HW3_soln-1

HW3_soln-1 - ECE 2100 Spring 2010 Homework assignment 3...

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ECE 2100 Spring 2010 Homework assignment 3 Solution Due March 11, 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 b = 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 LC circuit, what its time constant? =L/R=·1200 μ H /560 = 2.14 μ s (alternate, R=585 = 2.05 μ s ) d. What will the LC 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) 2) Prelab: Consider the following RLC circuit which is driven by the function generator as shown below. Note that the 50 load which appears on the output of the function generator is also placed in the circuit diagram. The square wave transitions from zero to +1.0 V. a. What is the equivalent parallel resistance in this RLC circuit? I(t) I(t)

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Equivalent parallel is 560 (or 585 accounting for the source) b. 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 c. Calculate ω d if the circuit is under damped and then, calculate the expected period of oscillation. ω d = ( ω 0 2 - 2 ) ½ = 90900 s -1 T = 2 π / ω d = 69 μ s d. 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(- 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 A- ω d B= - ω d B = = d RC V B ϖ 0 0.2V Vout(t) = d RC V 0 exp(- t)(sin( ω d t))= 0.2V·exp(-t·8930 s -1 )( sin(t·90900 s -1 )) e. 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: = ω 0 R||R 2 =(2C· ω 0 ) -1 =54.8 b R 2 =(1/( R||R 2 )-1/R))= 60.7
3) Book problem 5.5 a) + = t t t I d L V t I 0 ) ( ) ( ) ( 0 τ 0<t<2s: t 0 = 0, I(t 0 )=0, V(t)=-25V/s·t, I(t) = -5A/s 2 ·t 2 2s<t<6s : t 0 = 2s, I(t 0 )=-20A, V(t)=25V/s·(t - t 0 )-50A, I(t) = -20A-20A(t-2s) +5A/s 2 ·(t-2s) 2 6s<t<10s : t 0 = 6s, I(t 0 )=-20A, V(t)=-25V/s·(t - t 0 )+50A, I(t) = -20A+50A(t-6s)-5A/s 2 ·(t-6s) 2 10s<t<12s : t 0 =10s, I(t 0 )=-20A,V(t)=25V/s·(t - t 0 )-50A, I(t)=-20A+50A(t-10s)-5A/s 2 ·(t-10s) 2 b)the current is 0A at t=0, -40A at t=4s, 0A at t=8s, and -40A at t=12s.

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This note was uploaded on 11/21/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Fall '05 term at Cornell.

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HW3_soln-1 - ECE 2100 Spring 2010 Homework assignment 3...

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