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Unformatted text preview: k with a switching action (Example 6.10).
• E xample 6.10
In the circuit of Fig. 6.8a, the switch is in the closed position for a long time before
t = 0, when it is opened instantaneously. Find the inductor current y( t ) for t 2: o.
When the switch is in the closed position (for a long time), the inductor current is
2 a~ps and the capacitor voltage is 10 volts. When the switch is opened, the circuit is
eqU1v~l~nt to th~t depicted in Fig. 6.8b, with the initial inductor current y (O) = 2 and
the initial capacitor voltage v c(O) = 10. The input voltage is 10 volts, starting at t = 0,
and, therefore, can be represented by 1 0u(t). 6 394 6.3 Solution of Differential a nd I ntegroDifferential Equations. ContinuousTime S ystem Analysis Using t he L aplace or The loop equation of the circuit in Fig. 6.8b is
dy
dt + 2 y(t) + 5 yes) (G.46a) s Y(s) _ y (O) = s Y(s)  2 (6.46b) yet) then
{ => { => and [see Eq. (6.36)J
y es)
 s t 1 ( Dn = lOu(t) y (r) d r  00 If dy
dt jt y (r)dr { => f:~y(r)dr + 395 (6.46c) s + a n_1D n  1 + ... + a lD + a o)y(t) =
n
( bnDn + b n_1D  1 + '" + b 1D + b o)f(t) (6.49) W e s hall now nnd. t he g eneral expTess10n t m t he 'Zerostate r esponse of a n L TIC
system. Z erostate r esponse y et), b y definition, is t he s ystem r esponse t o a n i nput
w hen t he s ystem is initially relaxed (in zero s tate). T herefore, y et) satisfies t he
s ystem e quation (6.49) w ith zero initial conditions Moreover, t he i nput J (t) is causal, so t hat  00 Because yet) is t he capacitor current, the integral f~~ y (r) dr is q c(O), t he capacitor
charge a t t = 0 , which is given by C times the capacitor voltage a t t = 0 . Therefore L et
y et) ~ y es) (6.47) a nd J (t) ~ F (s) B ecause of zero initial conditions From Eq. (6.46c) it follows t hat j t y (r)dr { => y es)
 s 2 (6.48) +;  00 Taking the Laplace transform of Eq. (6.45) and using Eqs. (6.46a), (6.46b), and (6.48), we
obtain 5 Y(s)
s Y(s)  2 + 2 Y(s) +  s  10 Therefore, t he L aplace t ransform o f Eq. (6.49) yields 10 +  ; = ; or or and (6.50a) 2s y es) = s2 + 2s + 5 To find t he inverse Laplace transform of y es), we use Pair lOc (Table 6.1) with values
A = 2, B = 0, a = 1, and c = 5. This yields
r
Therefore = [ ii = v'5, b= ~ = 2 and () = t an W = 26.6° (6.50b) Q (s) B ut we have shown in Eq. (6.41) t hat y es) = H (s)F(s). C onsequently,
H ()=P(s) yet) = v '5e This response is shown in Fig. 6.8c. 6.31 1 = p es) F (s) t cos (2t + 26.6°)u(t) • ZeroState Response: The Transfer Function o f an LTIC System C onsider a n n thorder L TIC system specified by t he e quation
Q (D)y(t) = P (D)J(t) s Q(s) (6.51) T his e quation was derived earlier in timedomain [see Eq. (2.50)]. Thus, y es) = H (s)F(s) (6.52) We have a lready d erived t his r esult twice using o ther a pproaches. We now have a n
a lternate i nterpretation (or a n a lternate definition) for t he t ransfer function H (s). 396 6 6.3 Solution of Differential a nd IntegroDifferential Equations. Gontinuous...
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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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