt = 2 s
h(2) =

Answer key

  1. 1. y[2] = −8
  2. 2. y[2] = 21
  3. 3. y(2.5) = 1
  4. 4. D = 5 s
  5. 5. φ = −63.43 °
  6. 6. h(2) = 0.001677

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