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The following equalities are used on many occasions in this text. Prove their validity. in .NET
1.19. The following equalities are used on many occasions in this text. Prove their validity. QR Code JIS X 0510 Recognizer In VS .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. Print QR Code ISO/IEC18004 In VS .NET Using Barcode generation for VS .NET Control to generate, create QR Code image in .NET framework applications. SIGNALS AND SYSTEMS
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< 1, lim
a N= 0.
Then by Eq. (1.96) we obtain
N l
a n=
Nm 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. 126. Fig. 126 1.22. A continuoustime signal following signals.
is shown in Fig. 127. 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. 127 (a) By definition ( 1 . 1 9 ) and x(r)u(l  t ) is sketched in Fig. 128(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. 128(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. 128(c). Fig. 128 1.23. A discretetime signal x [ n ] is shown in Fig. 129. 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( 4321 1 2 3 Fig. 129 ( a ) By definition ( 1 . 4 4 ) and x[n]u[l n ] is sketched in Fig. 130(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. 130(b). (c) By definition (1.48) x [ n ] S [ n  11 = x [ l ] S [ n I ] =S[n  11 = which is sketched in Fig. 130(c). (4 Fig. 130 124 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.

