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Fig. 3-16

or Hence,

Note that i(O+) = 2 = i(0-); that is, there is no discontinuity in the inductor current before and after the switch is opened. Thus, we have

3.42. Consider the circuit shown in Fig. 3-17(a). The two switches are closed simultaneously at t = 0. The voltages on capacitors C, and C, before the switches are closed are 1 and 2 V, respectively.

t = 0'

Fig. 3-17

Taking the inverse Laplace transforms of I , ( s ) and 12(s),we get i l ( t )= 6 ( t ) + i e - ' l 4 u ( t ) i 2 ( t )= 6 ( t ) - f e - ' / 4 ~ ( t ) ( b ) From Fig. 3-17(b) we have

v , , ( O + ) = lim sVcl(s)= lim - 71 -1 =

S - 7

Note that ucl(O+)#u,1(0-) and ucz(O+)# ~ ~ $ 0 -This is due to the existence of a ). capacitor loop in the circuit resulting in a sudden change in voltage across the capacitors. This step change in voltages will result in impulses in i , ( t ) and i 2 ( t ) . Circuits having a circuits. capacitor loop or an inductor star connection are known as degener~ti~e

Find the Laplace transform of the following x(t 1:

2s ( c ) If a > 0 , X ( s ) = - - a < Re(s) < a. If a < 0, X ( s ) does not exist since X ( s ) does s z - ,2 '

not have an ROC. ( d ) Hint: x ( t ) = u ( t ) + u ( - t ) X ( s ) does not exist since X ( s ) does not have an ROC. ( e l Hint: x ( t ) = u ( t ) - u ( - t ) X ( s ) does not exist since X ( s ) does not have an ROC.

x(t) =

Show that if X(I) is a left-sided signal and X(s) converges for some value of s, then the ROC of X(s) is of the form

equals the minimum real part of any of the poles of X(s). where amin Hint:

Verify Eq. (3.21), that is,

Hint:

Show the following properties for the Laplace transform: ( a ) If x ( t ) is even, then X( -s) = X(s); that is, X(s) is also even. ) (b) If ~ ( t isodd, then X ( - s ) = -X(s); that is, X(s)is alsoodd. (c) If x ( t ) is odd, then there is a zero in X(s) at s = 0. Hint: Use Eqs. (1.2) and (3.17). ( b ) Use Eqs. (1.3) and (3.17). ( c ) Use the result from part (b) and Eq. (1.83~).

= (e-'cos21-

Se-*')u(t)

+ :e2'u(-t)

2s-2

Find the inverse Laplace transform of the following X(s);

( b ) X(s) =

x ( r ) = ( 1 - e-' - t e - ' M t ) x ( t )= - 4 - t ) -(l + t ) e - ' d t ) ~ ( t )( - 1 + e-' + t e - ' I d - t ) = x ( t ) = e - 2 ' ( ~ o s 3- f sin 3 t ) u ( t ) t x ( t ) = at sin 2tu(t) ( f ) x ( t ) = ( - $e-2'+ Acos3t-t & s i n 3 t ) u ( t ) (a) (b) (c) (d) (el

Using the Laplace transform, redo Prob. 2.46. Hint: Use Eq. (3.21) and Table 3-1.