signal processing and signal spectra

1. CONVOLUTION

2. CONVOLUTION
• The relationship between the input to a linear shift-invariant system, x(n), and the
output, y(n), is given by the convolution sum
y n = 𝑥 𝑛 ∗ ℎ 𝑛 =
𝑘=−∞
∞
𝑥 𝑘 ℎ(𝑛 − 𝑘)
• h(n)=impulse response
• x(n)=input
y n = ℎ 𝑛 ∗ 𝑥 𝑛 =
𝑘=−∞
∞
ℎ 𝑘 𝑥(𝑛 − 𝑘)
• x(n)=impulse response
• h(n)=input

3. CONVOLUTION PROPERTIES
• Convolution is a linear operator and, therefore, has a number of
important properties including the associative, commutative, and
distributive properties.

4. COMMUTATIVE PROPERTY
• 𝑥 𝑛 ∗ ℎ 𝑛 = ℎ 𝑛 ∗ 𝑥 𝑛 • 𝑥 𝑛 ∗ ℎ1 𝑛 ∗ ℎ2 𝑛 =
𝑥 𝑛 ∗ [ℎ1 𝑛 ∗ ℎ2 𝑛 ]
ASSOCIATIVE PROPERTY
CONVOLUTION PROPERTIES

5. DISTRIBUTIVE PROPERTY
• 𝑥 𝑛 ∗ ℎ1 𝑛 + ℎ2 𝑛 =
𝑥 𝑛 ∗ ℎ1 𝑛 + 𝑥 𝑛 ∗ ℎ2 𝑛
CONVOLUTION PROPERTIES

6. PERFORMING CONVOLUTIONS

7. DIRECT EVALUATION
• In performing convolutions directly, it is usually necessary to evaluate
finite or infinite sums involving terms of the form ‘an’ or ‘nan’.
• This is the mathematical calculation of a convolution

8. COMMONLY ENCOUNTERED SERIES

9. EXAMPLE
• 𝑥 𝑛 = 𝑎𝑛𝑢 𝑛
• ℎ 𝑛 = 𝑢(𝑛)
• What is 𝑥 𝑛 ∗ ℎ(𝑛)?

10. SOLUTION
𝑥 𝑛 ∗ ℎ 𝑛 =
𝑘=−∞
∞
𝑥 𝑘 ℎ 𝑛 − 𝑘
=
𝑘=−∞
∞
𝑎𝑘
𝑢 𝑘 𝑢 𝑛 − 𝑘
𝑘=0
𝑛
𝑎𝑘
=
1 − 𝑎𝑛+1
1 − 𝑎
𝑦 𝑛 =
1 − 𝑎𝑛+1
1 − 𝑎
𝑢(𝑛)
ℎ 𝑛 ∗ 𝑥 𝑛 =
𝑘=−∞
∞
ℎ 𝑘 𝑥 𝑛 − 𝑘
=
𝑘=−∞
∞
𝑢 𝑘 𝑎𝑛−𝑘
𝑢 𝑛 − 𝑘
𝑘=0
𝑛
𝑎𝑛−𝑘
= 𝑎𝑛
𝑘=0
𝑛
𝑎−𝑘
= 𝑎𝑛
𝑘=0
𝑛
1
𝑎
𝑘
𝑎𝑛
1 −
1
𝑎
𝑛+1
1 −
1
𝑎
=
𝑎𝑛
− 𝑎𝑛
𝑎−𝑛−1
𝑎 − 1
𝑎
=
𝑎𝑛
− 𝑎−1
𝑎 − 1
𝑎
𝑎𝑛
−
1
𝑎
𝑎 − 1
𝑎
=
𝑎𝑛+1
− 1
𝑎
𝑎 − 1
𝑎
=
𝑎𝑛+1
− 1
𝑎 − 1

11. EXAMPLE
• Determine the impulse response for the cascade of 2 linear time-
invariant systems having impulse responses
ℎ1 𝑛 =
1
2
𝑛
𝑢 𝑛
ℎ2 𝑛 =
1
4
𝑛
𝑢(𝑛)

12. SOLUTION
ℎ 𝑛 =
𝑘=−∞
∞
ℎ1(𝑘)ℎ2(𝑛 − 𝑘)
ℎ 𝑛 =
𝑘=−∞
∞
1
2
𝑘
𝑢 𝑘
1
4
𝑛−𝑘
𝑢(𝑛 − 𝑘)
ℎ 𝑛 =
𝑘=0
𝑛
1
2
𝑘
1
4
𝑛−𝑘
=
𝑘=0
𝑛
1
2
𝑘
1
2
2 𝑛−𝑘
=
𝑘=0
𝑛
1
2
𝑘
1
2
2𝑛−2𝑘
=
𝑘=0
𝑛
1
2
2𝑛−𝑘
ℎ 𝑛 =
1
2
2𝑛
𝑘=0
𝑛
1
2
−𝑘
=
1
2
2𝑛
𝑘=0
𝑛
2 𝑘
=
1
2
2𝑛
𝑘=0
𝑛−1
2 𝑘
+ 2𝑛
=
1
2
2𝑛
1 − 2𝑛
1 − 2
+ 2𝑛
ℎ 𝑛 =
1
2
2𝑛
1 − 2𝑛
− 2𝑛
−1
=
1
2
2𝑛
2𝑛+1 − 1 = 21−𝑛 −
1
4𝑛
ℎ 𝑛 = 21−𝑛
−
1
𝑛
𝑢(𝑛)

13. GRAPHICAL APPROACH
Steps:
1. Plot both sequences, x(k) and h(k), as functions of k
2. Choose one of the sequences, say h(k), and time-reverse it to form the sequence h(-k)
3. Shift the time reversed sequence by n. Note that if n>0, this corresponds to a shift to the right (delay), whereas if
n<0, this corresponds to a shift to the left (advance)
4. Multiply the sequences x(k) and h(n-k) and sum the product for all values of k. The resulting value will be equal to
y(n). This process is repeated for all possible shifts, n.
5. A useful fact to remember in performing the convolution of 2 finite-length sequences is that if x(n) is of length 𝐿1,
and h(n) is of length 𝐿2, 𝑦 𝑛 = 𝑥 𝑛 ∗ ℎ(𝑛) will be of length
𝐿 = 𝐿1 + 𝐿2 − 1
6. Furthermore, if the nonzero values of x(n) are contained in the interval [𝑀𝑥, 𝑁𝑥] and the nonzero values of h(n) are
contained in the interval [𝑀ℎ, 𝑁ℎ], the nonzero values of y(n) will be confined to the interval [𝑀𝑥 + 𝑀ℎ, 𝑁𝑥 + 𝑁ℎ]

14. EXAMPLE

15. SOLUTION

16. NUMERICAL CALCULATION
• 𝑥 𝑘 = [0,1,1,1]
• ℎ 𝑘 = [1, 2, 3]
• ℎ −𝑘 = [3, 2, 1]
• ℎ 1 − 𝑘 = [3, 2,1]
• ℎ 2 − 𝑘 = [0,3,2,1]
• ℎ 3 − 𝑘 = [0,0,3,2,1]
• ℎ 4 − 𝑘 = [0,0,0,3,2,1]
• ℎ 5 − 𝑘 = [0,0,0,0,3,2,1]
• ℎ −1 − 𝑘 = [3,2, 1]
𝑥 𝑘 = 0,0, 0, 1,1,1,0,0,0
ℎ −𝑘 = 0,3, 2, 1,0,0,0,0,0
ℎ 1 − 𝑘 = 0,0, 3, 2,1,0,0,0,0
ℎ 2 − 𝑘 = 0,0, 0, 3,2,1,0,0,0
ℎ 3 − 𝑘 = 0,0, 0, 0,3,2,1,0,0
ℎ 4 − 𝑘 = 0,0, 0, 0,0,3,2,1,0
ℎ 5 − 𝑘 = 0,0, 0, 0,0,0,3,2,1
ℎ −1 − 𝑘 = 3,2, 1, 0,0,0,0,0,0
𝑦 −1 = 𝑥 𝑘 ℎ −1 − 𝑘
= 0 × 3 + 0 × 2 + 0 × 1 + 1 × 0 + 1 × 0 + 1 × 0 + 0 × 0 + 0 × 0 + (0 × 0) = 0
𝑦 0 = 𝑥 𝑘 ℎ −𝑘 = 0 × 0 + 0 × 3 + 0 × 2 + 1 × 1 + 1 × 0 + 1 × 0 + (0 × 0) + (0 × 0) + (0 × 0)
= 1
𝑦 1 = 𝑥 𝑘 ℎ 1 − 𝑘
= 0 × 0 + 0 × 0 + 0 × 3 + 1 × 2 + 1 × 1 + 1 × 0 + (0 × 0) + (0 × 0) + (0 × 0) = 3
𝑦 2 = 𝑥 𝑘 ℎ 2 − 𝑘
= 0 × 0 + 0 × 0 + 0 × 0 + 1 × 3 + 1 × 2 + 1 × 1 + (0 × 0) + (0 × 0) + (0 × 0) = 6
𝑦 3 = 𝑥 𝑘 ℎ 3 − 𝑘
= 0 × 0 + 0 × 0 + 0 × 0 + 1 × 0 + 1 × 3 + 1 × 2 + (0 × 1) + (0 × 0) + (0 × 0) = 5

17. SLIDE RULE METHOD
1. Write the values of x(k) along a piece of paper, and the values of
h(-k) along the top of another piece of paper as illustrated
2. Line up the two sequence values x(0) and h(0), multiply each
pair of numbers, and add the products to form the value of y(0)
3. Slide the paper with the time-reversed sequence h(k) to the
right by one, multiply each pair of numbers, sum the products
to find the value of y(1), and repeat all shifts to the right by n>0.
Do the same, shifting the time-reversed sequence to the left, to
find the values of y(n) for n<0