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(a) To what analog frequency does the index k = 150 correspond What about k = 800 (b) What is the spacing between the spectral samples
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(a) With a sampling frequency R, = 2n/T, = 2rr(20 . I@), the discrete frequency w is related to the analog fre-
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quency R by w = QT,
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With an N-point DFT, the DTFT is sampled at the N frequencies 2rr wa=-k N k=0,1,
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Therefore, X(k) corresponds to an analog frequency of
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Thus, with N = 1000, the index k = 150 corresponds to f = 3 kHz. For k = 800, we need to be careful. Because X(eJ1*)is periodic,
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k = 800 corresponds to the frequency
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with N = 1000, this is wk = -0.41~. In analog frequency, this corresponds to
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CHAP. 61 (b) The spacing between spectral samples is
THE DFT
An important operation in digital signal processing is correlation. Correlations are used in applications such as target detection and frequency (spectrum) estimation. The correlation of two signals, xl(n) and x2(n),is defined by
r,,.,,(m) = which is the convolution of xl(m) with .r;(-m),
x~(n).r;(m + n )
Given two finite-length sequences xl (n) and r2(n)of length M , the N-point circular correlation is
where i ,(n) and i;(n) are the periodic extensions of the finite-length sequences 2 (n) and xT(n), respectively. What is the DFT of the circular correlation r:l,,,(m)
The circular correlation of xl(n)and x;(n) is one period of the periodic convolution of i l ( n ) with I;(-n). This, in turn, is equal to one period of the periodic convolution of i l ( n ) with P*(N - n), which is the same as the circular convolution of x l ( n ) with x,*(N- n):
Consequently, the DFT of ~-:~,,,(rn)is equal to the producl of the DFTs of xl(m)and x;(N - m):
R:l ,,(k) = DFT[xl(m)lDFT [X;(N
Because the DFT of x;(N - rn) is
- m)]
then
r.:1,,2(m)
x,(k)XX~:k)
Consider the two sequences
Find the N -point circular convolution of
X I (n)
with xz(n).
Find the N-point circular correlation of x ~ ( nwith itself (this is referred to as the autocorrelation). ) Find the N-point circular correlation of x l (n) with x2(n).
We may perform the circular convolution of xl(n)with xz(n) in any one of several different ways. However, because there are only two coefficients in the DFT of xl(n)and xz(n)that are nonzero, the easiest approach is to find the inverse DFT of the product Xl(k)X2(k). With
THE DFT
[CHAP. 6
the N-voint DFTs are
x I( k ) =
k = Iandk=N-1
else
Therefore,
and the inverse DFT is a sinusoid with an amplitude of N / 2 :
x(n) =xl(n)
o x2(n) =
N sin
(2Y)
(b) For the DFT of the circular autocorrelation of x l ( n ) , we have
and it follows that
r, ( n ) = N cos
else
(c) Finally, for the DFT of the circular correlation of x l ( n ) with x2(n), we have
else
Therefore,
r . r , x 2 ( n= - i N sin )
(2;)
It is known that y(n) is the output of a stable LSI system that has a system function
H(z) =
1 -c u r '
The input x ( n ) is completely unknown, and we would like to recover x ( n ) from y ( n ) . The following procedure is proposed for recovering part of x ( n ) from y(n).
Using N values of y(n) for 0 5 n < N , calculate the N -point DFT
Form the sequence
V ( k ) = -[I
-U W ~ ] Y ( ~ )
CHAP. 61
THE DF'T
Invert V ( k ) to obtain v(n) u(n) = IDFT(V(k)J
For what values of n in the range n = 0, 1 , . . . , N - 1 is it true that x(n) = u(n)
To recover x ( n ) , ideally we would take y ( n ) and convolve it with the inverse system
Therefore, x ( n ) may be recovered exactly by convolving y ( n ) with the FIR filter
which is of length L = 2. Therefore, because multiplying the DFT of g ( n ) with the DFT of y ( n ) is equivalent to performing the circular convolution,
v ( n ) = ~ ( nO y ( n ) ) v ( n ) will be equal to x ( n ) only for I 5 n 5 N - 1.
The N-point circular convolution of two sequences x(n) and h(n) of length N may be written in matrix form as follows: y=Hx where H is an N x N circulantmatrix, and xand y are vectors that contain the signal valuesx(O), x(l), . . . , x(N - I) and y(O), y(l), . .. y(N - l), respectively. Determine the form of the matrix H.
The circular convolution of x ( n ) with h ( n ) is
For example, ,y(O)is the sum of the products of x ( k ) with the circularly time-reversed sequence h ( ( - k ) ) N :
Next, for y(l), we circularly shift h ( ( - k ) ) N to the right by 1 and multiply by the sequence values x ( k ) :
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