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−10−11−22−234h[n]n−10−1112−234

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

x[n] = {−1, −2, −2}, n = 0…2h[n] = {−1, 1, −2}
Σy[n] =
2.Easy
x[n]n−10−21−12234h[n]n−10−3112−234

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

x[n] = {−2, −1, 2}, n = 0…2h[n] = {−3, 1, −2}, n = 0…2
y[3] =
3.Medium
x(t)t11h(t)t31

The pulse x(t) = 1 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 1 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(1.5) at t = 1.5 s.

x(t) = 1, 0 ≤ t < 1 sh(t) = 1, 0 ≤ t < 3 st = 1.5 s
y(1.5) =
4.Medium
x(t)t14h(t)t1

The pulse x(t) = 4 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with h(t) = e−5tu(t). Find the output y(2) at t = 2 s, after the pulse has ended.

x(t) = 4, 0 ≤ t < 1 sh(t) = e−5tu(t)t = 2 s
y(2) =
5.Hard
x(t)t (ms)510204−4

A triangle wave rises and falls linearly between −4 and +4 with period T = 10 ms. Find the amplitude c5 of its Fourier component at 500 Hz.

±4T = 10 msf = 500 Hz
c5 =
6.Hard
δ[n]3+h[n]−0.8·z⁻¹+

A system obeys y[n] = −0.8·y[n − 1] + 3·x[n] and starts at rest. Find its impulse response sample h[4].

y[n] = −0.8y[n − 1] + 3x[n]x[n] = δ[n]
h[4] =

Answer key

  1. 1. Σy[n] = 10
  2. 2. y[3] = 4
  3. 3. y(1.5) = 1
  4. 4. y(2) = 0.005354
  5. 5. c5 = 0.1297
  6. 6. h[4] = 1.229

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