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−10−11−22−33h[n]n−1012−233425

The sequence x[n] is nonzero for n = 0…2 and h[n] for n = 2…4, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 0…2h[n]: n = 2…4
nlast =
2.Easy
x[n]n−102132−23h[n]n−2−1−10−3132

The sequence x[n] is nonzero for n = 0…2 and h[n] for n = -1…1, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 0…2h[n]: n = -1…1
nlast =
3.Medium
x(t)t0.55h(t)t1

The pulse x(t) = 5 for 0 ≤ t < 0.5 s (zero elsewhere) is applied to a system with h(t) = e−5tu(t). Find the output y(1.5) at t = 1.5 s, after the pulse has ended.

x(t) = 5, 0 ≤ t < 0.5 sh(t) = e−5tu(t)t = 1.5 s
y(1.5) =
4.Medium
x(t)t2h(t)t1

The input x(t) = 2·e−3tu(t) is applied to a system with h(t) = e−1tu(t). Find the output y(1.5) at t = 1.5 s.

x(t) = 2·e−3tu(t)h(t) = e−1tu(t)t = 1.5 s
y(1.5) =
5.Hard
x(t)t (ms)15104

A pulse train has height 4, pulse width 1 ms and period 5 ms. Find the amplitude c1 of harmonic number 1, at 200 Hz.

A = 4τ = 1 msT = 5 ms
c1 =
6.Hard
10cos(20t)h(t) = 5e−20t​u(t)y(t)

The input x(t) = 10·cos(20t) 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(20t)
Ym =

Answer key

  1. 1. nlast = 6
  2. 2. nlast = 3
  3. 3. y(1.5) = 0.006185
  4. 4. y(1.5) = 0.212
  5. 5. c1 = 1.497
  6. 6. Ym = 1.768

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