Signal Processing and Linear Systems-B.P.Lathi copy

Signal Processing and Linear Systems-B.P.Lathi copy

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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 ntegro-Differential Equations. Continuous-Time 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 'Zero-state r esponse of a n L TIC system. Z ero-state 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. 6.3-1 1 = p es) F (s) t cos (2t + 26.6°)u(t) • Zero-State Response: The Transfer Function o f an LTIC System C onsider a n n th-order 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 time-domain [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 Integro-Differential 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.

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