Analysis of Second Order Circuits

in this video we're going to examine the solution to a second-order system I find that when you're first starting out solving these it's very useful to create a little cheat sheet to kind of help you keep all of the information that you need about second-order differential equations in one convenient spot this is the cheat sheet that I've created and you can pause the video and take a look at it if you use it to solve a dozen or more problems you'll find that a lot of this stuff you kind of end up memorizing just from using it so frequently we'll come back to it throughout the video okay so here we have a circuit and we know that it's a second-order system we know that it's a second-order circuit because it has two energy storage devices okay and we're further tipped off to this being a circuit that's going to require some sort of differential equation because of the quantum event when the switch is flipped open here so our ultimate solution is going to be to find an equation that describes the current I of T at all points greater than T equals zero when you're solving these circuits this is the algorithm that I use first I'll determine the initial conditions the initial conditions are the current through the inductor at T equals zero the voltage across the capacitor at T equals zero the variable that you're solving for in this case I of T at T equals zero plus and the the terminal value of that parameter that you're looking for so I at infinity notice that I don't care about whether I'm saying I the current through the inductor at zero minus or zero plus or the voltage across the capacitor at zero minus or zero plus because the energy in a capacitor and inductor cannot change instantaneously so those values will be the same before the quantum event and just after the quantum event ii will write an equation to describe the circuit if we have a second-order circuit like this one we are guaranteed to have a second-order differential equation we're going to write that differential equation in standard form and use that to extract the characteristic equation the characteristic equation will give us all of these variables we'll use these variables to solve the rest of the circuit and the case is going to give us the general equation in my cheat sheet I have a general equation written for each of the cases here it is here here it is here so on and so forth we're going to use that general equation and the derivative of that general equation to solve the circuit and it works like this the general we'll be able to solve the general equation with one of those initial conditions that's on the energy one of the energy storage device then we'll be able to use the derivative of the general solution and the initial condition on the other energy storage device to come up with some way to generate a solution to that finally that's going to leave us with two equations two unknowns and we'll solve that system of linear equations using any method that we like so we begin by determining the initial conditions what are those initial conditions well we need to find I at the moment after the quantum event the current through the inductor the voltage across the capacitor and the terminal condition for the parameter that we're looking for I at infinity in this case let's go ahead and also recognize right off the bat that I add I naught or I of T is the same as I've L so once we found one we found the other in this particular circuit that's going to make things a little bit simplified for us okay so we have our circuit here and we want to imagine that it was invented you know a million years ago and it's just been sitting around here existing and right before T equals zero it's been doing its thing for a long time and that means that the inductor is treated like a short and the capacitor is treated like an open so we can just redraw our circuit like this as a short and we can imagine the capacitor as being open so because the capacitor is open there's no current flowing through this 5 ohm resistor here no current flowing through the capacitor and so at the point right before the switch is flipped the only thing that we have is this single KVL loop to look at okay and so we can write a little KVL equation and determine what I L is so it's going to be starting from right here we get minus 20 combine the two resistors and use the passive sign convention plus 10 IL minus 10 equals zero and that will tell you that IL equals 3 and that is at the moment that the before the switch is flipped now we know that the current through an inductor cannot change instantly so the current through the inductor before just before and just after stays the same and that's three now we know that because the current through the inductor at I add 0 plus is 3 amps that the current I of T at 0 plus is also 3 amps ok don't get confused by this that works nicely in this circuit it won't always work like that remember that I of T could change instantly the voltage across a resistor and the voltage and the current through a resistor are bound by Ohm's law and therefore they can change instantaneously this is just a happy coincidence ok the next thing that we want to do is determine what is the voltage that accumulates on the capacitor and that's we know that there's a voltage accumulated on the capacitor because that's what causes it to act like an open and we can simply solve that by assigning a V of C like this and doing a KVL equation notice that there is no current through this 5 ohm resistor so we're really just looking at V of C equaling the voltage drop across the 10 volt and the 5 ohm resistor so that's 10 plus 5i L plus VC equals zero now be careful when you do this my il right here is going in this direction so I've actually have not followed the passivity rule here this should be a minus okay and so and we know il is 3 so this is 10 minus 15 and so V of C equals 5 right before and right after the quantum event when the switch is flipped okay and just to keep that in my memory I want to make sure that I follow that polarity all the way through my problem so that I get the proper signs on my later equations okay so now imagine that the switch is flipped open when the switch is flipped open at T now we want to know what happens an infinity of time later well we've broken this entire connection to the circuit so it doesn't do anything and we only have this loop so the capacitor which had a bunch of charge on it starts to discharge and this battery causes the capacitor to charge incidentally if you look at it you'll see that the charge that on the capacitor will actually flip its sign from the way that we've assigned it here and what ends up happening is that the capacitor gets fully charged so that no current flows through it no current flows through it meaning that I of T at infinity is actually zero volts I'm sorry zero amps okay so those are our initial conditions the next thing that we want to do is determine what our characteristic equation is to do this we're going to write an equation to describe the circuit after the switch has been flipped there's going to be a second order differential equation that we will write in standard form and you can look at the circuit and see that when the switch is flipped we simply have an RC L in series with a voltage source and you can look that up in the book if you want however you'll be better served to practice writing these equations by looking at the circuit because we have in our book a quick reference for series and parallel RCL circuits but not all second-order circuits are going to be series in parallel so let's go through the process of writing the equation for this particular circuit and this one's easy the we have a single loop so it makes sense to do a KVL equation sometimes they'll be more complicated we'll begin by examining KVL in that loop right there and actually what I'll do is I'm going to draw the loop the other way just to make it a little bit easier on myself because then I of T my kurt my KVL mesh loop is going in the same direction as my I of T so we'll start at the battery it's going to be minus 10 and now we're going to go through the capacitor the capacitor remember the equation is i equals c dv/dt but we want to write this in terms of the voltage so that's going to be we're entering the negative terminal minus 1 over c integral of I DT now pay attention here it's easy to get confused by this this equate this part right here is assuming some current I we need to obey the passivity rule and according to the polarity that we've assigned we do not obey the passivity rule if we're going to use I of T as our current or the mesh current so this is going to be minus I okay and the minus and the minus will make this a plus okay and then I'll combine the two resistors and I'm just going to leave the I'm going to use symbolic values for now so this is going to be R times I and then finally plus L di DT and all of that equals zero so now I want to write this in standard form we don't want to deal with the integral so we're going to take the derivative of everything so we get and I'm going to up rearrange this as I go so here's my I'm going to take the derivative of this and that's going to be my second order term and that's that's going to be d i d squared I DT squared and I'm going to divide by L to put it in standard form I'm going to divide everything through by L as I go this is going to be my second order term so that's going to be R over L di DT and then this is going to be 1 over CL I and then the derivative of 10 is 0 so that's just 0 okay so from this we can extract the characteristic equation that says s squared plus R over L so that's a plus there plus plus 1 over CL equals 0 keep in mind I on the cheat sheet here that our characteristic equation is given by this so we can solve now for Sigma and Omega Sigma equals R over 2 L and if we plug in our actual values for that you'll see that it equals 5 Omega naught equals 1 over the square root of CL if you look back at that characteristic equation you'll see why that becomes 1 over the square root of CL because the term this term is equal to Omega squared okay and so that also happens to equal 5 right after I solve for Omega and Sigma I write the inequality in this case Sigma equals Omega naught and this means that the circuit is critically damped okay what does that mean well that means that it's going to be case three where Sigma equals Omega not critically damped if we were to look at a graph of the particular characteristic that we're looking for it would look like one of these three different lines we know our s values are given and how did we get these s values we use the original quadratic equation to solve for them this is a simplified version of that that you can use to make things a little bit quicker but notice that if Sigma and Omega are not an Omega not are both the same that this whole term just goes to zero and so as just equals minus Sigma and finally we have our characteristic equation right here so I'm sorry our general the general form of our equation there's notice that there are two unknowns we know Sigma we presumably no exit infinity and we just need to solve for B 1 and B 2 so let's write down our characteristic equation and move on to the next step I'm sorry our general equation and move on to the next step so our general equation is going to be b1 e to the minus Sigma T plus b2 T e to the minus Sigma T plus X at infinity okay and every part of our circuit will follow this general equation at this point we're looking for the particular solution of I of T so we want to use this general equation but we want to write it in terms of what it is that we're looking for here so I'm just rewriting this where X is I and I'm also going to place in my Sigma I of T equals B 1 e to the minus 5t plus B 2 e to the minus 5t and exit infinity is I at infinity which we already solved to be 0 so that is our equation now what we want to do is using one of the energy storage using one of the initial conditions of the energy storage device one of the energy storage devices we want to solve this equation so we know from our initial conditions we were able to use I've al the current through the inductor to tell us what I of T at 0 plus was equal to so what we're going to do now is use that to fill in the blanks here this is I this says that I of 0 equals 3 amps and that's equal to B 1 e to the minus 5 times 0 plus B 2 times 0 times e to the minus 5 times 0 and so that tells us that B 1 equals 3 now the next thing that we need to do is we need to take the derivative of our general solution and look for a way to relate that to what we know about the circuit so I'm going to write it right here di DT so I'm just taking this equation right here and rewriting and writing its derivative di DT equals minus 5 B 1 e to the minus 5t plus B 2 e to the minus 5t minus b2 let's do it like this minus 5b 2t e to the minus 5t so let's see what we did before was we use the current we used what we knew about the initial conditions of the current through the inductor to inform our first equation so we're going to use the voltage on the capacitor to help us figure out what's going on in the second equation and what we can do is redraw or re-examine this circuit knowing that this voltage right here is equal to 5 volts okay how is that going to help me well it turns out that when I look at this acquit right a mesh equation here and we'll put the inductor back in okay because we're looking at an equation for all time after T equals zero so we can't treat the inductor and the capacitor as shorts at the instant the switch is flipped they act like an inductor and a capacitor so I can write a KVL equation for that loop and I'll do that right here we get kind of hard to do without looking at it right we get minus 10 minus 5 and then we get 10 IL following the passivity rule and we have to follow the passivity rule with the inductor as well and that's l di/dt and all of that equals zero so look at that we now have an equation you plugging in what we knew about the initial conditions on the capacitor we were able to write an equation that gave us something with the ID T in it okay and if you if you solve this you'll see that di DT equals minus fifteen okay so now we can look at and this is at the moment T equals zero so recall then I should have been more explicit here this is the function of T so di DT at 0 equals minus 15 which is equal to we know this is we know B 1 is 3 right so this is just minus 15 and B 1 e to the minus 5t I'm sorry e to the minus 5t just becomes times 1 ok so that's times 1 B 2 e to the minus 5t is simply B 2 and -5 b2t all of this is 0 because of this tea right here and so what you end up with is this equation right here and when you bring the 15 over you find out that be - in fact equals 0 okay so in summation now we'll bring all that together and write our final equation the final equation is 3 sorry I of T equals 3 e to the minus 5t amps for T greater than 0 and that's our final solution ok good luck

Channel: Matthew Araujo

Views: 165,356

Rating: 4.8589983 out of 5

Keywords: Electrical Engineering, Circuit analysis, critically damped, second order circuits, differential equations, circuits, electronics

Id: DZDxhq4ITu0

Length: 27min 34sec (1654 seconds)

Published: Thu Mar 28 2013