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Unformatted text preview: ECE 2100 Spring 2011 Homework Assignment 3 Solution Due March 3, 2010 Covered: Chapters 57: Inductors and capacitors, Natural and step responses of RC and RL circuits, natural and step responses of RLC circuits, under, over and critically damped responses. 1) Prelab: Consider the following RC circuit and RL circuits: a. For the RC circuit, what its time constant? τ =RC=560 Ω ·0.1 μ F= 56 μ s (alternate: accounting for 50 Ω in source and shunt, R=585 Ω b τ = 58.5 μ s ) b. What will the RC step response be? Hint: what are I( ∞ ) and Vout( ∞ ) if Vin(t)=1V·u(t)? Vout( ∞ )=1V, so Vout(t)= u(t)1V(1exp(t/56 μ s) c. For the LR circuit, what its time constant? τ =L/R=·1200 μ H /560 Ω = 2.14 μ s (alternate, R=585 Ω b τ = 2.05 μ s ) d. What will the LR step response be? Hint: what are I( ∞ ) and Vout( ∞ ) if Vin(t)=1V·u(t)? Vout( ∞ )=0V, I( ∞ )=1V/560 Ω , Vout(t)= u(t)1V(exp(t/2.14 μ s) RC circuit RL circuit I(t) I(t) 2) Prelab: Consider the following RLC circuit which is driven by the function generator as shown below. The square wave transitions from zero to +1.0 V. a. Calculate the parameters α and ω for this circuit and determine the type of response which v(t) will have. Is it under damped, over damped, or critically damped? α =1/(2RC)= 8930s1 (8550s1 accounting for source) ω =(LC)½ = 91300 s1 b. Calculate ω d , if the circuit is under damped, then calculate the expected period of oscillation. ω d = ( ω 2 α 2 ) ½ = 90900 s1 T = 2 π / ω d = 69 μ s c. Find an equation for v(t) for the step response. This corresponds to the transition where the function generator “switches" from zero volts up to V at time t = 0. Form of the equation: Vout(t)=exp( α t)(Acos( ω d t)+Bsin( ω d t)) Steady state: Vout( ∞ )=0V, I L ( ∞ )=1V/560 Ω Solve for Vout(t)’= Vout(t) Vout( ∞ ), I L (t)’= I L (t)I L ( ∞ ) Initial conditions: Vout(0)’=0, I L (0)’= V /560 Ω A= Vout(0)’=0 I L (0)’/C= = dt dVout α A ω d B=  ω d B = = d RC V B ϖ 0.2V Vout(t) = d RC V ϖ exp( α t)(sin( ω d t))= 0.2V·exp(t·8930 s1 )( sin(t·90900 s1 )) d. How small of a resistor (call it R 2 ) would have to be placed in parallel with the LC combination to achieve a critically damped response? Critical damping: α = ω RR 2 =(2C· ω )1 =54.8 Ω b R 2 =(1/( RR 2 )1/R))= 60.7 Ω 3) Book problem 5.2 a) di/dt = Vs/L, for 0<t<1ms, Vs(t) = 6mV, I = t(6mV/L) +I(o) = t(6/.3mH) = 20A/s*t for 1ms<t<2ms, Vs(t) = 12mVt(6mV/ms), I = (12t3000t 29)/L + I(1ms) I = 10kA/s 2 *t 2 +40A/s*t30mA+20mA = 10mA(t 2 /ms 2 +4t/ms1) For t>2ms, I = I(2ms) = 30mA b) 5 10 15 20 25 30 35 0.5 1 1.5 2 2.5 3 t, ms I, A 4) Book problem 5.14 Note that I = Cdv/dt b For 0<t<1, I(t) = 0.5 μ F*(120t 2 ) For 1<t<2, I(t) = 0.5 μ F*(120(2t) 2 ) 15010050 50 100 150 0.5 1 1.5 2 time, s V, A Vin I, A 5) Book problem 5.28 a)Ceq=2nF b Vb(0)=100V, t t t t Ve e V V Vb Adt e Ceq Vb 250 250 250 100 ) 1 ( 100 100 ) ( 50 1 = = +...
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This note was uploaded on 06/10/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Spring '05 term at Cornell.
 Spring '05
 KELLEY/SEYLER

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