chapter 11 - the discrete-time fourier...
TRANSCRIPT
M. J. Roberts - 10/15/06
Solutions 11-1
Chapter 11 - The Discrete-Time FourierTransform
Solutions
DTFT Direct from Definition
1. From the definition, find the DTFT of
x n = 10rect
4n .
and compare with the Fourier transform table in Appendix E.
X F( ) = x n ej2 Fn
n=
= 10rect4
n ej2 Fn
n=
= 10 ej2 Fn
n= 4
4
X F( ) = 10 ej2 F m 4( )
m=0
8
= 10ej8 F
ej2 Fm
m=0
8
= 10ej8 F 1 e
j18 F
1 ej2 F
X F( ) = 10ej8 F e
j9 F
ej F
ej9 F
ej9 F
ej F
ej F
= 10sin 9 F( )sin F( )
= 90drcl F ,9( )
From the table,
rect
Nw
nF
2Nw
+ 1( )drcl F ,2Nw
+ 1( )
rect
4n
F9drcl F ,9( )
10rect
4n
F90drcl F ,9( )
Check
2. From the definition, derive a general expression for the F and forms of theDTFT of functions of the form,
x n = Asin 2 F
0n( ) = Asin
0n( ) .
M. J. Roberts - 10/15/06
Solutions 11-2
(It should remind you of the CTFT of x t( ) = Asin 2 f
0t( ) = Asin
0t( ) .) Compare
with the Fourier transform table in Appendix E.
X F( ) = x n ej2 Fn
n=
= Asin 2 F0n( )e
j2 Fn
n=
= Ae
j2 F0n
ej2 F
0n
j2e
j2 Fn
n=
X F( ) =A
j2e
j2 F0
F( )ne
j2 F0
+ F( )n
n=
Then, using
ej2 xn
n=
=1
x( )
we get
X F( ) = A / j2( ) 1
F0
F( ) 1F
0F( ) = A / j2( ) 1
F0
F( ) +1
F0
F( )
X F( ) = jA / 2( ) 1
F + F0( ) 1
F F0( )
The form can be found by the transformation, F / 2 .
X j( ) = jA / 2( ) 1
/ 2 +0
/ 2( ) 1/ 2
0/ 2( )
X j( ) = j A
2+
0( ) 2 0( )
DTFT Using Tables and Transforms
3. A DT signal is defined by
x n = sinc n / 8( ) .
Sketch the magnitude and phase of the DTFT of x n 2 .
From the table of transform pairs,
sinc n / w( ) F
wrect wF( ) 1F( )
sinc n / 8( ) F
8rect 8F( ) 1F( )
sincn 2
8
F8rect 8F( ) 1
F( ) ej4 F
M. J. Roberts - 10/15/06
Solutions 11-3
sincn 2
8
F8e
j4 Frect 8 F k( )( )
k =
n-32 32
x[n]
1
F -1 1
|X( F )|8
F -1 1
Phase of X( F )|
- p
p
4. A DT signal is defined by
x n = sin n / 6( ) .
Sketch the magnitude and phase of the DTFT of x n 3 and
x n + 12 .
From the table,
sin 2 F
0n( ) F
j / 2( ) 1F + F
0( ) 1F F
0( )
sinn
6
Fj / 2( ) 1
F + 1 / 12( ) 1F 1 / 12( )
sinn 3( )6
Fj / 2( ) 1
F + 1 / 12( ) 1F 1 / 12( ) e
j6 F
sinn 3( )6
Fj / 2( ) e
j6 FF + 1 / 12 k( ) e
j6 FF 1 / 12 k( )
k =
sinn 3( )6
Fj / 2( ) e
j6 k 1/12( )
= j
F + 1 / 12 k( ) ej6 k +1/12( )
= j
F 1 / 12 k( )k =
sinn 3( )6
F1 / 2( ) F + 1 / 12 k( ) + F 1 / 12 k( )
k =
sinn 3( )6
F 1
21
F + 1 / 2( ) +1
F 1 / 2( )
M. J. Roberts - 10/15/06
Solutions 11-4
Similarly,
sinn + 12( )
6
Fj / 2( ) e
+ j24 k 1/12( )F + 1 / 12 k( ) e
+ j24 k +1/12( )F 1 / 12 k( )
k =
sinn + 12( )
6
Fj / 2( ) e
+ j 24 k 2( )F + 1 / 12 k( ) e
+ j 24 k +2( )F 1 / 12 k( )
k =
sinn + 12( )
6
Fj / 2( ) F + 1 / 12 k( ) F 1 / 12 k( )
k =
sinn + 12( )
6
Fj / 2( ) 1
F + 1 / 2( ) 1F 1 / 2( )
This is the same as the transform of the unshifted function because a shift of 12 is ashift over exactly one period.
n-12 12
x[n]
-1
1
F -1 1
|X( F )|0.5
F -1 1
Phase of X( F )|
- p
p
Phase of X( F )|n
-12 12
x[n]
-1
1
F -1 1
|X( F )|0.5
F -1 1- p
p
5. The DTFT of a DT signal is defined by
X j( ) = 8 rect 2 /( ) / 2( )( ) + rect 2 /( ) + / 2( )( ) 2 ( ) .
Sketch x n .
M. J. Roberts - 10/15/06
Solutions 11-5
From the table,
sinc n / w( ) F
wrect w / 2( ) 2 ( )
sinc n / 4( ) F
4rect 2 /( ) 2 ( )
ej n / 2
sinc n / 4( ) F4rect
2
22 ( )
ej n / 2
sinc n / 4( ) F4rect
2+
22 ( )
sinc n / 4( ) ej n / 2
+ ej n / 2( ) F
4 rect2
2+ rect
2+
22 ( )
2sinc n / 4( )cos n / 2( ) F4 rect
2
2+ rect
2+
22 ( )
Therefore
x n = 2sinc n / 4( )cos n / 2( )
n-16 16
x[n]
-2
2
6. Sketch the magnitude and phase of the DTFT of
x n = rect
4n cos 2 n / 6( ) .
Then sketch x n .
From the table,
M. J. Roberts - 10/15/06
Solutions 11-6
rect
Nw
nF
2Nw
+ 1( )drcl F ,2Nw
+ 1( )and
cos 2 F
0n( ) F
1 / 2( ) 1F F
0( ) +1
F + F0( )
X F( ) = 9drcl F ,9( ) 1 / 2( ) 1
F 1 / 6( ) +1
F + 1 / 6( )
Since both functions are periodic with period, one, at every impulse in the combfunction the value of the Dirichlet function will be the same.
X F( ) = 9 / 2( ) drcl 1 / 6,9( ) 1
F 1 / 6( ) + drcl 1 / 6,9( ) 1F + 1 / 6( )
X F( ) = 9 / 2( )drcl 1 / 6,9( )sin 3 / 2( )9 sin /6( )
1F 1 / 6( ) +
1F + 1 / 6( )
X F( ) =
1F 1 / 6( ) +
1F + 1 / 6( )
Then, using
cos 2 F
0n( ) F
1 / 2( ) 1F F
0( ) +1
F + F0( )
2cos 2 n / 6( ) F
1F 1 / 6( ) +
1F + 1 / 6( )
and, therefore,
x n = 2cos 2 n / 6( )
n-12 12
x[n]
-2
2
F -1 1
|X( F )|1
F -1 1
Phase of X( F )|
- p
p
7. Sketch the inverse DTFT of
X F( ) = 1 / 2( ) rect 4F( ) 1
F( ) 1/ 2F( ) .
From the table,
M. J. Roberts - 10/15/06
Solutions 11-7
sinc n / w( ) F
wrect wF( ) 1F( ) and
N
0
nF 1
N0
1/ N0
F( ) = F0 F
0
F( )
Therefore
sinc n / 4( ) F
4rect 4F( ) 1F( ) and
2
nF 1
21/ 2
F( )
1 / 4( )sinc n / 4( ) F
rect 4F( ) 1F( )
Therefore using multiplication-convolution duality,
x n = 1 / 4( )sinc n / 4( ) 2
n .
n-16 16
x[n]
-0.1
0.25
8. Find the numerical values of the constants.
(a)
ArectW
n ejB n F
10sin 5 F + 1( )( )sin F + 1( )( )
Find A, W and B.
rectW
nF
sin F 2W + 1( )( )sin F( )
10rect2
nF
10sin 5 F( )sin F( )
10rect2
n ej2 n F
10sin 5 F + 1( )( )sin F + 1( )( )
A = 10 , W = 2 , B = -2
(b) 2
15n 3 rect
3n
FAe
jB F Find A and B.
2
15n 3 rect
3n = 2 n 3
2 n 3
F2e
j6 F
M. J. Roberts - 10/15/06
Solutions 11-8
A = 2 , B = -6
(c) 2 / 3( )
n
u n + 2F Ae
jB F
1 ej2 F
Find A, B and .
nu n
F 1
1 ej2 F
, < 1
2 / 3( )n
u nF 1
1 2 / 3( )ej2 F
2 / 3( )n+2
u n + 2F e
j4 F
1 2 / 3( )ej2 F
2 / 3( )2
2 / 3( )n
u n + 2F e
j4 F
1 2 / 3( )ej2 F
2 / 3( )n
u n + 2F
3 / 2( )2 e
j4 F
1 2 / 3( )ej2 F
A = 9 / 4 , B = -4, = 2 / 3
(d) 4sinc n / 10( ) F
Arect BF( ) 1F( ) Find A and B.
sinc n / N
w( ) FN
wrect N
wF( ) 1
F( )
sinc n / 10( ) F
10rect 10F( ) 1F( )
4sincn 1
10
F40rect 10F( ) 1
F( )
A = 40 , B = 10
9. Find the numerical values of these functions.
(a) x n = 4 2 / 3( )
n
u n X j( )
=
X j( ) =4
1 2 / 3( )ej
X j( ) =4
1 2 / 3( )ej
=4
1+ 2 / 3=
12
5= 2.4
(b) x n = 2rect
3n 2 Find the numerical value of
X F( )
F =1/8.
X F( ) = 2sin 7 F( )sin F( )
ej4 F
X 1/ 8( ) = 2sin 7 / 8( )sin / 8( )
ej / 2
= j2
M. J. Roberts - 10/15/06
Solutions 11-9
(c) X F( ) = rect 10F( ) comb F( ) 1 / 2( ) comb F 1 / 4( ) + comb F + 1 / 4( )
Find the numerical value of x 2 .
x n = 1 / 10( )sinc n / 10( )cos 2 n / 4( )
x 2 = 1 / 10( )sinc 2 / 10( )cos( ) = 1 / 10( )
sin / 5( )/ 5
cos( ) = 0.09355
10. Using the differencing property of the DTFT and the transform pair,
tri n / 2( ) F
1+ cos 2 F( ) ,
find the DTFT of 1 / 2( ) n + 1 + n n 1 n 2( )( ) . Compare it with
Fourier transform found using the table in Appendix E.
T h e f i r s t b a c k w a r d d i f f e r e n c e o f tri n / 2( ) i s
1 / 2( ) n + 1 + n n 1 n 2( )( ) . Applying the differencing
property,
trin
2tri
n 1
2
F1 e
j2 F( ) 1+ cos 2 F( )( )
1 / 2( ) n + 1 + n n 1 n 2( ) F
1 ej2 F( ) 1+ cos 2 F( )( )
1 / 2( ) n + 1 + n n 1 n 2( ) F
1 ej2 F
+e
j2 F+ e
j2 F
2e
j2 F ej2 F
+ ej2 F
2
1 / 2( ) n + 1 + n n 1 n 2( ) F
1 ej2 F
+e
j2 F
2+
ej2 F
2
1
2
ej4 F
2
1 / 2( ) n + 1 + n n 1 n 2( ) F
1 / 2( ) ej2 F
+ 1 ej2 F
ej4 F( )
Other route to the DTFT:
M. J. Roberts - 10/15/06
Solutions 11-10
1 / 2( ) n + 1 + n n 1 n 2( ) F
1 / 2( ) ej2 F
+ 1 ej2 F
ej4 F( )
Check.
11. Using Parseval’s theorem, find the signal energy of
x n = sinc n / 10( )sin 2 n / 4( ) .
Ex
= x n2
n=
= X F( )2
dF1
From the table,
sinc n / w( ) F
wrect wF( ) 1F( )
and
sin 2 F
0n( ) F j
21
F + F0( ) 1
F F0( )
Using the multiplication-convolution duality of the DTFT,
sinc n / 10( )sin 2 n / 4( ) F
10rect 10F( ) 1F( ) j / 2( ) 1
F + 1 / 4( ) 1F 1 / 4( )
Periodic convolution is the same as aperiodic convolution with one period.
sinc n / 10( )sin 2 n / 4( ) F
j5rect 10F( ) 1F( ) F + 1 / 4( ) F 1 / 4( )
sinc n / 10( )sin 2 n / 4( ) F
j5rect 10F( ) 1F + 1 / 4( ) 1
F 1 / 4( )
sinc n / 10( )sin 2 n / 4( ) F
j5 rect 10F( ) 1F + 1 / 4( ) rect 10F( ) 1
F 1 / 4( )
E
x= j5 rect 10F( ) 1
F + 1 / 4( ) rect 10F( ) 1F 1 / 4( )
2
dF1
Since we are integrating only over a range of one, only one impulse in each combis significant.
E
x= 25 rect 10F( ) F + 1 / 4( ) rect 10F( ) F 1 / 4( )
2
dF1
M. J. Roberts - 10/15/06
Solutions 11-11
E
x= 25 rect 10 F + 1 / 4( )( ) rect 10 F 1 / 4( )( )
2
dF1
The square of the sum equals the sum of the squares because there is no crossproduct; the two rectangles do not overlap.
E
x= 25 rect 10 F + 1 / 4( )( )dF
1+ rect 10 F 1 / 4( )( )dF
1{ }
Ex
= 25 dF1/ 4 1/ 20
1/ 4+1/ 20
+ dF1/ 4 1/ 20
1/ 4+1/ 20
= 251
10+
1
10= 5
Fourier Method Relations
12. Sketch the magnitude and phase of the CTFT of
x
1t( ) = rect t( )
and of the CTFS of
x
2t( ) = rect t( ) 8
t( ) .
For comparison purposes, sketch X
1f( ) versus f and
T
0X
2k versus
kf
0 on the
same set of axes. ( T
0 is the period of
x
2t( ) and
T
0= 1 / f
0.)
X
1f( ) = sinc f( )
Using the relationship between an the CTFT of an aperiodic signal and the CTFSof a periodic extension of that signal,
X
2k = f
sX
1f
sk( ) = 1 / 8( )sinc k / 8( )
M. J. Roberts - 10/15/06
Solutions 11-12
f -4 4
|X1( f )|
1
f -4 4
X1( f )
-
kf0
-4 4
|T0X
2[k]|
1
kf0
-4 4
T0X
2[k]
-
13. Sketch the magnitude and phase of the CTFT of
x
1t( ) = 4cos 4 t( )
and of the DTFT of
x
2n = x
1nT
s( )where
T
s= 1 / 16 . For comparison purposes sketch
X
1f( ) and
T
sX
2T
sf( ) versus
f on the same set of axes.
X
1f( ) = 2 f 2( ) + f + 2( )
x
2n = 4cos 4 nT
s( ) = 4cos n / 4( )
X
2F( ) = 2
1F 1 / 8( ) +
1F + 1 / 8( )
X2
F( ) = 2 F 1 / 8 k( ) + F + 1 / 8 k( )k =
X2
Tsf( ) = 2 f / 16 1 / 8 k( ) + f / 16 + 1 / 8 k( )
k =
X2
Tsf( ) = 32 f 2 16k( ) + f + 2 16k( )
k =
TsX
2T
sf( ) = 2 f 2 16k( ) + f + 2 16k( )
k =
M. J. Roberts - 10/15/06
Solutions 11-13
f -16 16
|X1( f )|
2
f -16 16
Phase of X1( f )
- p
p
f
f
-16 16
|TsX
2(T
sf )|
2
-16 16
Phase of TsX
2(T
sf )
- p
p
......
......
14. Sketch the magnitude and phase of the DTFT of
x
1n =
sinc n / 16( )4
and of the DTFS of
x
2n =
sinc n / 16( )4
32n .
For comparison purposes sketch X
1F( ) versus F and
N
0X
2k versus
kF
0 on
the same set of axes.
From the table,
sinc n / w( ) F
wrect wF( ) 1F( )
sinc n / 16( )4
F4rect 16F( ) 1
F( )
M. J. Roberts - 10/15/06
Solutions 11-14
therefore
X1
F( ) = 4rect 16F( ) 1F( ) = 4 rect 16 F q( )( )
q=
.
The fundamental frequency of the periodic signal is the reciprocal of its period,
F
0= 1 / N
0= 1 / 32 .
Using the results of the analysis of periodic extensions of aperiodic DT signals,
X
2k = F
0X
1kF
0( )
X2
k = 1 / 32( )X1
k / 32( ) =1
8rect 16 k / 32 q( )( )
q=
= 1 / 8( ) rect k / 2 16q( )q=
F -1 1
|X1(F )|
4
F -1 1
Phase of X1(F )
- p
p
kF0
-1 1
N0|X
2[k]|
4
kF0
-32 32
Phase of X2[k]
- p
p
DTFT
M. J. Roberts - 10/15/06
Solutions 11-15
15. Find the DTFT of each of these signals:
(a) x n = 1 / 3( )
n
u n 1
X j( ) = x n ej n
n=
= 1 / 3( )n
u n 1 ej n
n=
= 1 / 3( )n
ej n
n=1
X j( ) = 1 / 3( )m+1
ej m+1( )
m=0
=e
j
3
m+1
m=0
X j( ) =e
j
3
ej
3
m
m=0
=e
j
3
1
1 ej
/ 3=
ej
3 ej
Alternate Solution:
x n = 1 / 3( )
n
u n 1 = 1 / 3( )n
u n n
Using
nu n
F 1
1 ej
and
nF
1
x n = 1 / 3( )n
u n 1 / 3( ) n =1
1e
j
3
1
x n =
1 1e
j
3
1 ej
/ 3=
ej
3
1 ej
/ 3=
ej
3 ej
Second Alternate Solution:
x n = 1 / 3( ) 1 / 3( )
n 1
u n 1
X j( ) =
1
3
1
1 ej
/ 3e
j=
ej
3 ej
(b) x n = sin n / 4( ) 1 / 4( )
n
u n 2
sin n / 4( ) = sin 2 n / 8( ) = cos 2 n 2( ) / 8( )
M. J. Roberts - 10/15/06
Solutions 11-16
x n = 1 / 4( )
2
cos 2 n 2( ) / 8( ) 1 / 4( )n 2
u n 2
ncos
0n( )u n
F1 cos
0( )ej
1 2 cos0( )e
j+
2e
j2, < 1
X j( ) = 1 / 4( )2
ej2
1 1 / 4( )cos / 4( )ej
1 1 / 2( )cos / 4( )ej
+ 1 / 16( )ej2
=e
j2
16
1 2 / 8( )ej
1 2 / 4( )ej
+ 1 / 16( )ej2
Alternate Solution:
x n =e
j n / 4e
j n / 4
j21 / 4( )
n
u n 2 =1
j2e
j / 4/ 4( )
n
ej / 4
/ 4( )n
( )u n 2
x n =1
j2e
j / 4/ 4( )
2
ej / 4
/ 4( )n 2
ej / 4
/ 4( )2
ej / 4
/ 4( )n 2
( )u n 2
ej / 4
4
n
u nF 1
1 ej / 4
/ 4( )ej
ej / 4
4
n 2
u n 2F e
j2
1 ej / 4
/ 4( )ej
X j( ) =1
j2
ej / 4
4
2
ej2
1 ej / 4
/ 4( )ej
ej / 4
4
2
ej2
1 ej / 4
/ 4( )ej
X j( ) =e
j2
j2
ej / 4
/ 4( )2
1 ej / 4
/ 4( )ej( ) e
j / 4/ 4( )
2
1 ej / 4
/ 4( )ej( )
1 ej / 4
/ 4( )ej( ) 1 e
j / 4/ 4( )e
j( )
X j( ) =e
j2
j32
ej / 4( )
2
ej / 4( )
2
ej / 4
/ 4( )ej
ej / 4( )
2
+ ej / 4( )
2
ej / 4
/ 4( )ej
1 ej / 4
/ 4 + ej / 4
/ 4( )ej
+ ej2
/ 16
X j( ) =e
j2
j32
ej / 2
ej / 2
ej / 4
/ 4( )ej
+ ej / 4
/ 4( )ej
1 1 / 2( )cos / 4( )ej
+ 1 / 16( )ej2
M. J. Roberts - 10/15/06
Solutions 11-17
X j( ) =e
j2
j32
j2 1 / 4( ) ej / 4
ej / 4( )e
j
1 1 / 2( )cos / 4( )ej
+ 1 / 16( )ej2
X j( ) =e
j2
16
1 1 / 4( )sin / 4( )ej
1 1 / 2( )cos / 4( )ej
+ 1 / 16( )ej2
=e
j2
16
1 2 / 8( )ej
1 2 / 4( )ej
+ 1 / 16( )ej2
(c) x n = sinc 2 n / 8( ) sinc 2 n 4( ) / 8( )
Using
sinc n / w( ) F
wrect wF( ) 1F( )
X F( ) = 8 / 2( ) rect 8F / 2( ) 1
F( ) 8 / 2( ) rect 8F / 2( ) 1F( ) e
j8 F
X F( ) = 4 /( )
2
rect 4 f /( ) 1F( ) e
j8 F
x n = 4 /( )sinc 2 n 4( ) / 8( )
(d) x n = sinc
22 n / 8( )
Using
sinc n / w( ) F
wrect wF( ) 1F( )
X F( ) = 8 / 2( )rect 8F / 2( ) 1
F( ) 8 / 2( )rect 8F / 2( ) 1F( )
X F( ) = 8 / 2( )
2
rect 8F / 2( ) 1F( ) rect 8F / 2( ) 1
F( )
X F( ) = 8 / 2( )
2
rect 8F / 2( ) rect 8F / 2( ) 1F( )
X F( ) = 8 / 2( )
2
2 / 8( ) tri 8F / 2( ) 1F( ) = 8 / 2( ) tri 8F / 2( ) 1
F( )
16. Sketch the magnitudes and phases of the DTFT’s of the following functions:
(a) rect
2n
M. J. Roberts - 10/15/06
Solutions 11-18
Using rect
Nw
nF
2Nw
+ 1( )drcl F ,2Nw
+ 1( ) ,
rect
2n
F5drcl F ,5( )
F -1 1
|X( F )|
5
F -1 1
Phase of X( F )
- p
p
(b) rect
2n 5 n( )
Using
nF
1 and x n y n
FX F( )Y F( )
rect
2n 5 n( ) F
5drcl F ,5( ) 5( ) = 25drcl F ,5( )
F -1 1
|X( F )|
25
F -1 1
Phase of X( F )
- p
p
(c) rect
2n 3 n + 3
Using x n n
0
Fe
j2 Fn0 X F( )
M. J. Roberts - 10/15/06
Solutions 11-19
rect
2n 3 n + 3
F15drcl F ,5( )e
j6 F
F -1 1
|X( F )|
15
F -1 1
Phase of X( F )
- p
p
(d) rect
2n 5 4n( ) = rect
2n 5 n( )
rect
2n 5 4n( ) F
25drcl F ,5( )
F -1 1
|X( F )|
25
F -1 1
Phase of X( F )
- p
p
(e) rect
2n
8n
Using
N0
nF 1
N0
1/ N0
F( ) = F0 F
0
F( ) ,
rect
2n
8n
F5 / 8( )drcl F ,5( ) 1/8
F( )
M. J. Roberts - 10/15/06
Solutions 11-20
rect2
n8
nF
5 / 8( ) drcl F ,5( ) F k / 8( )k =
rect2
n8
nF 5
8drcl k / 8,5( ) F k / 8( )
k =
F -1 1
|X( F )|
0.625
F -1 1
Phase of X( F )
- p
p
(f) rect
2n
8n 3
Using the result of (e) and x n n
0
Fe
j2 Fn0 X F( ) ,
rect2
n8
n 3F
5 / 8( )ej6 F
drcl m / 8,5( ) F m / 8( )m=
F -1 1
|X( F )|
0.625
F -1 1
Phase of X( F )
- p
p
(g)
rect2
n8
2n = rect2
n 2n 8mm=
= rect2
n 2 n 4m( )m=
M. J. Roberts - 10/15/06
Solutions 11-21
rect2
n8
2n = rect2
n n 4mm=
= rect2
n4
n
Therefore
rect
2n
82n
F5 / 4( )drcl F ,5( ) 1/ 4
F( )
rect2
n8
2nF
5 / 4( ) drcl F ,5( ) F m / 4( )m=
rect2
n8
2nF
5 / 4( ) drcl m / 4,5( ) F m / 4( )m=
F -1 1
|X( F )|
1.25
F -1 1
Phase of X( F )
- p
p
(h) rect
2n
5n
rect
2n
5n
F5drcl F ,5( ) 1 / 5( ) 1/5
F( )
rect2
n5
nF
drcl F ,5( ) F m / 5( )m=
rect2
n5
nF
drcl m / 5,5( ) F m / 5( )m=
Then, using
drcln
2m + 1,2m + 1 =
2m+1n
M. J. Roberts - 10/15/06
Solutions 11-22
rect2
n5
nF
5m F m / 5( )
m=
5
m F m / 5( )m=
has an impulse at every 5th integer value of m and
therefore has an impulse at every integer value of F. Therefore
rect
2n
5n
F
1F( )
and, since 1
F
1F( ) , that implies that
rect
2n
5n = 1 .
F -1 1
|X( F )|
1
F -1 1
Phase of X( F )
- p
p
17. Sketch the inverse DTFT’s of these functions.
(a) X F( ) =
1F( ) 1
F 1 / 2( )
Using 1
F
1F( ) and
e
j2 F0n
x nF
X F F0( )
1 e
j n F
1F( ) 1
F 1 / 2( )
e
j n / 2e
j n / 2e
+ j n / 2( ) F
1F( ) 1
F 1 / 2( )
j2e
j n / 2sin n / 2( ) F
1F( ) 1
F 1 / 2( )
2e
j n+1( )/ 2sin n / 2( ) F
1F( ) 1
F 1 / 2( )
2 cos n + 1( ) / 2( ) + j sin n + 1( ) / 2( ) sin n / 2( ) F
1F( ) 1
F 1 / 2( )
M. J. Roberts - 10/15/06
Solutions 11-23
2 cos n + 1( ) / 2( )sin n / 2( )
sin n / 2( ) + j sin n + 1( ) / 2( )sin n '2( )=0
F
1F( ) 1
F 1 / 2( )
2sin
2n / 2( ) F
1F( ) 1
F 1 / 2( )
n-12 12
x[n]
2
(b) X F( ) = j
1F + 1 / 8( ) j
1F 1 / 8( )
Using 1
F
1F( ) and
e
j2 F0n
x nF
X F F0( )
je
j n / 4je
j n / 4 Fj
1F + 1 / 8( ) j
1F 1 / 8( )
j j2sin n / 4( )( ) F
j1
F + 1 / 8( ) j1
F 1 / 8( )
2sin n / 4( ) F
j1
F + 1 / 8( ) j1
F 1 / 8( )
n-12 12
x[n]
-2
2
(c) X F( ) = sinc 10 F 1 / 4( )( ) + sinc 10 F + 1 / 4( )( ) 1
F( )
Using
sinc t / w( ) T
0
t( ) = wf0
cos t / w( ) + T0
/ w 1( )drcl f0t,T
0/ w 1( )
from Appendix A (because T
0/ 2w is an integer),
M. J. Roberts - 10/15/06
Solutions 11-24
X F( ) = 1 / 10( )cos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )
+cos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )
cos 10 F( )e
j2 FndF
1
Fcos 10 F( )
1
2e
j10 F+ e
j10 F( )ej2 Fn
dF1
Fcos 10 F( )
1
2e
j2 F n+5( )dF
1+ e
j2 F n 5( )dF
1
Fcos 10 F( )
1
2
cos 2 F n + 5( )( ) + j sin 2 F n + 5( )( ) dF1
+ cos 2 F n 5( )( ) + j sin 2 F n 5( )( ) dF1
Fcos 10 F( )
These integrals are zero unless n = ±5 . Therefore
1 / 2( ) n + 5 + n 5( ) F
cos 10 F( ) .
Then using rect
Nw
nF
2Nw
+ 1( )drcl F ,2Nw
+ 1( ) ,
rect
4n
F9drcl F ,9( )
Combining inverse transforms,
1 / 2( ) n + 5 + n 5( ) + rect
4n
Fcos 10 F( ) + 9drcl F ,9( ) .
Then, using e
j2 F0n
x nF
X F F0( )
ej n / 2
1 / 2( ) n + 5 + n 5( ) + rect4
n{ }
Fcos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )
and
ej n / 2
1 / 2( ) n + 5 + n 5( ) + rect4
n{ }
Fcos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )
Then, finally
M. J. Roberts - 10/15/06
Solutions 11-25
1
10
ej n / 2
1 / 2( ) n + 5 + n 5( ) + rect4
n{ }+e
j n / 21 / 2( ) n + 5 + n 5( ) + rect
4n{ }
F 1
10
cos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )
+cos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )
The impulses on the left side cancel and we get
1 / 5( )rect4
n cos n / 2( ) F 1
10
cos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )
+cos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )
n-12 12
x[n]
-0.2
0.2
(d) X F( ) = F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1
2F( )
X F( ) = F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1 / 2( ) 1
F( ) +1
F 1 / 2( )
X F( ) =1
2
F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1F( )
+ F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1F 1 / 2( )
X F( ) =1
2
1F 1 / 4( ) +
1F 3 / 16( ) +
1F 5 / 16( )
+1
F 1 / 4 1 / 2( ) +1
F 3 / 16 1 / 2( ) +1
F 5 / 16 1 / 2( )
Then, using 1
F
1F( ) and
e
j2 F0n
x nF
X F F0( ) we get
1
2
ej n / 2
+ ej3 n /8
+ ej5 n /8
+ej3 n / 2
+ ej11 n /8
+ ej13 n /8
F 1
2
1F 1 / 4( ) +
1F 3 / 16( ) +
1F 5 / 16( )
+1
F 3 / 4( ) +1
F 11 / 16( ) +1
F 13 / 16( )
or
M. J. Roberts - 10/15/06
Solutions 11-26
cos n / 2( ) + cos 3 n / 8( ) + cos 5 n / 8( )
F 1
2
1F 1 / 4( ) +
1F 3 / 16( ) +
1F 5 / 16( )
+1
F 3 / 4( ) +1
F 11 / 16( ) +1
F 13 / 16( )
n-16 16
x[n]
-3
3
18. A signal, x n , has a DTFT ,
X F( ) = 10sinc 5F( ) 1
F( ) . What is its signal
energy?
Using rect
Nw
nF
2Nw
+ 1( )sinc 2Nw
+ 1( ) F( ) 1F( ) ,
x n = 2rect
2n .
Ex
= 2rect2
n2
n=
= 4 1n= 2
2
= 20
19. A signal, x n , has a DTFT,
X F( ) =
1F 1 / 4( ) +
1F + 1 / 4( ) + j
1F + 1 / 3( ) j
1F 1 / 3( ) .
What is the fundamental period, N
0, of
x n ?
Using cos 2 F
0n( ) F
1 / 2( ) 1F F
0( ) +1
F + F0( ) and
sin 2 F
0n( ) F
j / 21
F + F0( ) 1
F F0( ) ,
x n = 2cos2 n
4+ 2sin
2 n
3 .
The least common multiple of the two periods of the two sinusoids, 4 and 3, is 12.Therefore
N
0= 12 .
20. The DTFT of x n = 2 n + 3 3 n 3 can be expressed in the form,
X F( ) = Asin bF( ) + Ce
dF . Find the numerical values of A, b, C and d.
X F( ) = 2e
+ j6 F3e
j6 F= 2e
+ j6 F2e
j6 Fe
j6 F= j4sin 6 F( ) e
j6 F
M. J. Roberts - 10/15/06
Solutions 11-27
A = j4 , b = 6 , C = -1 , d = j6
21. Let x n be a DT signal and let
y n = x m
m=
n
. If Y j( ) = cos 2( ) ,
x n consists of exactly four DT impulses. What are their numerical strengths
and locations?
Y j( ) = cos 2( ) =
X j( )1 e
jX j( ) = cos 2( ) 1 e
j( )
X j( ) =
1
2e
j2+ e
j2( ) 1 ej( ) =
1
2e
j2+ e
j2e
je
j3( )
x n =
1
2n + 2 + n 2 n + 1 n 3( )
Impulse #1. Strength = 1/2 Located at n = -2
Impulse #2. Strength = 1/2 Located at n = 2
Impulse #3. Strength = -1/2 Located at n = 3
Impulse #4. Strength = -1/2 Located at n = -1
22. A signal, x n = 4cos 2 n / 15( ) + 2cos 2 n / 9( ) is the excitation of a system
whose impulse response is h n = rect
Nw
n ( N
w is an integer). When
N
w= 22 the response,
y n , of the system is zero. The response of the system is
also zero for some larger values of N
w. Find the next smallest positive numerical
integer value of N
w greater than 22 that makes the response zero. (Hint: The
zeros of drcl F , N( ) occur where F is an integer multiple of 1/N, except when F
itself is an integer.)
X F( ) = 2
1F 1 / 15( ) +
1F + 1 / 15( ) + j
1F + 1 / 9( ) 1
F 1 / 9( )
H F( ) = 2N
w+ 1( )drcl F ,2N
W+ 1( )
The response is the product of X and H. Therefore, for the response to be zero,the impulses in X must occur at zeros of H. The zeros of H occur at integer
multiples of F
0= 2N
w+ 1( )
1
. To have zeros at the impulse locations, this F
0
M. J. Roberts - 10/15/06
Solutions 11-28
must be a common divisor of 1/15 and 1/9. The greatest common divisor of thosetwo numbers is 1/45. That yields
2N
w+ 1 = 45 and
N
W= 22 . The next greatest
common divisor of 1/15 and 1/9 is 1/90 but that does not yield an integer value of
N
W. The next greatest common divisor of 1/15 and 1/9 is 1/135 yielding
N
W= 67 .
23. In Figure E-23 are some DT signals, numbered 1-14. Below them are some DTFTmagnitude plots. For each DTFT magnitude plot, identify the corresponding DTsignal by writing its number in the space provided. If there is no match, write“none”.
1. 3sinc n( ) 2.
5sinc n / 4( ) 2sinc n / 4( ) 3.
7cos 2 n / 8( )
4.
n + 1 n 1 5. 3sinc n / 4( ) 6.
4sin 2 n / 8( )
7. 2 / 3( )
n
u n 8. 2rect
2n 3 9.
4
4n
10.
42
n 11.
3sinc2
n / 4( ) 12.
n + 1 + n 1
13. 2rect
3n 3 14.
1 / 3( )
n
u n
6 10 9 8 3 7
F 1 1
|X(F)|
2
F 1 1
|X(F)|
2
F 1 1
|X(F)|
1
F 1 1
|X(F)|
10
F 1 1
|X(F)|
3.5
F 1 1
|X(F)|
3
5 12 14 1 4 2
F 1 1
|X(F)|
12
F 1 1
|X(F)|
2
F 1 1
|X(F)|
1.5
F 1 1
|X(F)|3
F 1 1
|X(F)|
2
F 1 1
|X(F)|
160
13 11
F 1 1
|X(F)|
14
F 1 1
|X(F)|
12
M. J. Roberts - 10/15/06
Solutions 11-29
Figure E-23
Fourier Method Relations
24. Using the relationship between the CTFT of a signal and the CTFS of a periodicextension of that signal, find the CTFS of
x t( ) = rect t / w( ) T
0
t( )and compare it with the table entry.
X f( ) = wsinc wf( )
X k = f
0X kf
0( ) = f0wsinc wkf
0( ) = w / T0( )sinc wk / T
0( )
25. A CT signal, x t( ) , has a CTFT,
X f( ) = 4rect f / 2( ) . A new signal,
x
pt( ) , is
formed by periodically repeating x t( ) with a period of 8. The CTFS harmonic
function of x
pt( ) is
X
pk . Find
X
pk .
X
pk = f
pX kf
p( ) where fp
= 1 / Tp
= 1 / 8
X
pk = 1 / 8( )X k / 8( ) = 1/2( )rect k / 16( )
X
p3 = 1 / 2( )rect 3 / 16( ) = 1 / 2 .
26. An aperiodic signal, x t( ) , has a CTFT,
X f( ) =
0 , f > 4
4 f , 2 < f < 4
2 , f < 2
= 4 tri f / 4( ) 2 tri f / 2( ) .
Let
xp
t( ) = x t( ) 2t( ) = x t 2k( )
k =
and let X
pk be its CTFS harmonic
function.
(a) Sketch the magnitude of X f( ) .
M. J. Roberts - 10/15/06
Solutions 11-30
4
4
-4
-4
f
|X( f )|
(b) Find an expression for X
pk .
Xp
k = f0
X kf0( ) =
1
2
0 , k / 2 > 4
4 k / 2 , 2 < k / 2 < 4
2 , k / 2 < 2
= 2 trik / 2
4tri
k / 2
2.
Xp
k =1
2
0 , k > 8
4 k / 2 , 4 < k < 8
2 , k < 4
= 2 tri k / 8( ) tri k / 4( )
(c) Find the numerical values of X
p3 ,
X
p5 ,
X
p10 .
X
p3 = 1 / 2( ) 2 = 1 or
X
p3 = 2 tri 3 / 8( ) tri 3 / 4( ) = 10 / 8 1 / 4 = 1
X
p5 = 1 / 2( ) 4 5 / 2( ) = 3 / 4 or
X
p5 = 2 tri 5 / 8( ) tri 5 / 4( ) = 6 / 8 0 = 3 / 4
X
p10 = 1 / 2( ) 0 = 0 or
X
p10 = 2 tri 10 / 8( ) tri 10 / 4( ) = 0 0 = 0
27. Using the relationship between the DTFT of a signal and the DTFS of a periodicextension of that signal, find the DTFS of
rect
Nw
nN
0
n
and compare it with the table entry.
M. J. Roberts - 10/15/06
Solutions 11-31
X F( ) = 2N
w+ 1( )drcl F ,2N
w+ 1( )
From the text,
X
pk = 1 / N
p( )X kFp( ) .
Therefore,
Xp
k =2N
w+ 1
N0
drcl k / N0,2N
w+ 1( ) .
28. A DT signal, x n , is formed by sampling a CT signal,
x t( ) = 12sinc 5t( ) , with a
time between samples, T
s= 0.1. The DTFT of
x n is
X
DTFTF( ) . Find the
numerical value of X
DTFT0.2( ) .
XDTFT
F( ) = fs
XCTFT
fs
F k( )( )k =
,
XCTFT
f( ) =12
5rect
f
5
XDTFT
F( ) = 24 rect10 F k( )
5k =
= 24 rect 2 F k( )( )k =
XDTFT
0.2( ) = 24 rect 2 0.2 k( )( )k =
= 24rect 0.4( ) = 24