Digital Signal Processing Reference
In-Depth Information
k
Re
(X
[
k
]
)
Im
(X
[
k
]
)
0
,
3
1
0
1
,
4
1
1
.
7321
i
2
,
5
1
−
1
.
7321
i
−
4
,
−
1
1
−
1
.
7321
i
3.9.3 FREQUENCY MULTIPLICATION IN TIME DOMAIN
Suppose that the time domain sequence we have been considering is multiplied by a periodic sequence
with a certain frequency
k
0
. Let's compare the DFT of the original (unmultiplied) sequence and the DFT
of the sequence after it has been multiplied (before sampling) by some frequency
k
0
.
Before proceeding, let's try to see what to expect. From trigonometry, we know
1
2
cos
(α
1
2
cos
(α
sin
(α)
sin
(β)
=
−
β)
−
+
β)
or
1
2
cos
(α
1
2
cos
(α
cos
(α)
cos
(β)
=
−
β)
+
+
β)
and so on for various combinations of cosine and sine. When you multiply two frequencies in the time
domain, the resultant signal contains the sum and difference of the original frequencies.
With this in mind, consider Fig. 3.9, which shows at (a) a cosine of frequency five; its DFT (real
only, since the time domain signal is a cosine) at (c) shows spikes at frequencies 5 and -5 (Bin 31 is
equivalent to Bin -1, Bin 30 is equivalent to Bin -2, etc.).
At (e), the cosine of five cycles has been multiplied by a cosine of one cycle. Thus, the expected
frequencies in the new time domain signal would be 5
1, and indeed, we see spikes in the DFT (plot
(g)) at frequencies 4, 6, and -4 and -6 (the two latter are in their aliased positions).
In Fig. 3.10, we initially show the time domain signal as a cosine of one cycle at (a); the DFT (at
(c)) shows frequencies of 1 and -1, as expected. At (e), the time domain signal is now the product of a
cosine of one cycle and a cosine having five cycles. We would thus expect frequencies of 1
±
±
5 =6
;−
4.
The DFT (all real) at plot (g) shows frequencies at 4, 6, 26 (= -6), and 28 (= -4), as expected.
Note that the DFT bin values in Figs. 3.9 and 3.10 were computed using MathScript, which does
not scale the forward (i.e., DFT) transform by 1
/N
.
3.10 COMPUTATION OF DF T VIA MATRIX
By setting up a matrix
W
each row of which is a DFT complex correlator, and multiplying by the signal
(in column vector form), we can get the DFT.
Example 3.11.
Compute the DFT of the sequence
x
[
n
]
=
[
1
,
2
,
3
,
4
]
using the matrix method and
check using the MathScript function fft.
We construct the matrix as
x = [1 2 3 4]; N=length(x); nkvec = 0:1:N-1;
W = exp(nkvec'*nkvec).ˆ(-j*2*pi/N); dft = W*x',
mfft = fft(x)
