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−10−2142−3314h[n]n−10−21−12434

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

x[n] = {−2, 4, −3, 1}, n = 0…3h[n] = {−2, −1, 4}, n = 0…2
y[3] =
2.Easy
x[n]n−10−11−12334h[n]n−10414234

The sequences x[n] = {−1, −1, 3} and h[n] = {4, 4} are convolved. Find the sum Σy[n] of all samples of y[n] = x[n] ∗ h[n].

x[n] = {−1, −1, 3}, n = 0…2h[n] = {4, 4}
Σy[n] =
3.Medium
x(t)t11h(t)t21

The pulse x(t) = 1 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 1 for 0 ≤ t < 2 s (zero elsewhere). Find the output y(2.5) at t = 2.5 s.

x(t) = 1, 0 ≤ t < 1 sh(t) = 1, 0 ≤ t < 2 st = 2.5 s
y(2.5) =
4.Medium
x(t)t22h(t)t1

The pulse x(t) = 2 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−1tu(t). Find the output y(2.5) at t = 2.5 s, after the pulse has ended.

x(t) = 2, 0 ≤ t < 2 sh(t) = e−1tu(t)t = 2.5 s
y(2.5) =
5.Hard
x(t)t (ms)0.4125

A pulse train has height 5, pulse width 0.4 ms and period 1 ms. Find the amplitude c1 of harmonic number 1, at 1000 Hz.

A = 5τ = 0.4 msT = 1 ms
c1 =
6.Hard
x(t)t (ms)510203−3

A square wave switches between +3 and −3 with period T = 10 ms (half a period at each level). Find the amplitude c5 of its Fourier component at 500 Hz.

±3T = 10 msf = 500 Hz
c5 =

Answer key

  1. 1. y[3] = 17
  2. 2. Σy[n] = 8
  3. 3. y(2.5) = 0.5
  4. 4. y(2.5) = 1.049
  5. 5. c1 = 3.027
  6. 6. c5 = 0.7639

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