homework eight part one i am okay with the rest but dont understand the first bit
MIT 6.002 Circuits and Electronics, Spring 2007
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The impulse response is a concept often used in electrical engineering, because it fully characterizes the reaction of a linear system as a function of time. As a first step, we will learn the techniques used to calculate the impulse response of a given system.
Consider the circuit shown below. Assume that both the capacitor and the inductor have been left in their relaxed states for a very long time. The voltage source is driven by a signal of vin=δ(t), that is a unit impulse at time t=0. R1=10Ω, L=5mH, R2=10kΩ, C=33nF.
(a) What is the value of vR1(t) at t=0−?
(b)Derive a differential equation expression for vR1(t), the voltage across the resistor in series with our inductor. What is the value of vR(t) at t=0+? Hint: it may be helpful to know that ∫0+0−δ(t)dt=1
(c)Assuming an exponential decay, what is the value for vR(t) at t=1ms?
(d)Similarly, derive a differential equation expression for vC(t), the voltage across our capacitor. What is the value of vC(t) at t=0+?
(e)What is the value for vC(t) at t=1ms?
Well, I can answer H8P1 (a), which is simply "0V".
The time is t = 0-s, so JUST before the Impulse will start. So the supplied voltage is zero and therefore all voltages in the network are 0. Still working on Question (b)-(e).
ok, I got all of them now. I do all calcs with MatLab since I like scripting :)
R1 = 10;
R2 = 10000;
C = 33*10^-9;
L = 5 * 10^-3;
t0 = 0;
t0plus = 1;
t0min = 0;
tms = 0.001;
v_R1_0min = R1/L * exp(-t0*R1/L) * t0min
% Q2 (see: http://en.wikipedia.org/wiki/RL_circuit#Impulse_response )
v_r1_0plus = R1/L * exp(-t0*R1/L) * t0plus
% Q3 (see: http://en.wikipedia.org/wiki/RL_circuit#Impulse_response )
v_r1_1ms = R1/L * exp(-tms*R1/L) * t0plus
% Q4 (see: http://en.wikipedia.org/wiki/RC_circuit#Impulse_response )
v_c_0plus = 1/(R2*C) * exp(-t0/(R2*C)) * t0plus
% Q5 (see: http://en.wikipedia.org/wiki/RC_circuit#Impulse_response )
v_c_0plus = 1/(R2*C) * exp(-tms/(R2*C)) * t0plus
Values then are:
v_R1_0min = 0
v_r1_0plus = 2000
v_r1_1ms = 270.6706
v_c_0plus = 3.0303e+003
v_c_0plus = 146.3667