COMPUTATION OF THE DISCRETE FOURIER TRANSFORM - PART 1
Solutions 18.1
The flow-graph of Fig. 9.3 of the text is based on the decomposition of
X(k) in the form of equation 9.14 of text. For N =16, the corresponding
flow-graph expressing X(k) as a combination of two eight-point DFT's
is shown in Figure S18.1-1 below.
x
(0) c0
W16
1
x (2) o
16
x
(6)
0
2
W16
3
16
8-point
DFT
x(8) 0
16
x (10 ),
.x (12)
-
0
W1\5
16
6
x (14)0
W7
x (1)
0
x(3)
a
x(5)
-
x(7)
-
10
16
x (9)
-
W11
W16
x (11)0
16
8-point
DFT
W9
W16
W 12
W16
x(13)­
W13
W16
x (15)G
W16
W 15
16
Figure S18.1-1
S18.1
Solution 18.2
From the expression for the inverse DFT it follows that
N-1
-j(2 1)kn
N x
(n) =
X
(k) e
k=0
Thus by using as the input to the DFT program the complex conjugate of
X(k), the output sequence will be N times the complex conjugate of x(n).
Solution 18.3
(a)
(b) In proceeding from array (m - 1) to array m we are combining
Thus the coefficient are
2 (m-l) point DFT's to form 2 m point DFT's.
these coefficients are
Thus
2
M.
=
M
where
WM
of
powers
successive
W
M
where k
0,1,2,... (M - 1) or, since
=
WN) N/M
WM
th
The powers of WN involved in computing the m- array from the (m - 1)st
array are
k = 0,1,2,... (M/2 -
WNNk/M
1)
(m-1)
(c)
2
(d)
2m
for 1 < m < (log2 N)-l-
For the last array
(m=log 2 N) there
are no two butterflies utilizing the same coefficients.
Solution 18.4
With g(n) = x 1 (n) +
G(k) = X 1 (k)
+
j
j
x 2 (n),
X 2 (k)
With X (k) and X 2 (k) expressed in terms of their real and imaginary parts,
X (k) = XlR (k) +
j X1 (k)
+
j X, (k)
X 2 (k) =X
2
R (k)
G(k) can be written as
G(k)
S18.2
=
[XlR (k)
-
X2I(k)] +
j
[X 2 R(k)
+ X1
(k)}
Now, since x1 (n) and x2 (n) are real, XlR (k) and X 2R (k) are even and
x1 1 (k) and X 2 1 (k) are odd.
GER (k) =
GOR (k)
Thus,
XlR (k)
= -
X I (k)
GEI (k) = X 2 R(k)
GOI (k) = Xy
(k)
S18.3
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Resource: Digital Signal Processing
Prof. Alan V. Oppenheim
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