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−1011221344h[n]n−102132−234

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

x[n] = {1, 2, 1, 4}, n = 0…3h[n] = {2, 3, −2}
Σy[n] =
2.Easy
x[n]n−3−2−2−11041h[n]n−10123−343526

The sequence x[n] is nonzero for n = -2…0 and h[n] for n = 3…5, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -2…0h[n]: n = 3…5
nlast =
3.Medium
x(t)t01h(t)t15

The signal x(t) is nonzero only between t = 0 s and t = 1 s, and h(t) only between t = 1 s and t = 5 s. Find the time tstart at which y(t) = x(t) ∗ h(t) starts.

x: 0 … 1 sh: 1 … 5 s
tstart = s
4.Medium
x(t)t1h(t)t1

The input x(t) = 1·e−2tu(t) is applied to a system with h(t) = e−5tu(t). Find the output y(0.5) at t = 0.5 s.

x(t) = 1·e−2tu(t)h(t) = e−5tu(t)t = 0.5 s
y(0.5) =
5.Hard
x(t)t (ms)2.510201

A pulse train has height 1, pulse width 2.5 ms and period 10 ms. Find the amplitude c1 of harmonic number 1, at 100 Hz.

A = 1τ = 2.5 msT = 10 ms
c1 =
6.Hard
u[n]0.5+y[n]0.8·z⁻¹+

A system obeys y[n] = 0.8·y[n − 1] + 0.5·x[n] and starts at rest. The input is the unit step x[n] = u[n]. Find y[2].

y[n] = 0.8y[n − 1] + 0.5x[n]x[n] = u[n]
y[2] =

Answer key

  1. 1. Σy[n] = 24
  2. 2. nlast = 5
  3. 3. tstart = 1 s
  4. 4. y(0.5) = 0.09526
  5. 5. c1 = 0.4502
  6. 6. y[2] = 1.22

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