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−1011−223334h[n]n−10−3142334

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

x[n] = {1, −2, 3, 3}, n = 0…3h[n] = {−3, 4, 3}
Σy[n] =
2.Easy
x[n]n−10413233−145h[n]n−100.510.2520.534

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

b = {0.5, 0.25, 0.5}x[n] = {4, 3, 3, −1}, n = 0…3
y[3] =
3.Medium
x(t)t21h(t)t32

The pulse x(t) = 1 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with the impulse response h(t) = 2 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(1) at t = 1 s.

x(t) = 1, 0 ≤ t < 2 sh(t) = 2, 0 ≤ t < 3 st = 1 s
y(1) =
4.Medium
x(t)t45h(t)t41

The pulse x(t) = 5 for 0 ≤ t < 4 s is convolved with h(t), a triangle that falls linearly from 1 at t = 0 to 0 at t = 4 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 5, 0 ≤ t < 4 sh(0) = 1, h(4) = 0
Ay =
5.Hard
x(t)t (ms)1245−5

A triangle wave rises and falls linearly between −5 and +5 with period T = 2 ms. Find the amplitude c5 of its Fourier component at 2500 Hz.

±5T = 2 msf = 2500 Hz
c5 =
6.Hard
x(t)t (ms)15102

A pulse train has height 2, pulse width 1 ms and period 5 ms. Find the amplitude c1 of harmonic number 1, at 200 Hz.

A = 2τ = 1 msT = 5 ms
c1 =

Answer key

  1. 1. Σy[n] = 20
  2. 2. y[3] = 1.75
  3. 3. y(1) = 2
  4. 4. Ay = 40
  5. 5. c5 = 0.1621
  6. 6. c1 = 0.7484

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