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−4−3−3−2−2−1−201h[n]n−103132−13

The sequence x[n] is nonzero for n = -3…-1 and h[n] for n = 0…2, as shown. Find the first index nfirst at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -3…-1h[n]: n = 0…2
nfirst =
2.Easy
x[n]n−10−114213345h[n]n−10210.2520.2534

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

b = {2, 0.25, 0.25}x[n] = {−1, 4, 1, 3}, n = 0…3
y[4] =
3.Medium
x(t)t14h(t)t33

The pulse x(t) = 4 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(3) at t = 3 s.

x(t) = 4, 0 ≤ t < 1 sh(t) = 3, 0 ≤ t < 3 st = 3 s
y(3) =
4.Medium
x(t)t1h(t)t0.251

A system has the impulse response h(t) = e−4tu(t). The input is a step x(t) = 1·u(t). Find the output y(0.05) at t = 0.05 s.

h(t) = e−4tu(t)x(t) = 1·u(t)t = 0.05 s
y(0.05) =
5.Hard
u[n]2+y[n]−0.5·z⁻¹+

A system obeys y[n] = −0.5·y[n − 1] + 2·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[3].

y[n] = −0.5y[n − 1] + 2x[n]x[n] = u[n]
y[3] =
6.Hard
x(t)t (ms)1245−5

A square wave switches between +5 and −5 with period T = 2 ms (half a period at each level). Find the amplitude c3 of its Fourier component at 1500 Hz.

±5T = 2 msf = 1500 Hz
c3 =

Answer key

  1. 1. nfirst = −3
  2. 2. y[4] = 1
  3. 3. y(3) = 12
  4. 4. y(0.05) = 0.04532
  5. 5. y[3] = 1.25
  6. 6. c3 = 2.122

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