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[CHAP. 2
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Derive the up-sampling property of the DTFT, which states that if x (ej") is the DTFT of x ( n ) , the DTFT of
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y(n) = n = O , & L , f 2 L , ...
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From the definition of the DTFT, we have
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Because y(n) is equal to zero except when n is an integer multiple of L,
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Thus, Y(eJW) formed by scaling X(eJW) frequency. is in
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Find the inverse DTFT of
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For this problem, the direct approach of performing the integration
is not easy. However, a simple approach is to recall that the inverse DTFT of
y(n) = ( f ) " u ( n )
and to note that ~ ( e j " is related to X(eju) by scaling in frequency, ) ~ ( e j " = Y (eJIOw ) 1 Therefore, it follows from the up-sampling property in Prob. 2.3 1 that
otherwise
In other words, the sequence x(n) is formed by inserting nine zeros between each value of y ( n ) .
Let x ( n ) be a sequence with a DTFT ~ ( e ; " ) .For each of the following sequences that are formed from x ( n ) , express the DTFT in terms of X ( e J W ) :
( a ) x*(-n>
(b) x ( n ) * x * ( - n ) (c) x ( 2 n
+ 1)
CHAP. 21 (a) The DTFT of x*(-n) is
FOURIER ANALYSIS
Bringing the conjugate outside, we have
which leads to the DTFT pair x*(-n)
Dg x*(ejw)
(b) For y(n) = x(n) * x*(-n), note that because y(n) is the convolution of two sequences, the DTFT of y(n) is the product of the DTFTs of x(n) and x*(-n). As shown in part (a), the DTFT of x*(-n) is X*(ejW).Therefore, we have the DTFT pair x(n) * x*(-n) ~(ej")~*(e'") 1 ~ ( e j " ) 1 ~ =
n=-m
(c) For x(2n
+ 1) we have
DTFT(x(2n
+ I)) =
r(2n
+ 1)e-In" =
n odd n even
n odd
x(n)e-jnY
To evaluate this sum, a "trick" is to use the identity 1This allows us to write the DTFT as follows: DTFT(x(2n
+ 1)) =
n odd
x(n)e-jn" =
n=-m
- (-lr]x(n)e-jnw
Because the first sum is simply X(eJU),and the second is the DTFT of the modulated signal
then
DTFT(x(2n
+ 1)) = f [x(eiw) - x(eicw-") )I
Let x ( n ) be the sequence
which has a DTFT
~ ( e j " )= xR(eJ")
~(eJo) ,
where x R ( e j w )and x l ( e j w ) are the real part and the imaginary part of x ( e j W ) ,respectively. Find the sequence y(n) that has a DTFT given by
The key to solving this problem is to recall that if x(n) is real, and if X(eJW) written in terms of its real and is imaginary parts, XR(ejw) the DTFT of the even part of x(n), and Xl(eJW) the DTFT of the oddpart: is is x,(n> = [x(n)
+ x(-n)] D xr(eiw) B
DTFT
x&) = f [x(n) - x(-n)]
IXl(eJw)
FOURIER ANALYSIS
Therefore, the DTFT of -j x , ( n ) is X l ( e i " ) ,
[CHAP. 2
and the DTFT of j x e ( n + 2 ) is
jxe(n
+ 2 ) DTF j .X R ( e j W ) e j z W c-T
y ( e j W )= X , ( e J W ) j X R ( e i W ) e J h +
Thus, and it follows that
ixe(n
+ 2 ) - jx,(n) & D
y(n) = j x J n
+ 2 ) - jx&)
Finally, with x e ( n ) and x,(n) as tabulated below.
it follows that y ( n ) , which is formed from these two sequences, is as shown below:
Let x(n ) be the sequence
Evaluate the following quantities without explicitly finding X(eJo):
(4 x (ejo)lo=O
( b ) 9Mw)
(c) x(eJo)dw x(ejo)lo=r IX(ejw)l2d~
(a) Because the DTFT of x ( n ) is
x ( e j W )=
)7 x(n)e-jnw
n=-m
note that if we evaluate X ( e j W )at w = 0, we have
which is simply the sum of the values of x ( n ) . Therefore, for the given sequence it follows that
(b) To evaluate the phase, note that because x ( n ) is real and even, X ( e J W is real and even and, therefore, the phase ) is equal zero or n for all o.
CHAP. 2 1
( c ) From the inverse DTFT,
FOURIER ANALYSIS
note that when n = 0 :
x(0) =
Therefore, it follows that
- -,( e l Y ) d w X 2n
x(eM)do = 2nx(0) =6 n
( d ) Evaluating the DTFT of x ( n ) at o = n,we have
which, for the given values of x ( n ) , evaluates to
( e ) From Parseval's theorem, we know that
Therefore,
~ x ( e ~ ~ ) l = 271 'do
n=-w
lr(n)12 = 3871
The center of gravity of a sequence x ( n ) is defined by
and is used as a measure of the time delay of a sequence. Find an expression for c in terms of the DTFT of x ( n ) , and find the value of c for the sequence x ( n ) that has a DTFT as shown in the figure below.
To find the value of c in terms of X ( e J W ) first note that the denominator is simply the value of X ( e J o ) evaluated at ,
o=O:
For the numerator, recall the DTFT pair
nx(n)
" j d do ~ ~ m , W
FOURIER ANALYSIS
[CHAP.2
Therefore, and c may be evaluated in terms of X ( e J m )as follows:
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