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−4−32−2−1−1101h[n]n−3−24−1−2021

The sequence x[n] is nonzero for n = -3…-1 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 = -3…-1h[n]: n = -2…0
nlast =
2.Easy
x[n]n−105152535…h[n]n−10−3142−234

A system with the impulse response h[n] = {−3, 4, −2} (n = 0…2) receives the step input x[n] = 5·u[n]. Find the output y[4].

h[n] = {−3, 4, −2}x[n] = 5·u[n]
y[4] =
3.Medium
x(t)t10h(t)t1

The input x(t) = 10·e−5tu(t) is applied to a system with h(t) = e−1tu(t). Find the output y(1) at t = 1 s.

x(t) = 10·e−5tu(t)h(t) = e−1tu(t)t = 1 s
y(1) =
4.Medium
x(t)t11h(t)t23

The pulse x(t) = 1 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 3 for 0 ≤ t < 2 s (zero elsewhere). Find the output y(1.5) at t = 1.5 s.

x(t) = 1, 0 ≤ t < 1 sh(t) = 3, 0 ≤ t < 2 st = 1.5 s
y(1.5) =
5.Hard
x(t)t (ms)1241

A square wave switches between 1 and 0 with period T = 2 ms (half a period at each level). Find the amplitude c1 of its Fourier component at 500 Hz.

0 … 1T = 2 msf = 500 Hz
c1 =
6.Hard
x(t)t (ms)25105

A pulse train has height 5, pulse width 2 ms and period 5 ms. Find its DC component (mean value) c0.

A = 5τ = 2 msT = 5 ms
c0 =

Answer key

  1. 1. nlast = −1
  2. 2. y[4] = −5
  3. 3. y(1) = 0.9029
  4. 4. y(1.5) = 3
  5. 5. c1 = 0.6366
  6. 6. c0 = 2

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