Thus, by inspection, we see that

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with the other DFS coefficients from k = 0 to k = 19 equal to zero.

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The Discrete Fourier 'Ikansform

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Compute the N-point DFT of each of the following sequences:

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xl(n) = &n)

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(b) x2(n) = S(n - no), where 0 < no

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THE DFT (c) x 3 ( n ) = c r n

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[CHAP. 6

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O s n c N

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( d ) x&)

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= u(n) - u(n - no),where 0 < no < N

(a) The DFT of the unit sample may be easily evaluated from the definition of the DFT:

Another approach, however, is to recall that the DFT corresponds to samples of the z-transform X l ( z ) at N equally spaced points around the unit circle. Because X I ( z ) = I, i t follows that X l ( k ) = I .

(h) For the second sequence, we may again evaluate the DFT directly from the definition of the DFT. Let us instead, however, sample the z-transform. We know that X 2 ( z ) = z - " U . Therefore, sampling X 2 ( z )at the points z = ~ , o r k = O , I ...., N-1,wefind

( c ) For x 3 ( n ) ,the DFT may be found directly as follows:

iV-I

(d) The DFT of the pulse. x 4 ( n ) = u ( n ) - u ( n - no), may be evaluated directly as follows:

Factoring out a complex exponential w l2 denominator, the DFT may be written as

from the numerator and a complex exponential w;l2 from the

Find the 10-point inverse DFT of

To find the inverse DFT, note that X ( k ) may be expressed as follows:

Written in this way, the inverse DFT may be easily determined. Specifically, note that the inverse DFT of a constant is a unit sample: x l ( n )= S ( n ) Similarly. the DFT of a constant is a unit sample:

x l ( k )= 1

Therefore, i t follows that

x(n) =

+6(n)

CHAP. 61

THE DFT

Find the N -point DFT of the sequence

Compare the values of the DFT coefficients X ( k ) when wo = 2 n k o l N to those when wo # 2 r r k o l N . Explain the difference.

To find the N-point DFT of this sequence, it is easier if we write the cosine in terms of complex exponentials:

Evaluating the DFT of each of these terms, we find

~ ( k= C x ( n ) e - i % n i = )

At this point, note that if 9 = 2 r r k o / N ,

N- I e - ~ n ( % ~ - r n+ ~

rV - I

e-~nliJi+ul)

Because the first term is a sum of a complex exponential of frequency wo = 2 z ( k - k o ) / N , the sum will be equal to i zero unless k = ko, in which case the sum is equal to N . Similarly, the second sum is equal to zero unless != N - k,,, in which case the sum is equal to N . Therefore. if on= 2 z k o / N . the DFT coefficients are

X(k) =

k = ko and k == N - ko

otherwise

In the general case, when wo # 2 n k n / N , we must use the geometric series to evaluate Eq. (6.18):

Factoring out a complex exponential from the numerator and one from the denominator, we have

Note that. unless % is an integer multiple of 2 n / N , X ( k ) is, in general, nonzero for each k . The reason for this difference belween these two cases comes from the fact that X ( k ) corresponds to samples of the DTFT of x ( n ) , which is

When sampled at N equally spaced points over the interval [ O , 2 n ] , the sample values will, in general, be nonzero. However, if wo = 2 z k o / N , all of the samples except those at k = kc and k = N - ko occur at the zeros of the sine function.

Find the N -point DFT of the sequence

THE DFT The DFT of this sequence may be evaluated by expanding the cosine as a sum of complex exponentials:

=4 +

[CHAP. 6

[ej2nnlN

+e-j2~nlN

]2 = 4 + ;+ $ p ~ 4 n " l N + ,e - j 4 ~ n l N I

Using the periodicity of the complex exponentials, we may write x ( n ) as follows: