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−3−2−2−1−20−2132h[n]n−102112

The sequence x[n] is nonzero for n = -2…1 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…1h[n]: n = 0…1
nfirst =
2.Easy
x[n]n−102122232…h[n]n−10−313243245

A system with the impulse response h[n] = {−3, 3, 4, 2} (n = 0…3) receives the step input x[n] = 2·u[n]. Find the output y[3].

h[n] = {−3, 3, 4, 2}x[n] = 2·u[n]
y[3] =
3.Medium
x(t)t41h(t)t21

The pulse x(t) = 1 for 0 ≤ t < 4 s is convolved with h(t), a triangle that falls linearly from 1 at t = 0 to 0 at t = 2 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 1, 0 ≤ t < 4 sh(0) = 1, h(2) = 0
Ay =
4.Medium
x(t)t15h(t)t1

The pulse x(t) = 5 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with h(t) = e−0.5tu(t). Find the output y(2.5) at t = 2.5 s, after the pulse has ended.

x(t) = 5, 0 ≤ t < 1 sh(t) = e−0.5tu(t)t = 2.5 s
y(2.5) =
5.Hard
δ(t)y′ + 4y = 2xh(t)

A system obeys y′(t) + 4·y(t) = 2·x(t). Find its impulse response at t = 1 s, h(1).

y′ + 4y = 2xt = 1 s
h(1) =
6.Hard
x(t)t (ms)310204

A pulse train has height 4, pulse width 3 ms and period 10 ms. Find the amplitude c1 of harmonic number 1, at 100 Hz.

A = 4τ = 3 msT = 10 ms
c1 =

Answer key

  1. 1. nfirst = −2
  2. 2. y[3] = 12
  3. 3. Ay = 4
  4. 4. y(2.5) = 1.859
  5. 5. h(1) = 0.03663
  6. 6. c1 = 2.06

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