Sunday, May 29, 2016

Circuits and Electronics - Damped Second Order Systems (Lecture 18)

Video and lecture notes:- edX    MITx: 6.002.2x Circuits and Electronics 3:  week 1

 An inductor here, a resistor and a capacitor is connected in series.The voltage across the capacitor is V(t), and the current in the series circuit is i(t).

With node voltage analysis at nodes V and V_A, we will be able to come up with some equations.

We will need to get rid of V_A. By equating , differentiating and arranging the equations, we will be able to get the second order equation.

The node method always works but there is another cute little way of getting the equation. And that is the KVL method.

To solve the differential equations, there are  three steps.The first step is to find the particular solution, vP. We will call this solution vP. The second step is to find the homogeneous solution.
And, as you recall from our solution of the LC circuit, the solution to the homogeneous equation will
result in vH. And we will do this in a four-step process. So step two of the solution involves finding the homogeneous solution, which itself has four steps. The third step of the solution is to find the total solution as the sum of the particular and homogeneous solutions.

Input will step from 0 to capital V_I at time t equal to 0.  And this input can be expressed as V_I, u(t). The u(t) is rhe  unit step. Let's pick the following initial conditions. Let's go ahead and solve it for the ZSR,  OK, so initial conditions will be these. Since it's a ZSR, v (0) would be 0. And similarly, i(t) will also be equal to 0.

If we try try v_p equals v_i, we will find that  v_p equals v_i is a particular solution.

v_H is the solution to the homogeneous equation which is the original equation with the drive set to zero.

And as you recall, I've been using a four-step method to solve the homogeneous equation.
OK, so let's write down what those four steps are. So there's a step 2A, there's a step 2B, there's a step 2C, and finally a step 2D.

(A) Assume a solution of the form vH is Ae raised to st, where A and s are unknowns.
(B) Form the characteristic equation
(C) Solve   the characteristic equation for the roots s
(D) Form v_H by summing up the two terms. A1 e raised to s1 t plus A2 e raised to s2 t.

For step (A) take v_H = Ae raised to st, and  substitute that into the differential equation.
For step (B), after some simplification we are left with a  characteristic equation and .compare the  characteristic equation with the canonic form
For step (C) we would need to find the roots s of the   characteristic equation.
For step (D) we would need to write the general solution to the  homogeneous equation which is A1 e raised to s1 t plus A2 e raised to s2 t.

With that we can write out the total solution and solve for the copnstants using initial conditions v(0)=0 and i(0)=0. Take note that i(t)= C dv/dt.

At first glance, the solutions do not seem  to be what we want so we need to take a look at the total solution again.

If we look at square root of alpha squared minus omega naught squared.  When alpha is greater than omega naught, the stuff inside the square root sign is positive, so I get a real value as a square root.
However, if that quantity, alpha squared minus omega naught squared, or alpha versus omega is such that it is negative-- in other words, if omega is bigger than alpha-- then I get a square root of a negative number so I get the imaginary value.

Let's look at the  three cases

The first case, as we discussed, is the pretty straightforward case when alpha is greater than omega naught. This case is called the overdamped case. Damping relates to R, and when R is very large, there is more damping, for whatever that means for now.

The second case is when alpha is less than omega naught, in this case the quantity under the square root sign becomes negative, and very interesting things happen. We call this the underdamped case.

And then, of course, the obvious third case, alpha equals omega naught. This is called the critically damped case.

For overdamping, when alpha is greater than omega naught, I can rewrite the total solution v of t. When the input and output waveforms are plotted, we can see for the ouput, there are no sinusoids. This case occurs when the resistance is large.

For underdamping, when alpha is less  than omega naught, we need to put in j as some the part of total solution would seem to be imaginary. Let's pick  omega d. as omega naught squared minus alpha squared.Omega d is called the damped natural frequency of the circuit. Omega naught is called the undamped natural frequency. As v(t) is real, the RHS must be real.  Now, notice that e to the j omega t and e to the minus j omega t I know are here, but the good news is that they are complex conjugates of each other.  OK, so there's a likelihood that, as I add things up, the imaginary parts will cancel out giving me a real number. Now it turns out that, given that these two are complex conjugates of each other, it turns out the only way that I can get a real number out of it is if A1 and A2 are complex
conjugates of each other. So A1 and A2 must be complex conjugates of each other as well. I see a VI there, plus notice that I have these cosines and sines floating around, and if A1 and A2 are complex
conjugates of each other, the final expression will be some constant K1 times e raised to minus alpha t, times cosine of omega dt plus some other constant K2 times e raised to minus alpha t times sine omega dt. Next, we would need to solve for the initial conditions where v(0)=0 and i(0)=0. When we check the expression for R=0, we will find that it is correct for the LC circuit.  Writing  v( t) based on the scaled sum of sines formula, we will get an equation for which we can plot the graphs. We will need to see what the graph of  v(t) against t looks like for a step input. When plotted we will notice that the sinusoid is decaying.

In the critically damped case when alpha is equal to  omega naught, we will; see that the waveform ius somewhere between that of the overdamped and underdamped case. ( a slight blip will be seen)

For the inverter pair RLC, we will see that the case is actually an underdamped one when a 50 ohm resistor is used.

The characteristic equation tells you what the waveform looks like. When connected to the canonic form of the characteristic equation. The quality factor  Q is approximately the number of cycles of ringing.

Another way of observing how the waveform looks like is by intuition. When i(0) is negative, the capacitor is discharging so the capacitor voltage is falling down.

It is possible to find V_L by finding out the rest of the voltages for the other components first or by KVL.

For the RLC circuit, the same method more or less applies but current is dealt with instead of voltage.

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