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= cos-n
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( b ) x[n]=cos-n+sin-n 3 4
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FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS
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[CHAP. 6
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The fundamental period of x [ n ] is No = 8, and Ro = 277/N0 = n / 4 . Rather than using Eq. ( 6 . 8 )to evaluate the Fourier coefficients c,, we use Euler's formula and get
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Thus, the Fourier coefficients for x [ n ] are c l = f , c - , = c - , + , = c 7 = $, and all other c, = 0. Hence, the discrete Fourier series of x(n1 is
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( b ) From Prob. 1.16(i) the fundamental period of x [ n ] is No = 24, and R, Again by Euler's formula we have
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= 277/N0 = 77/12.
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I Thus, c3 = -j(4),c4 = $ , c - , = c - ~ =+ Z~= ~ , c P 3 c - ~ = c 2 ] = I ( I ) , and all other c O = + ~ ~ c, = 0. Hence, the discrete Fourier series of x [ n ] is
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From Prob. l.l6( j ) the fundamental period of x [ n ] is No = 8, and Ro = 277/No = n / 4 . Again by Euler's formula we have
I Thus, c0 = f , c 1= a, c Fourier series of x[nl is
= c - + n = c7 =
1 a, and all other c,
= 0.
Hence, the discrete
Let x[n] be a real periodic sequence with fundamental period N, and Fourier coefficients ck = a k + jb,, where a, and b, are both real.
( a ) Show that a - , = a k and b-,= -bk. (b) Show that c,,,,, is real if No is even. (c) Show that x [ n ] can also be expressed as a discrete trigonometric Fourier series of the form
( N u - 1)/2
x[n]=co+2
k= 1
(a,coskfl,n -b,sinkfl,n)
(6.123)
( N o - 2)/2
if No is odd or x[n]
= c,
+ (-
1)'ch/,
(a, cos kR,n - bk sin k f l o n ) (6.124)
if N,, is even.
CHAP. 61
FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS
( a ) If x [ n l is real, then from Eq. ( 6 . 8 ) we have
Thus,
=a_,+jb-, a-,=a,
=(a,
+ j b k ) *= a , b-,=-bk
and we have and
( 6 ) If No is even, then from Eq. ( 6 . 8 )
1 No-'
NO n = o
( - l ) " x [ n ] = real
If No is odd, then (No - 1) is even and we can write x [ n ] as
Now, from Eq. (6.17)
Thus,
= c,
(No-1)/2
( a , cos kRon - bk sin k R o n )
If No is even, we can write x [ n ] as
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
[CHAP. 6
Then
x [ n ] = c0
+ ( - I ) ~ C , ~ /+,
(Nu-2)/2
2Re(ckejkRun)
= c,
+ ( -l)nc,v/2 + 2
(Nu-2)/2
k =1
( a , cos kfl,,n - b, sin k R o n )
Let x , [ n ] and x , [ n ] be periodic sequences with fundamental period No and their discrete Fourier series given by
Show that the sequence x [ n ] = x , [ n ] x , [ n ] is periodic with the same fundamental period No and can be expressed as
where ck is given by
Now note that
Thus, x [ n ] is periodic with fundamental period No.Let
Nu-'
Then
Ck =
N .=o o
X[n]e-ikfM
No-'
No n = o
z x l [ n ] x 2 [ n ]e-',"on
since and the term in parentheses is equal to ek-,.
CHAP. 61
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
Let x,[n] and x,[n] be the two periodic signals in Prob. 6.8. Show that
Equation (6.127) is known as Parseval's relation for periodic sequences.
From Eq. (6.126) we have
N0-I
No- 1
ck = Setting k
NO n = o
x , [ n ] x 2 [ n ]e-jknon =
m =O
ddk-m
in the above expression, we get
6.10. (a) Verify Parseval's identity [Eq. (6.19)] for the discrete Fourier series, that is,
(6) Using x [ n ] in Prob. 6.3, verify Parseval's identity [Eq. (6.19)].
( a ) Let
1 No-'
Then
d --
- NO n = O
X*[n~e-~k~on= -
] No-'
X[n]e~kRon
(6.128)
NO n = o
Equation (6.128) indicates that if the Fourier coefficientsof x [ n ] are c , , then the Fourier coefficientsof x * [ n ] are c ,. Setting x , [ n ]= x [ n ] and x 2 [ n ]= x * [ n ] in Eq. (6.1271, we have d k = c k and ek = c , (or e - , = c : ) and we obtain
( b ) From Fig. 6-7 and the results from Prob. 6.3, we have
and Parseval's identity is verified.
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
FOURIER TRANSFORM
6.11. Find the Fourier transform of
x[n]
-anu[-n - 1 1
a real
From Eq. (4.12) the z-transform of x[n] is given by X(z) =
I -az-'
Izl< lal
Thus, X(eJf') exists for la1 > 1 because the ROC of X( z ) then contains the unit circle. Thus,
6.12. Find the Fourier transform of the rectangular pulse sequence (Fig. 6-10)
x [ n ] =u [ n ] -u [ n - N ]
Using Eq. (1.90), the z-transform of x[nl is given by N- 1 1-ZN X(Z) = C zn= n=O 1-2
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