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−10−3122−234h[n]n−1011−2234

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

x[n] = {−3, 2, −2}, n = 0…2h[n] = {1, −2}
Σy[n] =
2.Easy
x[n]n−10123−24−3536h[n]n−1012−134425

The sequence x[n] is nonzero for n = 3…5 and h[n] for n = 2…4, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 3…5h[n]: n = 2…4
nlast =
3.Medium
x(t)t5h(t)t0.51

A system has the impulse response h(t) = e−2tu(t). The input is a step x(t) = 5·u(t). Find the output y(0.1) at t = 0.1 s.

h(t) = e−2tu(t)x(t) = 5·u(t)t = 0.1 s
y(0.1) =
4.Medium
x(t)t−21h(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 tstart at which y(t) = x(t) ∗ h(t) starts.

x: −2 … 1 sh: 0 … 2 s
tstart = s
5.Hard
5cos(2t)h(t) = 5e−2t​u(t)y(t)

The input x(t) = 5·cos(2t) 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) = 5cos(2t)
φ = °
6.Hard
x(t)t (ms)510202−2

A square wave switches between +2 and −2 with period T = 10 ms (half a period at each level). Find the amplitude c1 of its Fourier component at 100 Hz.

±2T = 10 msf = 100 Hz
c1 =

Answer key

  1. 1. Σy[n] = 3
  2. 2. nlast = 9
  3. 3. y(0.1) = 0.4532
  4. 4. tstart = −2 s
  5. 5. φ = −45 °
  6. 6. c1 = 2.546

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