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−102132−13445h[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] = {2, 3, −1, 4} for n = 0…3 and zero elsewhere. Find the output y[2].

b = {2, 0.25, 0.25}x[n] = {2, 3, −1, 4}, n = 0…3
y[2] =
2.Easy
x[n]n−1041−1223h[n]n−3−2−1−120−21

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

x[n]: n = 0…2h[n]: n = -2…0
nlast =
3.Medium
x(t)t2h(t)t0.21

A system has the impulse response h(t) = e−5tu(t). The input is a step x(t) = 2·u(t). Find the output y(0.4) at t = 0.4 s.

h(t) = e−5tu(t)x(t) = 2·u(t)t = 0.4 s
y(0.4) =
4.Medium
x(t)t11h(t)t22

The pulse x(t) = 1 for 0 ≤ t < 1 s is convolved with h(t), a triangle that falls linearly from 2 at t = 0 to 0 at t = 2 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 1, 0 ≤ t < 1 sh(0) = 2, h(2) = 0
Ay =
5.Hard
x(t)t (ms)1241−1

A triangle wave rises and falls linearly between −1 and +1 with period T = 2 ms. Find the amplitude c3 of its Fourier component at 1500 Hz.

±1T = 2 msf = 1500 Hz
c3 =
6.Hard
5cos(5t)h(t) = 10e−5t​u(t)y(t)

The input x(t) = 5·cos(5t) is applied to a system with h(t) = 10·e−5tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

h(t) = 10e−5tu(t)x(t) = 5cos(5t)
φ = °

Answer key

  1. 1. y[2] = −0.75
  2. 2. nlast = 2
  3. 3. y(0.4) = 0.3459
  4. 4. Ay = 2
  5. 5. c3 = 0.09006
  6. 6. φ = −45 °

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