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−10−21−22334h[n]n−10−3112−234

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

x[n] = {−2, −2, 3}, n = 0…2h[n] = {−3, 1, −2}, n = 0…2
y[2] =
2.Easy
x[n]n−102142−33−14h[n]n−10114234

The sequences x[n] = {2, 4, −3, −1} and h[n] = {1, 4} both start at n = 0. Find the sample y[2] of the convolution y[n] = x[n] ∗ h[n].

x[n] = {2, 4, −3, −1}, n = 0…3h[n] = {1, 4}, n = 0…1
y[2] =
3.Medium
x(t)t11h(t)t33

The pulse x(t) = 1 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 3 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(1.5) at t = 1.5 s.

x(t) = 1, 0 ≤ t < 1 sh(t) = 3, 0 ≤ t < 3 st = 1.5 s
y(1.5) =
4.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−2tu(t). Find the output y(3) at t = 3 s, after the pulse has ended.

x(t) = 1, 0 ≤ t < 2 sh(t) = e−2tu(t)t = 3 s
y(3) =
5.Hard
2cos(15t)h(t) = 10e−10t​u(t)y(t)

The input x(t) = 2·cos(15t) is applied to a system with h(t) = 10·e−10tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 10e−10tu(t)x(t) = 2cos(15t)
Ym =
6.Hard
x(t)t (ms)0.8245

A pulse train has height 5, pulse width 0.8 ms and period 2 ms. Find the amplitude c2 of harmonic number 2, at 1000 Hz.

A = 5τ = 0.8 msT = 2 ms
c2 =

Answer key

  1. 1. y[2] = −7
  2. 2. y[2] = 13
  3. 3. y(1.5) = 3
  4. 4. y(3) = 0.06643
  5. 5. Ym = 1.109
  6. 6. c2 = 0.9355

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