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−101132334h[n]n−10−2142−234

The sequences x[n] = {1, 3, 3} and h[n] = {−2, 4, −2} both start at n = 0. Find the sample y[2] of the convolution y[n] = x[n] ∗ h[n].

x[n] = {1, 3, 3}, n = 0…2h[n] = {−2, 4, −2}, n = 0…2
y[2] =
2.Easy
x[n]n−10−211213445h[n]n−10110.2520.534

An FIR filter computes y[n] = x[n] + 0.25·x[n − 1] + 0.5·x[n − 2]. The input is x[n] = {−2, 1, 1, 4} for n = 0…3 and zero elsewhere. Find the output y[3].

b = {1, 0.25, 0.5}x[n] = {−2, 1, 1, 4}, n = 0…3
y[3] =
3.Medium
x(t)t4h(t)t0.251

A system has the impulse response h(t) = e−4tu(t). The input is a step x(t) = 4·u(t). Find the output y(0.25) at t = 0.25 s.

h(t) = e−4tu(t)x(t) = 4·u(t)t = 0.25 s
y(0.25) =
4.Medium
x(t)t23h(t)t−11

The signal x(t) is nonzero only between t = 2 s and t = 3 s, and h(t) only between t = −1 s and t = 1 s. Find the time tstart at which y(t) = x(t) ∗ h(t) starts.

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

The input x(t) = 5·cos(3t) is applied to a system with h(t) = 2·e−2tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

h(t) = 2e−2tu(t)x(t) = 5cos(3t)
φ = °
6.Hard
u[n]3+y[n]0.9·z⁻¹+

A system obeys y[n] = 0.9·y[n − 1] + 3·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[3].

y[n] = 0.9y[n − 1] + 3x[n]x[n] = u[n]
y[3] =

Answer key

  1. 1. y[2] = 4
  2. 2. y[3] = 4.75
  3. 3. y(0.25) = 0.6321
  4. 4. tstart = 1 s
  5. 5. φ = −56.31 °
  6. 6. y[3] = 10.32

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