Homework 3_soln

# Homework 3_soln - ECE 2100 Spring 2011 Homework Assignment...

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ECE 2100 Spring 2011 Homework Assignment 3 Solution Due March 3, 2010 Covered: Chapters 5-7: 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 ·0.1 μ F= 56 μ s (alternate: accounting for 50 in source and shunt, R=585 barb2right 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(1-exp(-t/56 μ s) c. For the LR circuit, what its time constant? τ =L/R=·1200 μ H /560 = 2.14 μ s (alternate, R=585 barb2right 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 , Vout(t)= u(t)1V(exp(-t/2.14 μ s) RC circuit RL circuit I(t) I(t)

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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 0 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)= 8930s -1 (8550s -1 accounting for source) ω 0 =(LC) = 91300 s -1 b. Calculate ω d , if the circuit is under damped, then calculate the expected period of oscillation. ω d = ( ω 0 2 - α 2 ) ½ = 90900 s -1 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 0 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 Solve for Vout(t)’= Vout(t)- Vout( ), I L (t)’= I L (t)-I L ( ) Initial conditions: Vout(0)’=0, I L (0)’= -V 0 /560 A= Vout(0)’=0 I L (0)’/C=- = dt dVout α A- ω d B= - ω d B = = d RC V B ϖ 0 0.2V Vout(t) = d RC V ϖ 0 exp(- α t)(sin( ω d t))= 0.2V·exp(-t·8930 s -1 )( sin(t·90900 s -1 )) 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 0 R||R 2 =(2C· ω 0 ) -1 =54.8 barb2right R 2 =(1/( R||R 2 )-1/R))= 60.7
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) = 12mV-t(6mV/ms), I = (12t-3000t 2 -9)/L + I(1ms) I = -10kA/s 2 *t 2 +40A/s*t-30mA+20mA = 10mA(-t 2 /ms 2 +4t/ms-1) For t>2ms, I = I(2ms) = 30mA b) 0 5 10 15 20 25 30 35 0 0.5 1 1.5 2 2.5 3 t, ms I, A 4) Book problem 5.14 Note that I = Cdv/dt barb2right For 0<t<1, I(t) = 0.5 μ F*(120t 2 ) For 1<t<2, I(t) = 0.5 μ F*(-120(2-t) 2 ) -150 -100 -50 0 50 100 150 0 0.5 1 1.5 2 time, s V, A Vin I, A

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5) Book problem 5.28 a)Ceq=2nF barb2right Vb(0)=100V, t t t t Ve e V V Vb Adt e Ceq Vb 250 250 0 250 100 ) 1 ( 100 100 ) 0 ( 50 1 - - - = - - = + - = μ b) t t t t Ve e V V Va Adt e nF Va 250 250 0 250 16 1 ) 1 ( 16 15 ) 0 ( 50 5 . 12 1 - - - + - = - - = + - = μ c) t t t Ve V Vc Adt e nF Vc 250 0 250 4 41 ) 0 ( 50 50 1 - - - - = + = μ d) t t t Ve V Vd Adt e nF Vd 250 0 250 80 40 ) 0 ( 50 5 . 2 1 - - + - = + - = μ e) A e nF nF i i t b μ 250 1 20 5 . 2 1 - = = f) A e nF nF i i t b μ 250 2 30 5 . 2 5 . 1 - = = 6) Book problem 5.29 a) W(0)=sum(CV 2 /2)=12.5nF·(15V) 2 /2+2.5nF·(40V) 2 /2+50nF·(-45V) 2 /2= 54 μ J b) W( )=sum(CV 2 /2)=12.5nF·(1V) 2 /2+2.5nF·(-40V) 2 /2+50nF·(-41V) 2 /2= 44 μ J c) W delivered = W(0)- W( )=10 μ J d)
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