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−4−34−22−1101h[n]n−3−22−1301

The sequence x[n] is nonzero for n = -3…-1 and h[n] for n = -2…-1, as shown. Find the first index nfirst at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -3…-1h[n]: n = -2…-1
nfirst =
2.Easy
x[n]n−10−113213145h[n]n−100.51120.534

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

b = {0.5, 1, 0.5}x[n] = {−1, 3, 1, 1}, n = 0…3
y[2] =
3.Medium
x(t)t15h(t)t1

The pulse x(t) = 5 for 0 ≤ t < 1 s (zero elsewhere) is applied to a system with h(t) = e−4tu(t). Find the output y(3) at t = 3 s, after the pulse has ended.

x(t) = 5, 0 ≤ t < 1 sh(t) = e−4tu(t)t = 3 s
y(3) =
4.Medium
x(t)t24h(t)t1

The pulse x(t) = 4 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−0.5tu(t). Find the output y(4) at t = 4 s, after the pulse has ended.

x(t) = 4, 0 ≤ t < 2 sh(t) = e−0.5tu(t)t = 4 s
y(4) =
5.Hard
10cos(30t)h(t) = 5e−20t​u(t)y(t)

The input x(t) = 10·cos(30t) is applied to a system with h(t) = 5·e−20tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 5e−20tu(t)x(t) = 10cos(30t)
Ym =
6.Hard
δ(t)y′ + 4y = 5xh(t)

A system obeys y′(t) + 4·y(t) = 5·x(t). Find its impulse response at t = 0.1 s, h(0.1).

y′ + 4y = 5xt = 0.1 s
h(0.1) =

Answer key

  1. 1. nfirst = −5
  2. 2. y[2] = 3
  3. 3. y(3) = 0.0004116
  4. 4. y(4) = 1.86
  5. 5. Ym = 1.387
  6. 6. h(0.1) = 3.352

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