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−1031−3243−34h[n]n−1011−22234

The sequences x[n] = {3, −3, 4, −3} and h[n] = {1, −2, 2} are convolved. Find the sum Σy[n] of all samples of y[n] = x[n] ∗ h[n].

x[n] = {3, −3, 4, −3}, n = 0…3h[n] = {1, −2, 2}
Σy[n] =
2.Easy
x[n]n−10−111233−245h[n]n−10110.52−0.534

An FIR filter computes y[n] = x[n] + 0.5·x[n − 1] − 0.5·x[n − 2]. The input is x[n] = {−1, 1, 3, −2} for n = 0…3 and zero elsewhere. Find the output y[4].

b = {1, 0.5, −0.5}x[n] = {−1, 1, 3, −2}, n = 0…3
y[4] =
3.Medium
x(t)t10h(t)t1

The input x(t) = 10·e−3tu(t) is applied to a system with h(t) = e−0.5tu(t). Find the output y(1.5) at t = 1.5 s.

x(t) = 10·e−3tu(t)h(t) = e−0.5tu(t)t = 1.5 s
y(1.5) =
4.Medium
x(t)t14h(t)t31

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

x(t) = 4, 0 ≤ t < 1 sh(t) = 1, 0 ≤ t < 3 st = 1 s
y(1) =
5.Hard
x(t)t (ms)2.510204

A pulse train has height 4, pulse width 2.5 ms and period 10 ms. Find its DC component (mean value) c0.

A = 4τ = 2.5 msT = 10 ms
c0 =
6.Hard
x(t)t (ms)2482−2

A triangle wave rises and falls linearly between −2 and +2 with period T = 4 ms. Find the amplitude c5 of its Fourier component at 1250 Hz.

±2T = 4 msf = 1250 Hz
c5 =

Answer key

  1. 1. Σy[n] = 1
  2. 2. y[4] = −2.5
  3. 3. y(1.5) = 1.845
  4. 4. y(1) = 4
  5. 5. c0 = 1
  6. 6. c5 = 0.06485

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