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−10−111243245h[n]n−1011−120.534

An FIR filter computes y[n] = x[n] − x[n − 1] + 0.5·x[n − 2]. The input is x[n] = {−1, 1, 4, 2} for n = 0…3 and zero elsewhere. Find the output y[2].

b = {1, −1, 0.5}x[n] = {−1, 1, 4, 2}, n = 0…3
y[2] =
2.Easy
x[n]n−1031−32−134h[n]n−101122−334

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

x[n] = {3, −3, −1}, n = 0…2h[n] = {1, 2, −3}, n = 0…2
y[2] =
3.Medium
x(t)t12h(t)t42

The pulse x(t) = 2 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 2 for 0 ≤ t < 4 s (zero elsewhere). Find the output y(1.5) at t = 1.5 s.

x(t) = 2, 0 ≤ t < 1 sh(t) = 2, 0 ≤ t < 4 st = 1.5 s
y(1.5) =
4.Medium
x(t)t12h(t)t23

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

x(t) = 2, 0 ≤ t < 1 sh(t) = 3, 0 ≤ t < 2 st = 1 s
y(1) =
5.Hard
δ[n]4+h[n]−0.5·z⁻¹+

A system obeys y[n] = −0.5·y[n − 1] + 4·x[n] and starts at rest. Find its impulse response sample h[5].

y[n] = −0.5y[n − 1] + 4x[n]x[n] = δ[n]
h[5] =
6.Hard
δ(t)y′ + 5y = 3xh(t)

A system obeys y′(t) + 5·y(t) = 3·x(t). Find its impulse response at t = 0.2 s, h(0.2).

y′ + 5y = 3xt = 0.2 s
h(0.2) =

Answer key

  1. 1. y[2] = 2.5
  2. 2. y[2] = −16
  3. 3. y(1.5) = 4
  4. 4. y(1) = 6
  5. 5. h[5] = −0.125
  6. 6. h(0.2) = 1.104

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