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−101223−3445h[n]n−2−110−21

The sequence x[n] is nonzero for n = 2…4 and h[n] for n = -1…0, 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 = -1…0
nfirst =
2.Easy
x[n]n−10−3112234h[n]n−102132−234

The sequences x[n] = {−3, 1, 2} and h[n] = {2, 3, −2} both start at n = 0. Find the sample y[1] of the convolution y[n] = x[n] ∗ h[n].

x[n] = {−3, 1, 2}, n = 0…2h[n] = {2, 3, −2}, n = 0…2
y[1] =
3.Medium
x(t)t4h(t)t0.51

A system has the impulse response h(t) = e−2tu(t). The input is a step x(t) = 4·u(t). Find the output y(0.5) at t = 0.5 s.

h(t) = e−2tu(t)x(t) = 4·u(t)t = 0.5 s
y(0.5) =
4.Medium
x(t)t4h(t)t1

The input x(t) = 4·e−1tu(t) is applied to a system with h(t) = e−2tu(t). Find the output y(0.25) at t = 0.25 s.

x(t) = 4·e−1tu(t)h(t) = e−2tu(t)t = 0.25 s
y(0.25) =
5.Hard
x(t)t (ms)0.51210−10

A triangle wave rises and falls linearly between −10 and +10 with period T = 1 ms. Find the amplitude c5 of its Fourier component at 5000 Hz.

±10T = 1 msf = 5000 Hz
c5 =
6.Hard
δ[n]4+h[n]0.6·z⁻¹+

A system obeys y[n] = 0.6·y[n − 1] + 4·x[n] and starts at rest. Find its impulse response sample h[3].

y[n] = 0.6y[n − 1] + 4x[n]x[n] = δ[n]
h[3] =

Answer key

  1. 1. nfirst = 1
  2. 2. y[1] = −7
  3. 3. y(0.5) = 1.264
  4. 4. y(0.25) = 0.6891
  5. 5. c5 = 0.3242
  6. 6. h[3] = 0.864

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