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Solution manual 5th edition communication signals
Typology: Study notes
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Chapter 2
2.1 (a)
( ) cos(2 ) , 2 2
c
c
g t A f t t
f T
We can rewrite the half-cosine as:
cos(2 (^) c ) rect
t A f t T
Using the property of multiplication in the time-domain:
[ ]
1 sin( ) ( ) ( ) 2
c c
G f G f G f
fT f f f f AT fT
Writing out the convolution:
[ ]
sin( ) ( ) ( ( ) ( ( ) 2
2 2
cos( ) cos( )
2 1 1
2 2
c c
c c c c c
G f f f f f d T
A f f T f f T f f f f f T
A fT fT
f f T T
∞
−∞
∫
(b)By using the time-shifting property:
0 0 0 ( ) exp( 2 ) 2
cos( ) cos( ) ( ) exp( ) 2 1 1
2 2
g t t j ft t
A fT fT G f j fT
f f T T
( )( )
( )
( ) exp( )sin(2 )u( )
exp( )u( ) sin(2 )
c
c
c c
c c
g t t f t t
t t f t
G f f f f f j f j
j j f f j f f
2.3 (a)
[ ]
[ ]
( ) rect 2
( ) rect rect
e o
e
e
o
o
g t g t g t
g t g t g t
t g t A T
g t g t g t
t T t T
g t A T T
(b)
By the time-scaling property g(-t) R G(-f)
[ ]
[ ]
[ ]
[ ]
sinc( ) exp( 2 ) sinc( ) exp( 2 ) 2
sinc( ) cos( )
sinc( ) exp( 2 ) sinc( ) exp( 2 ) 2
sinc( ) sin( )
e
o
G f G f G f
fT j fT fT j fT
fT fT
G f G f G f
fT j fT fT j fT
j fT fT
π π
π
π π
π
2.6 (a)
If g(t) is even and real then
( ) is all real
G f G f G f
G f G f
G f G f
G f
If g(t) is odd and real then
( ) must be all imaginary
G f G f G f
G f G f G f
G f G f
G f G f
G f
(b)
The previous step can be repeated n times so:
( )
( )
But each factor ( 2 ) represents another differentiation.
Replacing with
n n n
n n n
n n n
d j ft G t g f df
j ft
j t G t g f
g h
j t h t H f
[ ]
and ( ) ( ) ( ) ( )
g t g t g t
g t g t G f G f
[ ]
and ( ) ( ) ( ) ( )
g t g t g t
g t g t G f G f
( 2 ) ( ) ( ) by duality
d j t G t g f df
j d t G t g f df
(c)
Let
( ) ( ) ( ) and ( ) ( ) 2
n n j n h t t g t H f G f
( ) ( ) (0) (0) 2
n j n h t dt H G π
∞
−∞
∫
(d)
1 1
2 2
g t G f
g t G − f
1 2 1 2
1 2 1 2
1 2
g t g t G G f d
g t g t G G f d
G G f d
∞
−∞ ∞
−∞ ∞
−∞
∫
∫
∫
(e)
1 2 1 2
1 2
1 2 1 2
1 2 1 2
g t g t G G f d
g t g t dt G
g t g t dt G G d
g t g t dt G G d
∞
−∞ ∞
−∞ ∞ ∞
−∞ −∞ ∞ ∞
−∞ −∞
∫
∫
∫ ∫
∫ ∫
2.7 c)
2 2 2
2 2 2 2
2
sin ( ) 4 ( )
sin ( )
j f G f f G f
fT f AT fT
fT T
The second derivative of the triangular pulse is plotted as:
Integrating the absolute value of the delta functions gives:
2
2
2 2 2
d g t A dt dt T
d g t j f G f dt dt
π
∞
−∞
∞
−∞
∫
∫
2.8. (a)
1 2 1 2
2 1
( ) ( ) by the commutative property of multiplication
g t g t G f G f
G f G f
b)
[ ] [ ]
[ ] [ ]
[ ] [ ]
1 2 3 1 2 3
1 2 3 1 2 3
1 2 3 1 2 3
Because multiplication is commutative, the order of the multiplication
doesn't matter.
g f g f g f G f G f G f
G f G f G f G f G f G f
G f G f G f g f g f g f
c)
Taking the Fourier transform gives:
1 [^2 3 ]
1 2 2 3 1 2 1 2
Multiplication is distributive so:
G f G f G f
G f G f G f G f g t g t g t g t
2.11 a) Given a rectangular function:
( ) rect
t g t T T
, for which the area under g(t) is
always equal to 1, and the height is 1/T.
rect sinc( )
t fT T T
Taking the limits:
0
0
lim rect ( )
lim sinc( ) 1
T
T
t t T T
fT T
→
→
b)
( ) sgn( ) 2 2
By duality:
G f f
G f t j t
j g t t t
( ) 2 sinc(2 )
2 sinc(2 ) rect 2
g t W Wt
f W Wt W
lim 2 sinc(2 ) ( )
lim rect 1 2
W
W
W Wt t
→∞
→∞
2.13. a) By the differentiation property:
( )
2
2 2
2 ( ) exp( 2 )
( ) exp( 2 ) 4
i i i
i i i
j f G f k j ft
G f k j ft f
∑
∑
b)the slope of each non-flat segment is:
b a
t t
[ ]
( )
[ ]
2 2
2 2
( ) exp( 2 ) exp( 2 ) exp( 2 ) exp( 2 ) 4
cos(2 ) cos(2 ) 2
b a a b b a
b a b a
G f j ft j ft j ft j ft f t t
ft ft f t t
But: (^) [ ]
sin( ( ))sin( ( )) cos(2 ) cos(2 ) 2
( ) (^2 2) [ sin( ( )) sin( ( ))] ( )
b a b a b a
G f f t t f t t f t t
2.14 a) let g(t) be the half cosine pulse of Fig. P2.1a, and let g(t-t 0 ) be its time-shifted
counterpart in Fig.2.1b
( )( )
( )( )
2
*^2 0 0 0 0
*^2 0 0
( ) exp( 2 ) ( ) exp( 2 ) ( ) exp( 2 ) exp( 2 )
( ) exp( 2 ) ( ) exp( 2 ) ( )
G f G f
G f
G f j ft G f j ft G f j ft j ft
G f j ft G f j ft G f
2.14 b)Cont’d
By rearranging the previous expression, and summing over a common denominator, we
get:
( )
( )
2 2 2
2 2 2 2 2
2 2 2
(^2 4 ) 2 2 4
2 2 2
2 2 2 2
cos ( )
cos ( )
cos ( )
A T fT
f T
A T fT
T f T
A T fT
T f
π π π π π π
2.15 a)The Fourier transform of
dg t j fG f dt
Let
dg t g t dt
By Rayleigh’s theorem:
2 2 g t ( ) dt G f ( ) df
∞ ∞
−∞ −∞
∫ ∫
( )
( )
( )
( )
( )
2 2 2 2 2 2 2 2
2 2 *
2 2 2
2 2 * *
2 2 2
2
2 2 *
t g t dt f G f df W T
g t dt
t g t dt g t g t dt
g t dt
t g t g t tg t g t dt
g t dt
d t g t g t dt dt
g t g t dt
π
π
π
∫ ∫
∫
∫ ∫
∫
∫
∫
∫
∫
Using integration by parts, we can show that:
2 2
2 2 2
d t g t dt g t dt
∞ ∞
−∞ −∞
∫ ∫
Given:
2 x t ( ) dt and h t ( ) dt , which implies that h t dt ( )
∞ ∞ ∞
−∞ −∞ −∞
∫ ∫ ∫
However, if
2 2 4 x t ( ) dt then X ( f ) df and X ( f ) df
∞ ∞ ∞
−∞ −∞ −∞
∫ ∫ ∫
. This result also
applies to h(t).
Y ( f ) = H ( f ) X ( f )
(^2) * *
2 2
2 2 4 4
2
Y f df X f H f X f H f df
X f H f df
Y f df X f df H f df
Y f df
∞ ∞
−∞ −∞ ∞
−∞
∞ ∞ ∞
−∞ −∞ −∞
∞
−∞
∫ ∫
∫
∫ ∫ ∫
∫
By Rayleigh’s theorem:
2 2 Y ( f ) df y t ( ) dt
∞ ∞
−∞ −∞
∫ ∫
2 y t ( ) dt
∞
−∞
∫
The transfer function of the summing block is: H (^) 1 ( f ) = (^) [ 1 − exp( − j 2 π fT )].
The transfer function of the integrator is: (^2)
H f
These elements are cascaded :
( ) ( )
( )
[ ]
( )
[ ]
1 2 1 2
2 2
2
1 exp( 2 ) 2
1 2 exp( 2 ) exp( 4 ) 2
H f H f H f H f H f
j fT f
j fT j fT f
