The following equalities are used on many occasions in this text. Prove their validity. in .NET

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1.19. The following equalities are used on many occasions in this text. Prove their validity.
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Then
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N- 1 aS=a
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C an=a+aZ+a"
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Subtracting Eq. (1.95) from Eq. (1.941, we obtain Hence if
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1, we have
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1, then by Eq. (1.94)
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( 6 ) For
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l1 a
< 1, lim
a N= 0.
Then by Eq. (1.96) we obtain
N- l
a n=
N-m n =O
a n=
lim -- N
]-aN
Using Eq. (1.911, we obtain
(d) Taking the derivative of both sides of Eq. (1.91) with respect to a,we have
Hence,
1.20. Determine whether the following signals are energy signals, power signals, or neither.
( a ) x(t)=e-"'u(t), a > O (c) x ( t ) = tu(t) ( e ) x [ n l = u[nl
(b) x(t)=Acos(w,t+8) ( d l x [ n ] = ( - 0.5)"u[n] (f x [ n ] = 2ej3"
SIGNALS AND SYSTEMS
[CHAP. 1
Thus, x ( t ) is an energy signal. ( b ) The sinusoidal signal x ( t ) is periodic with To = 2 7 r / o o . Then by the result from Prob. 1.18, the average power of x(t ) is
Thus, x ( t ) is a power signal. Note that periodic signals are, in general, power signals.
Thus, x ( t ) is neither an energy signal nor a power signal. ( d ) By definition ( 1 . 1 6 ) and using Eq. (1.91), we obtain
Thus, x [ n l is an energy signal. By definition ( 1 . 1 7 )
P = lim - N+%
C 1 ,=-N
lxb1I2
lim - 1 = lim (N+l)=-<a N + C = 2 N + 1 ,= , ~ + m N + 1 2 2
Thus, x [ n ] is a power signal. Since I x [ n ] l = I2eiJnI = 2IeJ3"l= 2,
P = lim
= lim
N-+= 2 N
--- C
l x [ n ] 1 2 = lim
n= -N
2N+1
-( 2 N + 1) = 4 < m 4
2N+ 1
Thus, x [ n ] is a power signal.
BASIC SIGNALS
1.21. Show that
CHAP. 11
SIGNALS AND SYSTEMS
- t . Then by definition (1.18)
Since
> 0 and
< 0 imply, respectively, that t < 0 and r > 0, we obtain
which is shown in Fig. 1-26.
Fig. 1-26
1.22. A continuous-time signal following signals.
is shown in Fig. 1-27. Sketch and label each of the
( a ) x ( t ) u ( l - t ) ; ( b ) x ( t ) [ u ( t )- u(t - I)]; (c) x ( t ) H t -
Fig. 1-27
(a) By definition ( 1 . 1 9 )
and x(r)u(l - t ) is sketched in Fig. 1-28(a). ( 6 ) By definitions (1.18) and (1.19)
-( t - 1
O<tll otherwise
and x ( t ) [ u ( r )- u(t - I ) ] is sketched in Fig. 1-28(b).
SIGNALS AND SYSTEMS
[CHAP. 1
( c ) By Eq. (1.26)
x(t)s(t -
3) = x ( $ ) s ( t - $) = 26(t - $)
which is sketched in Fig. 1-28(c).
Fig. 1-28
1.23. A discrete-time signal x [ n ] is shown in Fig. 1-29. Sketch and label each of the
following signals.
( a ) x [ n ] u [ l - n ] ; ( b ) x [ n ] ( u [ n 21 - u [ n ] } ; c ) x [ n ] 6 [ n(
-4-3-2-1
1 2 3
Fig. 1-29
( a ) By definition ( 1 . 4 4 )
and x[n]u[l- n ] is sketched in Fig. 1-30(a).
CHAP. 11
SIGNALS AND SYSTEMS
( b ) By definitions (1.43) and (1.44) u [ n + 21 - u [ n ] = -21n <0 otherwise
and x [ n ] ( u [ n+ 21 - u [ n ] )is sketched in Fig. 1-30(b). (c) By definition (1.48)
x [ n ] S [ n - 11 = x [ l ] S [ n- I ]
=S[n -
11 =
which is sketched in Fig. 1-30(c).
(4 Fig. 1-30
1-24 The unit step function u ( t ) can be defined as a generalized function by the following
relation:
+ ( t ) u ( t ) dt = j w 4 ( t ) dt
(1.98)
where & ( t ) is a testing function which is integrable over 0 < t < m. Using this definition, show that
Rewriting Eq. (1.98) as
SIGNALS AND SYSTEMS
[CHAP. 1
we obtain
This can be true only if
1 4 ( t ) u ( t )dt a
k w 4 ( t ) [ l- u ( t ) ] dt
These conditions imply that
b ( t ) u ( t )= 0 , t < 0
c $ ( t ) [ l- u ( t ) ] = 0 , t > 0
Since 4 ( t ) is arbitrary, we have
u(t)=O,t<O
1 -u(t)=O,t>O
that is,
1.25. Verify Eqs. (1.23) and (1.24), that is, 1 ( a ) 6 ( a t )= -6(t); ( b ) S ( - t ) = S ( t ) la1
The proof will be based on the following equiualence property: Let g , ( t ) and g 2 ( t ) be generalized functions. Then the equivalence property states that g , ( t ) = g t ( t ) if and only if
for all suitably defined testing functions 4 ( t ) . ( a ) With a change of variable, at = T , and hence following equations: If a > 0 ,
t =r/a,
= ( l / a )d r ,
we obtain the
Thus, for any a
CHAP. 1 1
SIGNALS AND SYSTEMS
Now, using Eq. (1.20) for 4(0), we obtain
for any 4(t). Then, by the equivalence property (1.991, we obtain
1 6(at)= -6(t) la1
(6) Setting a
- 1 in the above equation, we obtain
6( - t )
1 -6(t) I- 1 1
= S(t)
which shows that S(t) is an even function.
1.26. (a) Verify Eq. (1.26):
x ( t ) 8 ( t - t o ) =x(t,)S(l - 1,)
if x ( t ) is continuous at t = to.
( b ) Verify Eq. (1.25): x ( r ) S ( t ) =x(O)S(t)
if x ( t ) is continuous at t
= 0.
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