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

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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)
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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)
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern