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−10−11−1223145h[n]n−10110.2520.534

An FIR filter computes y[n] = x[n] + 0.25·x[n − 1] + 0.5·x[n − 2]. The input is x[n] = {−1, −1, 2, 1} for n = 0…3 and zero elsewhere. Find the output y[3].

b = {1, 0.25, 0.5}x[n] = {−1, −1, 2, 1}, n = 0…3
y[3] =
2.Easy
x[n]n−10123141546h[n]n−1012223

The sequence x[n] is nonzero for n = 3…5 and h[n] for n = 1…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 = 1…2
nfirst =
3.Medium
x(t)t10h(t)t0.251

A system has the impulse response h(t) = e−4tu(t). The input is a step x(t) = 10·u(t). Find the output y(0.25) at t = 0.25 s.

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

The input x(t) = 2·e−6tu(t) is applied to a system with h(t) = e−2tu(t). Find the output y(1.5) at t = 1.5 s.

x(t) = 2·e−6tu(t)h(t) = e−2tu(t)t = 1.5 s
y(1.5) =
5.Hard
u[n]2+y[n]−0.5·z⁻¹+

A system obeys y[n] = −0.5·y[n − 1] + 2·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[4].

y[n] = −0.5y[n − 1] + 2x[n]x[n] = u[n]
y[4] =
6.Hard
x(t)t (ms)2.510202

A pulse train has height 2, pulse width 2.5 ms and period 10 ms. Find the amplitude c3 of harmonic number 3, at 300 Hz.

A = 2τ = 2.5 msT = 10 ms
c3 =

Answer key

  1. 1. y[3] = 1
  2. 2. nfirst = 4
  3. 3. y(0.25) = 1.58
  4. 4. y(1.5) = 0.02483
  5. 5. y[4] = 1.375
  6. 6. c3 = 0.3001

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