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Unformatted text preview: i (t) = C!fit
a nd its Laplace t ransform, a ssuming zero initial c apacitor voltage, yields I (s) =
C sV(s); t hat is,
1
V (s) =  I(s)
Cs ( b) I deal D i iferentiator yet) 6.4 Analysis of Electrical Networks: T he T ransformed Network m L lj(8) = 0
j=l (6.58) T his r esult shows t hat if we represent all t he v oltages a nd c urrents in a n e lectrical
network by t heir L aplace transforms, we can t reat t he n etwork as if i t c onsisted
of t he "resistances" R, L s a nd l /Cs c orresponding t o a r esistor R, a n i nductor
L , a nd a c apacitor C , respectively. T he s ystem equations (loop or node) are now
algebraic. Moreover, t he s implification techniques t hat have been developed for
resistive c ircuitsequivalent series a nd p arallel impedances, voltage a nd c urrent
d ivider rules, Thevenin a nd N orton t heoremscan b e a pplied t o g eneral electrical
networks. T he following examples d emonstrate t hese concepts.
• E xample 6 .13
Find the loop current i(t) in the circuit shown in Fig. 6.10a if all the initial conditions
are zero.
In the first step, we represent the circuit in the frequency domain, as illustrated in
Fig. 6.lOb. All the voltages and currents are represented by their Laplace transforms. The
voltage 1 0u(t) is represented by 1 0/s and the (unknown) current i(t) is represented by
its Laplace transform 1(8). All the circuit elements are represented by their respective
impedances. The inductor of 1 henry is represented by s, the capacitor of 1 /2 farad is
represented by 2 / s, and the resistor of 3 ohms is represented by 3. We now consider the
frequencydomain representation of voltages and currents. The voltage across any element 400 6 ContinuousTime System Analysis Using t he Laplace Transform 6.4 Analysis of Electrical Networks: T he Transformed Network
i ( I) SJ 10 s
.! F 2 ( a) I ( s) 1 +
+ v et) v (O)
C v(O) C ( b) is I (s) t imes its i mpedance. T herefore, t he t otal v oltage d rop in t he lo~p is I (s) t imes t he
t otal l oop i mpedance, a nd i t m ust b e e qual t o V (s), ( transform of) t he m put voltage. T he
t otal i mpedance i n t he loop is
2 Z (s)=s+3+;= s2 (a) (el
i ( t) + I ( s) V (s) ¥ _ S2±!S±2  +
Ls + 3s + 2
s 10 82 V (I) 10 + 3s + 2 ~ (8 + l)(s + 2) 10
10
   s +1
8+2 T he inverse t ransform o f t his e quation yields t he d esired result: i (t) = 1 0(e t  I rs) +
y es) y es) T he i nput"voltage" is V (s) = 1 0/s. T herefore, t he "loop current" I (s) is = Z (s) = 1 (8) J, + F ig. 6 .10 A c ircuit a nd i ts transformed version (Example 6.13). I (s) 401 30 IH 2t e )u(t) • (el T he above discussion, where we assumed zero initial conditions, c an ~e. rea?ily
e xtended to t he case of nonzero initial conditions because t he initial condItIOn m a
capacitor or a n i nductor can be represented by an equivalent source. We now sho.w
t hat a c apacitor C w ith a n initial voltage v(O) (Fig. 6.11a) can be r~pres~nted. m
t he frequency d omain by a n uncharged capacitor of impedance l /Cs m sene~ wI:h
a voltage source of value v(O)/s (Fig. 6.11b) or as t he same ~n:harged c~pacltor m
parallel with a c urrent source of value Cv(O) (Fig. 6.11c). ~lmllarly, a n m ductor.L
with a n initial c urrent i(O) (Fig. 6.11d) can be represented m t he frequency do...
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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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