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−10−31−22−334h[n]n−10411234

The sequences x[n] = {−3, −2, −3} and h[n] = {4, 1} are convolved. Find the sum Σy[n] of all samples of y[n] = x[n] ∗ h[n].

x[n] = {−3, −2, −3}, n = 0…2h[n] = {4, 1}
Σy[n] =
2.Easy
x[n]n−103132333…h[n]n−102112−134

A system with the impulse response h[n] = {2, 1, −1} (n = 0…2) receives the step input x[n] = 3·u[n]. Find the output y[5].

h[n] = {2, 1, −1}x[n] = 3·u[n]
y[5] =
3.Medium
x(t)t4h(t)t0.21

A system has the impulse response h(t) = e−5tu(t). The input is a step x(t) = 4·u(t). Find the output y(0.08) at t = 0.08 s.

h(t) = e−5tu(t)x(t) = 4·u(t)t = 0.08 s
y(0.08) =
4.Medium
x(t)t410h(t)t11

The pulse x(t) = 10 for 0 ≤ t < 4 s is convolved with h(t), a triangle that falls linearly from 1 at t = 0 to 0 at t = 1 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 10, 0 ≤ t < 4 sh(0) = 1, h(1) = 0
Ay =
5.Hard
10cos(7.5t)h(t) = 5e−5t​u(t)y(t)

The input x(t) = 10·cos(7.5t) 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(7.5t)
φ = °
6.Hard
δ[n]0.5+h[n]−0.8·z⁻¹+

A system obeys y[n] = −0.8·y[n − 1] + 0.5·x[n] and starts at rest. Find its impulse response sample h[5].

y[n] = −0.8y[n − 1] + 0.5x[n]x[n] = δ[n]
h[5] =

Answer key

  1. 1. Σy[n] = −40
  2. 2. y[5] = 6
  3. 3. y(0.08) = 0.2637
  4. 4. Ay = 20
  5. 5. φ = −56.31 °
  6. 6. h[5] = −0.1638

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