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−10−21221334h[n]n−10121314

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

x[n]: n = 0…3h[n]: n = 2…3
nfirst =
2.Easy
x[n]n−1041−12134h[n]n−10−21−2234

The sequences x[n] = {4, −1, 1} and h[n] = {−2, −2} both start at n = 0. Find the sample y[1] of the convolution y[n] = x[n] ∗ h[n].

x[n] = {4, −1, 1}, n = 0…2h[n] = {−2, −2}, n = 0…1
y[1] =
3.Medium
x(t)t21h(t)t1

The pulse x(t) = 1 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−4tu(t). Find the output y(4) at t = 4 s, after the pulse has ended.

x(t) = 1, 0 ≤ t < 2 sh(t) = e−4tu(t)t = 4 s
y(4) =
4.Medium
x(t)t−11h(t)t02

The signal x(t) is nonzero only between t = −1 s and t = 1 s, and h(t) only between t = 0 s and t = 2 s. Find the time tstart at which y(t) = x(t) ∗ h(t) starts.

x: −1 … 1 sh: 0 … 2 s
tstart = s
5.Hard
x(t)t (ms)0.5124

A square wave switches between 4 and 0 with period T = 1 ms (half a period at each level). Find the amplitude c3 of its Fourier component at 3000 Hz.

0 … 4T = 1 msf = 3000 Hz
c3 =
6.Hard
δ[n]0.5+h[n]−0.8·z⁻¹+

A system obeys y[n] = −0.8·y[n − 1] + 0.5·x[n] and starts at rest. Find its impulse response sample h[3].

y[n] = −0.8y[n − 1] + 0.5x[n]x[n] = δ[n]
h[3] =

Answer key

  1. 1. nfirst = 2
  2. 2. y[1] = −6
  3. 3. y(4) = 8.384 × 10⁻⁵
  4. 4. tstart = −1 s
  5. 5. c3 = 0.8488
  6. 6. h[3] = −0.256

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