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−104132−2314h[n]n−1012−33−24

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

x[n]: n = 0…3h[n]: n = 2…3
nfirst =
2.Easy
x[n]n−3−23−11031h[n]n−1011−32

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

x[n]: n = -2…0h[n]: n = 0…1
nfirst =
3.Medium
x(t)t210h(t)t1

The pulse x(t) = 10 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−5tu(t). Find the output y(3) at t = 3 s, after the pulse has ended.

x(t) = 10, 0 ≤ t < 2 sh(t) = e−5tu(t)t = 3 s
y(3) =
4.Medium
x(t)t31h(t)t14

The pulse x(t) = 1 for 0 ≤ t < 3 s is convolved with the pulse h(t) = 4 for 0 ≤ t < 1 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 1, 0 ≤ t < 3 sh(t) = 4, 0 ≤ t < 1 s
Ay =
5.Hard
5cos(20t)h(t) = 2e−20t​u(t)y(t)

The input x(t) = 5·cos(20t) is applied to a system with h(t) = 2·e−20tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

h(t) = 2e−20tu(t)x(t) = 5cos(20t)
φ = °
6.Hard
δ(t)y′ + 4y = 2xh(t)

A system obeys y′(t) + 4·y(t) = 2·x(t). Find its impulse response at t = 0.05 s, h(0.05).

y′ + 4y = 2xt = 0.05 s
h(0.05) =

Answer key

  1. 1. nfirst = 2
  2. 2. nfirst = −2
  3. 3. y(3) = 0.01348
  4. 4. Ay = 12
  5. 5. φ = −45 °
  6. 6. h(0.05) = 1.637

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