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−10−21−32−234h[n]n−102112134

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

x[n] = {−2, −3, −2}, n = 0…2h[n] = {2, 1, 1}
Σy[n] =
2.Easy
x[n]n−3−23−1−2011h[n]n−10123−14−35

The sequence x[n] is nonzero for n = -2…0 and h[n] for n = 3…4, 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 = 3…4
nfirst =
3.Medium
x(t)t13h(t)t32

The pulse x(t) = 3 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 2 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(3.5) at t = 3.5 s.

x(t) = 3, 0 ≤ t < 1 sh(t) = 2, 0 ≤ t < 3 st = 3.5 s
y(3.5) =
4.Medium
x(t)t2h(t)t1

The input x(t) = 2·e−3tu(t) is applied to a system with h(t) = e−1tu(t). Find the output y(0.25) at t = 0.25 s.

x(t) = 2·e−3tu(t)h(t) = e−1tu(t)t = 0.25 s
y(0.25) =
5.Hard
x(t)t (ms)15101

A pulse train has height 1, pulse width 1 ms and period 5 ms. Find the amplitude c3 of harmonic number 3, at 600 Hz.

A = 1τ = 1 msT = 5 ms
c3 =
6.Hard
δ(t)y′ + 10y = 3xh(t)

A system obeys y′(t) + 10·y(t) = 3·x(t). Find its impulse response at t = 0.25 s, h(0.25).

y′ + 10y = 3xt = 0.25 s
h(0.25) =

Answer key

  1. 1. Σy[n] = −28
  2. 2. nfirst = 1
  3. 3. y(3.5) = 3
  4. 4. y(0.25) = 0.3064
  5. 5. c3 = 0.2018
  6. 6. h(0.25) = 0.2463

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