Fundamentals of System Analysis and Synthesis - Digital Signal Processing in Power System Protection and Control - page 43

The inverse transform is then given by:

where n ¼ 0 ; 1 ; ... ; N 1

X

N 1

x ð n Þ¼ 1
N

jnk 2p
N

X ð k Þ exp ð 4 : 27 Þ

k ¼ 0

allowing to get the time series x(n) from given frequency samples X(k).
A pair of above equations describes DFT. It allows to pass between time and
frequency domains and has many applications to analysis of signals and systems,
to analysis and synthesis of digital filters, and others, [ 1 , 4 , 8 ].

Example 4.10 Determine the DFT of the signals: (a) x(n) = 1 for 0 B n B N - 1,
(b) x ð n Þ¼ cos n ð 2p = N Þ

ð Þ for the same range of samples as in example (a).

Solution

(a)

¼

X ð k Þ¼ X

N 1

exp jnk 2p
N

1 exp ð jk2p Þ
1 exp ð jk2p = N Þ

n ¼ 0

and thus

X ð k Þ¼ N

for k ¼ 0

:

0 for other k

(b) The considered signal can be decomposed as follows:

¼ 0 : 5 þ exp jn 2p

N

¼ 0 : 5 f x 1 ð n Þþ x 2 ð n Þg:

n 2p
N

jn 2p
N

x ð n Þ¼ cos exp

Now, calculating the DFT for both exponential components one obtains:

¼ 1 exp ½ j ð k 1 Þ 2p
1 exp j ð k 1 Þ 2 N

X 1 ð k Þ¼ DFT f x 1 ð n Þg ¼ X

N 1

exp jn 2p
N

ð k 1 Þ ;

n ¼ 0

which yields:

X 1 ð k Þ¼ N

for k ¼ 1

where 0 k N 1

0 for other k

and analogously:

X 2 ð k Þ¼ 1 exp ½ j ð k þ 1 Þ 2p
1 exp j ð k þ 1 Þ 2 N

: