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−101122134h[n]n−1021−1234

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

x[n] = {1, 2, 1}, n = 0…2h[n] = {2, −1}, n = 0…1
ymax =
2.Easy
x[n]n−102122134h[n]n−10−214234

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

x[n] = {2, 2, 1}, n = 0…2h[n] = {−2, 4}
Σy[n] =
3.Medium
x(t)t5h(t)t1

The input x(t) = 5·e−4tu(t) is applied to a system with h(t) = e−0.5tu(t). Find the output y(2) at t = 2 s.

x(t) = 5·e−4tu(t)h(t) = e−0.5tu(t)t = 2 s
y(2) =
4.Medium
x(t)t22h(t)t1

The pulse x(t) = 2 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−2tu(t). Find the output y(4) at t = 4 s, after the pulse has ended.

x(t) = 2, 0 ≤ t < 2 sh(t) = e−2tu(t)t = 4 s
y(4) =
5.Hard
x(t)t (ms)520403

A pulse train has height 3, pulse width 5 ms and period 20 ms. Find the amplitude c1 of harmonic number 1, at 50 Hz.

A = 3τ = 5 msT = 20 ms
c1 =
6.Hard
5cos(40t)h(t) = 2e−20t​u(t)y(t)

The input x(t) = 5·cos(40t) 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) = 5cos(40t)
Ym =

Answer key

  1. 1. ymax = 3
  2. 2. Σy[n] = 10
  3. 3. y(2) = 0.5251
  4. 4. y(4) = 0.01798
  5. 5. c1 = 1.35
  6. 6. Ym = 0.2236

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