Digital Signal Processing Reference
In-Depth Information
3.3.4 Time Reversal Property
Let us consider x(n)
= n = 0 a n e jωn . Next, to find
a n u(n) .ItsDTFT X(e )
=
the DTFT of x(
n) , if we replace n by
n , we would write the DTFT of x(
n)
as −∞
n
0 a n e jωn (but that is wrong), as illustrated by the following example:
=
X(e )
a n e jωn
ae
a 2 e j 2 ω
a 3 e j 3 ω
=
= 1 +
+
+
+···
n
=
0
ae
But the correct expression for the DTFT of x(
n) is of the form 1
+
+
a 2 e j 2 ω
a 3 e j 3 ω
+
+···
.
So the compact form for this series is n =−∞ a n e jωn . With this clarifica-
tion, we now prove the property that if x(n)
X(e ) then
X(e )
x(
n)
(3.24)
= n =−∞ x(
n)e jωn . We substitute (
Proof :DTFTof x(
n)
n)
=
m ,and
we get n =−∞ x(
= m =−∞ x(m)e jωm
= m =−∞ x(m)e j( ω)m
n)e jωn
=
X(e ) .
Example 3.3
Consider x(n)
=
δ(n) . Then, from the definition for DTFT, we see that δ(n)
X(e )
1 for all ω .
From the time-shifting property, we get
=
e jωk
δ(n
k)
(3.25)
The Fourier transform e jωk has a magnitude of one at all frequencies but a
linear phase as a function of ω that yields a constant group delay of k samples. If
we extend this result by considering an infinite sequence of unit impulses, which
can be represented by k =−∞ δ(n
k) , its DTFT would yield k =−∞ e jωk .
But this does not converge to any form of expression. Hence we resort to a
different approach, as described below, and derive the result (3.28).
Example 3.4
k) . Its DTFT is given by X(e )
e jωk
We consider x(n)
=
δ(n
+
k)
+
δ(n
=
+
e jωk
2cos (ωk) . In this example, note that the DTFT is a function of the
continuous variable ω whereas k is a fixed number. It is a periodic function of
ω with a period of 2 π , because 2 cos ((ω
=
= 2cos (ωk) ,where r is an
integer. In other words, the inverse DTFT of X(e )
+ 2 rπ)k)
= 2cos (ωk) is a pair of
impulse functions at n
=
k and n
=−
k , and this is given by
1
2 [ δ(n
cos (ωk)
+
k)
+
δ(n
k) ]
(3.26)
Search WWH ::




Custom Search