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−3−2−1−12031h[n]n−10−3112

The sequence x[n] is nonzero for n = -2…0 and h[n] for n = 0…1, as shown. Find the first index nfirst at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -2…0h[n]: n = 0…1
nfirst =
2.Easy
x[n]n−1031−1233145h[n]n−100.5122−0.534

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

b = {0.5, 2, −0.5}x[n] = {3, −1, 3, 1}, n = 0…3
y[3] =
3.Medium
x(t)t13h(t)t33

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

x(t) = 3, 0 ≤ t < 1 sh(t) = 3, 0 ≤ t < 3 st = 1.5 s
y(1.5) =
4.Medium
x(t)t1h(t)t1

The input x(t) = 1·e−1tu(t) is applied to a system with h(t) = e−4tu(t). Find the output y(2) at t = 2 s.

x(t) = 1·e−1tu(t)h(t) = e−4tu(t)t = 2 s
y(2) =
5.Hard
x(t)t (ms)0.3125

A pulse train has height 5, pulse width 0.3 ms and period 1 ms. Find the amplitude c3 of harmonic number 3, at 3000 Hz.

A = 5τ = 0.3 msT = 1 ms
c3 =
6.Hard
x(t)t (ms)2484−4

A square wave switches between +4 and −4 with period T = 4 ms (half a period at each level). Find the amplitude c7 of its Fourier component at 1750 Hz.

±4T = 4 msf = 1750 Hz
c7 =

Answer key

  1. 1. nfirst = −2
  2. 2. y[3] = 7
  3. 3. y(1.5) = 9
  4. 4. y(2) = 0.045
  5. 5. c3 = 0.3279
  6. 6. c7 = 0.7276

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