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−10213233−24h[n]n−10111234

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

x[n] = {2, 3, 3, −2}, n = 0…3h[n] = {1, 1}
Σy[n] =
2.Easy
x[n]n−105152535…h[n]n−10−11−2223445

A system with the impulse response h[n] = {−1, −2, 2, 4} (n = 0…3) receives the step input x[n] = 5·u[n]. Find the output y[3].

h[n] = {−1, −2, 2, 4}x[n] = 5·u[n]
y[3] =
3.Medium
x(t)t4h(t)t1

The input x(t) = 4·e−3tu(t) is applied to a system with h(t) = e−1tu(t). Find the output y(0.5) at t = 0.5 s.

x(t) = 4·e−3tu(t)h(t) = e−1tu(t)t = 0.5 s
y(0.5) =
4.Medium
x(t)t0.52h(t)t1

The pulse x(t) = 2 for 0 ≤ t < 0.5 s (zero elsewhere) is applied to a system with h(t) = e−5tu(t). Find the output y(1) at t = 1 s, after the pulse has ended.

x(t) = 2, 0 ≤ t < 0.5 sh(t) = e−5tu(t)t = 1 s
y(1) =
5.Hard
x(t)t (ms)2.510202

A pulse train has height 2, pulse width 2.5 ms and period 10 ms. Find the amplitude c2 of harmonic number 2, at 200 Hz.

A = 2τ = 2.5 msT = 10 ms
c2 =
6.Hard
10cos(10t)h(t) = 2e−5t​u(t)y(t)

The input x(t) = 10·cos(10t) is applied to a system with h(t) = 2·e−5tu(t). Find the phase φ of the steady-state output relative to the input, in degrees.

h(t) = 2e−5tu(t)x(t) = 10cos(10t)
φ = °

Answer key

  1. 1. Σy[n] = 12
  2. 2. y[3] = 15
  3. 3. y(0.5) = 0.7668
  4. 4. y(1) = 0.03014
  5. 5. c2 = 0.6366
  6. 6. φ = −63.43 °

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