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−10121334−15h[n]n−102112

The sequence x[n] is nonzero for n = 2…4 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…4h[n]: n = 0…1
nfirst =
2.Easy
x[n]n−101232435−16h[n]n−10−11−12−13

The sequence x[n] is nonzero for n = 3…5 and h[n] for n = 0…2, as shown. Find the first index nfirst at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 3…5h[n]: n = 0…2
nfirst =
3.Medium
x(t)t15h(t)t13

The pulse x(t) = 5 for 0 ≤ t < 1 s is convolved with h(t), a triangle that falls linearly from 3 at t = 0 to 0 at t = 1 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 5, 0 ≤ t < 1 sh(0) = 3, h(1) = 0
Ay =
4.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−0.5tu(t). Find the output y(3) at t = 3 s, after the pulse has ended.

x(t) = 10, 0 ≤ t < 2 sh(t) = e−0.5tu(t)t = 3 s
y(3) =
5.Hard
u[n]0.5+y[n]0.5·z⁻¹+

A system obeys y[n] = 0.5·y[n − 1] + 0.5·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[2].

y[n] = 0.5y[n − 1] + 0.5x[n]x[n] = u[n]
y[2] =
6.Hard
x(t)t (ms)2.510205

A pulse train has height 5, pulse width 2.5 ms and period 10 ms. Find the amplitude c1 of harmonic number 1, at 100 Hz.

A = 5τ = 2.5 msT = 10 ms
c1 =

Answer key

  1. 1. nfirst = 2
  2. 2. nfirst = 3
  3. 3. Ay = 7.5
  4. 4. y(3) = 7.668
  5. 5. y[2] = 0.875
  6. 6. c1 = 2.251

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