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−103132333…h[n]n−103122234

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

h[n] = {3, 2, 2}x[n] = 3·u[n]
y[5] =
2.Easy
x[n]n−10−2122−1334h[n]n−1021−32−134

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

x[n] = {−2, 2, −1, 3}, n = 0…3h[n] = {2, −3, −1}
Σy[n] =
3.Medium
x(t)t−2−1h(t)t02

The signal x(t) is nonzero only between t = −2 s and t = −1 s, and h(t) only between t = 0 s and t = 2 s. Find the time tend at which y(t) = x(t) ∗ h(t) ends.

x: −2 … −1 sh: 0 … 2 s
tend = s
4.Medium
x(t)t01h(t)t15

The signal x(t) is nonzero only between t = 0 s and t = 1 s, and h(t) only between t = 1 s and t = 5 s. Find the time tstart at which y(t) = x(t) ∗ h(t) starts.

x: 0 … 1 sh: 1 … 5 s
tstart = s
5.Hard
x(t)t (ms)0.251210

A pulse train has height 10, pulse width 0.25 ms and period 1 ms. Find its DC component (mean value) c0.

A = 10τ = 0.25 msT = 1 ms
c0 =
6.Hard
2cos(4t)h(t) = 5e−2t​u(t)y(t)

The input x(t) = 2·cos(4t) is applied to a system with h(t) = 5·e−2tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

h(t) = 5e−2tu(t)x(t) = 2cos(4t)
φ = °

Answer key

  1. 1. y[5] = 21
  2. 2. Σy[n] = −4
  3. 3. tend = 1 s
  4. 4. tstart = 1 s
  5. 5. c0 = 2.5
  6. 6. φ = −63.43 °

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