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−10−31−32−134h[n]n−1011−2234

The sequences x[n] = {−3, −3, −1} and h[n] = {1, −2} are convolved. Find the sum Σy[n] of all samples of y[n] = x[n] ∗ h[n].

x[n] = {−3, −3, −1}, n = 0…2h[n] = {1, −2}
Σy[n] =
2.Easy
x[n]n−10−11−3223−24h[n]n−102112334

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

x[n] = {−1, −3, 2, −2}, n = 0…3h[n] = {2, 1, 3}, n = 0…2
y[2] =
3.Medium
x(t)t23h(t)t41

The pulse x(t) = 3 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with the impulse response h(t) = 1 for 0 ≤ t < 4 s (zero elsewhere). Find the output y(3.5) at t = 3.5 s.

x(t) = 3, 0 ≤ t < 2 sh(t) = 1, 0 ≤ t < 4 st = 3.5 s
y(3.5) =
4.Medium
x(t)t11h(t)t35

The pulse x(t) = 1 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with the impulse response h(t) = 5 for 0 ≤ t < 3 s (zero elsewhere). Find the output y(0.5) at t = 0.5 s.

x(t) = 1, 0 ≤ t < 1 sh(t) = 5, 0 ≤ t < 3 st = 0.5 s
y(0.5) =
5.Hard
u[n]2+y[n]−0.8·z⁻¹+

A system obeys y[n] = −0.8·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] = −0.8y[n − 1] + 2x[n]x[n] = u[n]
y[4] =
6.Hard
δ(t)y′ + 2y = 2xh(t)

A system obeys y′(t) + 2·y(t) = 2·x(t). Find its impulse response at t = 0.25 s, h(0.25).

y′ + 2y = 2xt = 0.25 s
h(0.25) =

Answer key

  1. 1. Σy[n] = 7
  2. 2. y[2] = −2
  3. 3. y(3.5) = 6
  4. 4. y(0.5) = 2.5
  5. 5. y[4] = 1.475
  6. 6. h(0.25) = 1.213

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