Digital Signal Processing Reference
In-Depth Information
3
2
1
0
0
0.2
0.4
0.6
0.8
1
(a) Frequency, Units of
π
3
2
1
0
0
0.2
0.4
0.6
0.8
1
(b) Frequency, Units of
π
Figure 3.47:
(a) Samples of the DTFT of a filter, computed as a long DFT of the convolution of two
truncated IIR impulse responses and one FIR impulse response; (b) Samples of the DTFT of the same
filter, estimated by computing a long DFT of the truncated impulse response of the composite filter,
which was computed from the
z
-transform
b
and
a
coefficients.
We define the sequence and the
n
and
k
vectors, obtain the DFT coefficients using MathScript,
then compute the IDFT using the matrix method and check using the function
iff t
.
x = [(1+2*j),2,(3-2*j),4]; N = length(x); n = 0:1:N-1; k = 0:1:N-1;
MSfft = fft(x); CW = exp(n'*k).ˆ(j*2*pi/N);
idft = (1/N)*CW*conj(MSfft'), MSifft = ifft(MSfft)
A symbolic rendering of this would be
1
N
CW
x
=
·
D
(3.31)
where
CW
is the IDFT basis vector matrix
exp(n'*k).ˆ(j*2*pi/N)
where
n
=
k
= 0:1:
N
-1 and
D
is the DFT vector in column form. Here, note that we have scaled by 1
/N
since the DFT has not been so scaled. Note that the conjugate must be taken of the DFT coefficients
MSiff
when they are transposed into a column vector since MathScript automatically conjugates vectors
when they are transposed.
