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−1021−32−13−14h[n]n−10−21−3234

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

x[n] = {2, −3, −1, −1}, n = 0…3h[n] = {−2, −3}, n = 0…1
y[3] =
2.Easy
x[n]n−103112−23−145h[n]n−1021−120.534

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

b = {2, −1, 0.5}x[n] = {3, 1, −2, −1}, n = 0…3
y[3] =
3.Medium
x(t)t4h(t)t1

The input x(t) = 4·e−1tu(t) is applied to a system with h(t) = e−2tu(t). Find the output y(1.5) at t = 1.5 s.

x(t) = 4·e−1tu(t)h(t) = e−2tu(t)t = 1.5 s
y(1.5) =
4.Medium
x(t)t10h(t)t0.251

A system has the impulse response h(t) = e−4tu(t). The input is a step x(t) = 10·u(t). Find the output y(1.5) at t = 1.5 s.

h(t) = e−4tu(t)x(t) = 10·u(t)t = 1.5 s
y(1.5) =
5.Hard
x(t)t (ms)0.5243

A pulse train has height 3, pulse width 0.5 ms and period 2 ms. Find the amplitude c3 of harmonic number 3, at 1500 Hz.

A = 3τ = 0.5 msT = 2 ms
c3 =
6.Hard
2cos(15t)h(t) = 2e−20t​u(t)y(t)

The input x(t) = 2·cos(15t) is applied to a system with h(t) = 2·e−20tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 2e−20tu(t)x(t) = 2cos(15t)
Ym =

Answer key

  1. 1. y[3] = 5
  2. 2. y[3] = 0.5
  3. 3. y(1.5) = 0.6934
  4. 4. y(1.5) = 2.494
  5. 5. c3 = 0.4502
  6. 6. Ym = 0.16

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