EECE 290 Problems For Laplace, Spring 2008

L -9- Find : LYE (s-1)3+(52+25-8) 4} (13) — _ a a. 0.5e't?+(2/3)e'.Cosh(3t) <b._0,5e't2+(1/3)e" .Sinh(3t)’ c. 0.5e't2+(1/3)e’.Cosh(3t) 0.5e't2+(2/3)e*. Cosh(3t) e. None of the above Lo4- The Laplace transform of f(t)=Sin 5t is (pit) t a. (n/2)-tan*(s/2) b. (n/2)-tan“(s) c. (x/5)-tan"!(s/5) d. (n/2)-tan*(s/5)) e. None of the above L ~2- The inverse Laplace transform of F(s)=_s+3 , is (pie) s2-s-2 a. (2/3)e*+(5/3)et b. (2/3)e?"-(2/3)e* _ c. (5/3)e*"+(2/3)e* d. (2/3)e*-(5/3)e"/e)None of the above ~~, 6 (1-e j . *~5- Find the Laplace transform of f(t)= {t, Ost<2 {2, t>2 —_ )/s3_b. (1-s-e)/s?_¢. (1-25-€5)/s2 d~¢4--se**5)/s2_ e. None of the above (p12) a)e *(4Cost+2Sint) b) e(-4Cost+2Sint) << e*(4Cost-2Sint) d)-e7'(4Cost+2Sint) e)None of the above | 24.Find I(t) for t>0. yf L -2- In the following circuit, the switch is opened at t=0. Find I(s). (p/4] 300hm 60H > AAO 6 1F_| -T™ \ S$ 120V i I | 4 1} UL a. 240s+120 b. 120 60s2+30s+1 s+0.46 on 40s+2 d. 4s : (s+46)(S+0.36) ( s+0.46)(s+0.0036) e. None of the above L -3- In the circuit shown, the switch has been for a long time at (prs ) position 1. At t=0, it is moved to position 3. Find I(s). v ne eanoyw Y) 3 1 1000hm +--+ 2 0.4H i(o+)=0.5A . (0.552+1.5x10°)/(s2+10°)(s+1000) . (10°s2)/(s+10°)(s-1000) (s?+1.5x10°)/(s+10°)(s+1000) . (1.5x10°)/(s2+10°)(s+1000) None of the above L -4- Calculate I(t) for problem 3, for t>0. (ad a. [Sin(1000t+45°)-+e° 10} u(t) b. [(1/V2)Sin(1000t+452°) +7007, uct) c. [(1/V¥2)Sin(1000t-452)+e710y u(t) d. [V¥2.Sin(1000t-45°)-e710) u(t) e. None of the above L -5- In the circuit shown, the switch has been for a long time at position 3. At t=0, the switch is moved to position 1. Find V(s) P (1s) V4 1KV @) 1 imicroF “i -T~ 2microF a. 500/(s+0.1) c. ~-1020/(s+0.155) (g.None of the above L -6~- Find v(t) for t>0 for problem 5. a. 500(e°*) u(t) e. None of the above 5000K + b. 1020/(s?+0.8s+0.1) d. 500/(s2+0.8s-+0.1) (b) 1020(@°0:545t_9-0-155t) Y(t) Cc. -1020(e° 4°") U(t) dd. 500(e°-645t-@°0-155t) Y(t (py -—- \/-10-Consider the cascaded connection of 2-RC low-pass filters / and determine the impulse response at the output. Let RC=1/ 9 /6) R R a. t.e u(t) b. e‘U(t) OVW VV c. t2.e*U(t) Vd, t.e*U(t) Cc Cc e.None of the above # * 2H 200 Figure 22 L 23. The inductor and capacitor in figure 22 have no inital stored energy and the op amp is ideal. The switch is closed at t= 0. Determine v, for ¢>0. (p S>/) sa) Se! 2.567 2.5 b) 10e 1%. 5¢20_ 5 v c) Be! ge? _gy dy 6e! 367 _3y e) None of the above 10 1F 2V Q—~ II ime 6v © 1F 18 300 Figure 1. Derive V;(s) in Fig. 1, assuming zero initial conditions. ( P? 6¢ } AL (s+2)/(s? +s +1) B. 2s”/(s?+3s +1) C. (s+ 1)/Gs +2s+1) D. 2s/(2s’+2s+1) E. None of the above Figure 3. The voltage ratio V2(s) / V;(s) in Fig. 3 has a pole at -100 + j700 rad/s. If R = 500 ohms, find L and C. (p66) A. L=154H, C= 1.388 uF B. L=10H, C= 2.083 pF — C. L=2.5H, C=0.800 pF D. L=2.0H, C= 1.040 uF E. None of the above