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CHAP. 31
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LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
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Fig. 3-16
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From Fig. 3-16(b) the loop equation can be written as
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or Hence,
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Taking the inverse Laplace transform of I(s), we obtain
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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
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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.
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(a) Find the currents i , ( t ) and i,(t). ( b ) Find the voltages across the capacitors at
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t = 0'
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From the given initial conditions, we have uCl(O-) = 1 V and L!=~(O-) 2 V =
Thus, using Fig. 3-10, we construct a transform circuit as shown in Fig. 3-17(b). From
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
Fig. 3-17
Fig. 3-17(b) the loop equations can be written directly as
Solving for I,(s) and I,(s) yields
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
Substituting I,(s) and 12(s)obtained in part ( a ) into the above expressions, we get
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
Then, using the initial value theorem (3.971, we have
v , , ( O + ) = lim sVcl(s)= lim - 71 -1 =
s 4 m s-rm
+a +
+ 1= 2 V
ucJO+) = lim sVC-(s) = lim - + 2 7+ 2 = 3 V - =1 s-+m s-+m
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
Supplementary Problems
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.
3.44. Find the Laplace transform of x ( t ) given by
x(t) =
1 Am. X ( s ) = -[e-"I - e-"21, all s
t , _< t s t , otherwise
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
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:
Proceed in a manner similar to Prob. 3.4.
Verify Eq. (3.21), that is,
Hint:
Differentiate both sides of Eq. (3.3) with respect to s.
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~).
Find the Laplace transform of x(t) Ans. X ( s ) = s+l
= (e-'cos21-
Se-*')u(t)
+ :e2'u(-t)
( ~ + 1 ) ~ + 4 +2 S
2s-2
, - 1 < Re(s)<2
Find the inverse Laplace transform of the following X(s);
( b ) X(s) =
s + l (d) X(s) = , Re(s) > -2 s 2 + 4 s + 13
CHAP. 31
LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
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.
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