Then, using Eqs. (6.206)and (6.2121, we have in Visual Studio .NET

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Then, using Eqs. (6.206)and (6.2121, we have
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Fig. 6 3 8 Phase factors W: and W,".
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FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
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[CHAP. 6
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and by Eqs. ( 6 . 2 1 7 ~and (6.2176) we obtain )
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Noting that since x[n] is real and using Eq. (6.204),.X[7], X [ 6 ] , and X[5] can be easily obtained by taking the conjugates of X [ l ] , X[2], and X[3], respectively.
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6.58. Let x [ n ] be a sequence of finite length N such that
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x[n]=0
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n <0,n r N
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Let the N-point DFT X [ k ] of x [ n ] be given by [Eq. (6.92)]
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X[k]=
x [ n ]w,kn
wN -- e - ~ ( 2 7 7 / N )
k=0,1,
..., N - 1 (6.224)
Suppose N is even and let
(a) Show that the N-point DFT X [ k ] of x [ n ] can be expressed as
X[2k
+ 11 = Q [ k ]
p[nlW,k;z
k = 0 , 1 , ..., - - 1
(6.2266)
( N / 2 )- 1
where
P[kl=
k =0,1,
..., - 2
(6.227~) (6.2276)
(N/2)- 1
Q[k]=
4[nlW,k;2
k = 0 , 1 , ...,- - 1
(6) Draw a flow graph to illustrate the evaluation of X [ k ] from Eqs. (6.226~) and (6.2266) with N = 8.
( a ) We rewrite Eq. (6.224)as
Changing the variable n
+ N / 2 in the second term of Eq. (6.228),we have
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
Noting that [Eq. (6.223)l
Eq. (6.229) can be expressed as
For k even, setting k
= 2r
in Eq. (6.230), we have
where the relation in Eq. (6.220) has been used. Similarly, for k odd, setting k = 2r + 1 in Eq. (6.230). we get
Equations (6.231) and (6.232) represent the (N/2)-point DFT of ~ [ nand &I, respecl tively. Thus, Eqs. (6.231) and (6.232) can be rewritten as
(N/2)- 1
where
P[kl=
~[nlW,k;2
N k = O , l , ...,- - 1 2
The flow graph illustrating the steps involved in determining X[k] by Eqs. (6.227~) and (6.2276) is shown in Fig. 6-39. The method of evaluating X[k] based on Eqs. (6.227~) (6.2276) is known as the and decimation-in-frequencyfast Fourier transform (FFT) algorithm.
Fig. 6-39 Flow graph for an &point decimation-in-frequencyFFT algorithm.
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
[CHAP. 6
6.59. Using the decimation-in-frequency FFT technique, redo Prob. 6.57.
From Prob. 6.57 x[n]=(l,l,-1,-1,-l,l,l,-1)
By Eqs. (6.225a)and (6.225b)and using the values of W,"obtained in Prob. 6.57, we have
= (2,0, 12,O)
Then using Eqs. (6.206) and (6.212).we have
and by Eqs. (6.226a)and (6.2266)we get
X[0] X[1] X[2] X[3]
P[0] = 0 = Q[O] = 2 + j2 = P[1] = -j4 = Q[l] = 2 - j2
X[4] X[5] X[6] X[7]
= P[2]
=o = Q[2] = 2 + j2 = P[3] =j4 = Q[3] = 2 - j2
which are the same results obtained in Prob. 6.57.
6.60. Consider a causal continuous-time band-limited signal x ( t ) with the Fourier transform X(w). Let
where Ts is the sampling interval in the time domain. Let X [ k ] = X ( k Aw) where Aw is the sampling interval in the frequency domain known as the frequency resolution. Let T, be the record length of x ( t ) and let w, be the highest frequency of x ( t ) . Show that x [ n ] and X [ k ] form an N-point DFT pair if
TI T,
2% =N Aw
w ~ T 1
CHAP. 61 FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS
Since x(t ) = 0 for t < 0, the Fourier transform X(w) of x(t ) is given by [Eq.
Let T I be the total recording time of x ( t ) required to evaluate X(w). Then the above integral can be approximated by a finite series as
X(w) where tn = n At and T,
= NAt.
= At = w,
x ( t n ) e-;"'"
Setting w
in the above expression, we have x(t,)
e-'"'kln
N- 1
X ( w k ) = At
(6.237)
Next, since the highest frequency of x(t) is w,, by [Eq. (5.3211
the inverse Fourier transform of ~ ( w is given )
Dividing the frequency range -oMI w I w , integral can be approximated by
into N (even) intervals of length Aw, the above
where 2wM= NAw. Setting t
= t,
in the above expression, we have
Since the highest frequency in x(t) is w,, should sample x(t) so that
then from the sampling theorem (Prob. 5.59) we
where T, is the sampling interval. Since T, = At, selecting the largest value of A t (the Nyquist interval), we have At and Thus, N is a suitable even integer for which
T- ~ - , -- W M =
From Eq. (6.240) the frequency resolution A w is given by
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
Let t n = n A t and w k = k A w . Then
Substituting Eq. (6.243) into Eqs. (6.237) and (6.239), we get
N- 1
X(k Ao) x ( n At)
C A t x ( n A t ) e-j(2"/N)nk
(N/2)-1 and Rewrite Eq. (6.245) as X ( k A w ) e'(2"/N'nk+
( b w ) e(2"/N)nk k
-N/Z
X ( k A w ) ei(2"/N)nk
-N/2
Then from Eq. (6.244) we note that X ( k A w ) is periodic in k with period N. Thus, changing the variable k = m - N in the second sum in the above expression, we get
Multiplying both sides of Eq. (6.246) by At and noting that Aw At = 2,rr/N, we have 1 N-1 x ( n A t ) At = - C X ( k A w ) ei(2"/N)nk
Now if we define x [ n ]= A t x ( n A t ) X [ k ]= X ( k A w ) then Eqs. (6.244) and (6.247) reduce to the DFT pair, that is,
T,x(nT,)
(6.248)
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