Fourier Transform and Signal Spectrum - Digital Signal Processing: A Practitioner's Approach - page 89

Let us consider a periodic function x ð t Þ¼ A in the interval 0

< <

, otherwise
x ð t Þ¼ 0. For this signal, the values of a k and b k are obtained by evaluating the
integrals and are given as

t

sin T hi

2

A 2

T

2A

a k ¼ sin k and b k ¼ : ð 4 : 29 Þ

k k

We have taken a numerical example with 3; we use a

sampling time of 1 ms and a 512-point DFT. Figure 4.10 shows the theoretically
computed coefficients superimposed with the coefficients computed using the DFT.
There is a good match. Figure 4.10(a) shows the even nature of a k and Figure 4.10(b)
shows the odd nature of b k . As a complex spectrum it is conjugate symmetric.

¼ 0 : 1s,T ¼ 1 : 2 s and A ¼ 1 :

0.2

0.3

0.25

0.1

0.2

0.15

0

0.1

0.05

−0.1

0

− 0.05

− 0.2

−40 −20 0 20 40 60

− 40 − 20 0 20 40 60

Cosine

Sine

(a) (b)

Figure 4.10 Computing Fourier coefficients using a DFT

4.6 Convolution by DFT
Convolution can be best understood in a numerical sense by looking at it as
multiplication of two polynomials. We have taken this approach because poly-
nomial multiplication is very elementary. Consider

p 1 ð x Þ¼ 1 þ 2x ;

p 2 ð x Þ¼ 1 þ 3x þ 4x 2

ð 4 : 30 Þ

:

The product of the two polynomials p ð x Þ¼ p 1 ð x Þ p 2 ð x Þ is 1 þ 5x þ 10x 2 þ 8x 3 and

the coefficients in p ð x Þ are the convolution of the two sequences f 1 ; 2 g and

f 1 4 g . A convolution operation involves many computations. Typically, if an

m-sequence is to be convolved with an n-sequence, we need to perform OO(mn)
operations and the resulting sequence is of length m þ n 1. Taking the DFTs of
the two sequences, multiplying them point by point and taking an inverse DFT

; 3 ;