Let x l ( 1 7 ) be the sequence

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Perform the N-point circular convolution of x l ( n ) with lz(n) by forming the product H ( k ) X I(k) and then finding the inverse DFT, y l ( n ) . The first L - 1 values of the circular convolution are aliased, and the last

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Discard Fig. 6-5. Illustration of the overlap-save method of block convolution.

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THE DFT

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N - L 1 values correspond to the linear. convolution of x ( n ) with h ( n ) . Due to the zero padding at the start of x l ( n ) , these last N - L I values are the first N -- L 1 values of y ( n ) :

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3. Let x 2 ( n ) be the N-point sequence that is extracted from x ( n ) with the first L - 1 values overlapping with those of x, ( n ) . 4. Perform an N -point circular convolution of x 2 ( n ) with h ( n ) by forming the product H ( k ) X 2 ( k ) and taking the inverse DFT. The first L - I values of y 2 ( n ) are discarded and the final N - L + 1 values are saved and concatenated with the saved values of yl ( n ) :

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Steps 3 and 4 are repeated until all of the values in the linear convolution have been evaluated.

The reason for the name overlap-save is that x ( n ) is partitioned into overlapping sequences of length N and, after performing the N-point circular convolution, only the last N - L 1 values are saved.

Solved Problems

Discrete Fourier Series

Find the DFS expansion of the sequence

Because i ( n ) is periodic with a period N = 4, the DFS coefficients may be found by evaluating the sum

Alternatively, ~fwe express i ( n ) in terns of complex exponentials,

and use the fact that

we may pick efthe DFS coefficients directly as follows. Using the periodicity of the complex exponentials,

we may express f ( n ) as

A 2" + -,J$sn A i(n) = -,IT" 2 2 Comparing Eys. (6.15) and (6.16 ), we see that

THE DFT

[CHAP. 6

If i ( n ) is a periodic sequence with a period N ,

2 ( n ) is also periodic with period 2N. Let g ( k ) denote the DFS coefficients when f ( n ) is considered to be periodic with a period N , and let 2 z ( k ) be the DFS coefficients when the period is assumed to be 2N. Express the DFS coefficientsg 2 ( k ) in terms of g ( k ) .

If we considerP(n) to be periodic with a period 2 N , the DFS coefficients are

BecauseP(n) = P(n

+ N), this sum may be written as

Note that the term in brackets is equal to 2 when k is even, and it is equal to zero when k is odd. When k is even,

Therefore, the DFS coefficients g2(k) are

If i ! ( n )and i z ( n ) are periodic with period N with DFS coefficients2 ( ( k )and x 2 ( k ) ,respectively, show that the sequence with DFS coefficientsf ( k ) = f I ( k ) X z ( k )is equal to the periodic convolution of ,( n ) i I and i 2 ( n ) :

i(n)= xil(k),iz(n -k )

Given that X(k) = X I(k)z2(k),the sequence .P(n) is

Because we would like to express 3(n) in terms of l ( n ) and Pz(n), we begin by substituting

for Xl(k) into Eq. (6.17). With this substitution. we have

Rearranging the sums and combining the exponentials yields I f(n) =

EP( I ) ~=fOz ( k ) e 1 2 " ' n - ' ) k ~ N I=,, k

CHAP. 6 1 Note that

THE DFT

Therefore, we have

as was to be shown.

Let x,(t) be a periodic continuous-time signal

that is sampled with a sampling frequency f, = 1 kHz.Find the DFS coefficients of the sampled signal

Z ( n ) = xa(nTT).

With a sampling frequency f, = 1 kHz, the sampling period is T, = l/f, = IW3. and the sampled signal is J(n) = x,(nTv) = A cos - n

(3 ) + (3)

B cos

The first term is periodic with a period N I= 10, and the second is periodic with a period N2 = 4. Therefore, the sum is periodic with period N = 20, and we may write

Expressingi(n) in terms of complex exponentials, we have J(n) = t e i % 2 n + t e - i % 2 f l

+ L!J%sn

t e - i % ~ n

Using the periodicity of the complex exponentials, it follows that

e-j$$2n

- ej$18n

e-j%5n

== e,%15n

As a result, i ( n ) may be written as J ( ~= ~ )

~ l g 2 +l $ e ~ $ 1 8 " f +9%5n

+ej$$l~n

which is in the form of a DFS decomposition,