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by a n i nductor o f impedance L s in series with a voltage source of value LI(O) (F~g.
6.11e) or by t he same inductor in parallel with ~ c urrent ~ourc: of value 1(0)/ ~ (FI~.
6.11f). To prove this point, consider t he t ermmal relatIOnshIp of t he capacItor m
Fig. 6.11a
dv i (t) = C dt
T he Laplace transform of this equation yields 1(8) = C [sV(s)  v(O)] s (f) F ig. 6 .11 I nitial condition generators for a c apacitor a nd a n i nductor. T his equation can be rearranged as
1
V (s) =  1(8)
Cs Initial Condition Generators i tO) Ls L i(O) (d) + v(O)
+s (6.59a) Observe t hat V (s) is t he voltage (in t he frequency domain) across the charged
capacitor a nd I (s) / C s is t he voltage across t he s ame capacitor without any charge.
Therefore, t he above equation shows t hat t he c harged capacitor can be represented
by t he Same c apacitor (uncharged) in series with a voltage source of value v(O)/s,
as depicted in Fig. 6.11b. Equation (6.59a) can also be rearranged as
1
V (s) =  [I(s)
Cs + Cv(O)] (6.59b) This equation shows t hat t he charged capacitor voltage V (s) is equal to t he uncharged capacitor voltage caused by a current I (s) + Cv(O). T his result is reflected precisely in Fig. 6.l1c, where t he c urrent through t he uncharged capacitor
is I (s) + Cv(O).t
t In t he t ime d omain, a c harged c apacitor C w ith i nitial v oltage vIOl c an b e r epresented a s t he
s ame c apacitor u ncharged i n s eries w ith a v oltage Source v (O)u(t), o r i n p arallel w ith a c urrent
s ource C V(0)6(t). S imilarly, a n i nductor L w ith i nitial c urrent i(O) c an b e r epresented b y t he s ame
i nductor w ith z ero i nitial c urrent i n s eries w ith a v oltage s ource L i(0)6(t) o r w ith a p arallel c urrent
s ource i (O)u(t). 6 402 C ontinuousTime S ystem A nalysis U sing t he L aplace T ransform 403 6.4 A nalysis o f E lectrical N etworks: T he T ransformed N etwork 'C ( 0)= 16 F or t he i nductor i n Fig. 6 .lld, t he t erminal e quation is Y2 ( 0)=4 di vet) = L dt (a) a nd V (s) = L [sf(s)  itO)]
(6.60a) == L sf(s)  Li(O) 16 s i(O)) S (6.60b) = Ls [ f(8)  s W e c an v erify t hat F ig. 6.11e satisfies E q. ( 6.60a) a nd t hat F ig. 6 .1lf s atisfies Eq.
(6.60b).
2 il IH Z (s) b + (c) F ig. 6 .13 Solving Example 6.14 using initial condition generators and Thevenin equivalent
representation. 10 !O b
( b) S • (b) ( a) F ig. 6 .12 A c ircuit a nd its transformed version with initial condition generators (Example
6.12).
L et u s r ework E xample 6 .10 u sing t hese i deas. F igure 6 .12a s hows t he c ircuit
i n F ig. 6 .Sb w ith t he i nitial c onditions yeO) == 2 a nd v dO) ~ 10. Figu~e 6:1~b
s hows t he f requencydomain r epresentation ( transformed circUIt) o f t he c ircuit I II
F ig. 6.12a. T he r esistor is r epresented b y i ts i mpedance 2; t he in~uct~r w ith init.ial
c urrent o f 2 a mps is r epresented a ccording t o t he a rrangement I II F ig. 6 .lle w ith
a s eries v oltage s ource Ly(O) = 2. T he c apacitor w ith i nitial v ol...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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