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−101132434h[n]n−10411234

The sequences x[n] = {1, 3, 4} and h[n] = {4, 1} both start at n = 0. Find the largest sample ymax of y[n] = x[n] ∗ h[n].

x[n] = {1, 3, 4}, n = 0…2h[n] = {4, 1}, n = 0…1
ymax =
2.Easy
x[n]n−2−12031−22h[n]n−101232435−26

The sequence x[n] is nonzero for n = -1…1 and h[n] for n = 3…5, as shown. Find the first index nfirst at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -1…1h[n]: n = 3…5
nfirst =
3.Medium
x(t)t10h(t)t1

The input x(t) = 10·e−6tu(t) is applied to a system with h(t) = e−4tu(t). Find the output y(0.25) at t = 0.25 s.

x(t) = 10·e−6tu(t)h(t) = e−4tu(t)t = 0.25 s
y(0.25) =
4.Medium
x(t)t2h(t)t1

The input x(t) = 2·e−5tu(t) is applied to a system with h(t) = e−4tu(t). Find the output y(1.5) at t = 1.5 s.

x(t) = 2·e−5tu(t)h(t) = e−4tu(t)t = 1.5 s
y(1.5) =
5.Hard
δ(t)y′ + 10y = 3xh(t)

A system obeys y′(t) + 10·y(t) = 3·x(t). Find its impulse response at t = 0.5 s, h(0.5).

y′ + 10y = 3xt = 0.5 s
h(0.5) =
6.Hard
x(t)t (ms)1.255103

A pulse train has height 3, pulse width 1.25 ms and period 5 ms. Find the amplitude c1 of harmonic number 1, at 200 Hz.

A = 3τ = 1.25 msT = 5 ms
c1 =

Answer key

  1. 1. ymax = 19
  2. 2. nfirst = 2
  3. 3. y(0.25) = 0.7237
  4. 4. y(1.5) = 0.003851
  5. 5. h(0.5) = 0.02021
  6. 6. c1 = 1.35

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