 A warm welcome to the 19th session in the fourth module of signals and systems. We have now gained some maturity in the Laplace transform and calculating the inverse Laplace transform at least for the context of rational Laplace transforms and it is rational Laplace transforms that we shall encounter most frequently. So, we are now well set to deal with an example to understand what we have been learning all this while and for that example we will take the very same system with which we began our discussion of signals and systems. The very simple RC circuit resistive capacitive circuit and we would find its unit step response using the Laplace transform you know it is good to do the same thing in different ways. So, you see the connection and of course, then to extend it to other context. So, let us do the same thing to the RC circuit that we did earlier on in module 1, but now use the Laplace transform to affirm our understanding. So, here we go we take this circuit. Now, let us write down how each of these elements behaves. So, there is a resistance R and a capacitance of value C and of course, if you take the individual resistance of value R look at the voltage across it and the current in it. Then we have VRT divided by IRT is equal to R where upon VRT is equal to IRT times R and now we can take the Laplace transform on both sides. So, therefore, if we take a ratio of the Laplace transform of voltage and current across the resistance we get just R it is the same as the resistance. And this quantity this R is now also the ratio of the Laplace transforms of the voltage to the current just as it is the ratio of the voltage to the current and this is called the Laplace domain impedance of the resistance. In a similar manner we can take the capacitance and analyze it on the side. So, for the capacitance I would have again a voltage and a current associated with it. So, VCT and a current ICT flowing through it where upon ICT is equal to C times the derivative of VCT and we can take the Laplace transform on both sides giving us the Laplace transform of the current is CS. Now remember when you take a derivative you are multiplying the Laplace transform by S. So, therefore, VCS by ICS is 1 by CS and this now becomes the Laplace domain impedance of the capacitance. You see we study circuits in high school or in pre-college and largely we begin with circuits which have DC excitations, constant voltages and resistances. So, what is often easy to do and convenient to do is to use those principles of circuits carried to other elements as well. So, if I have inductances and capacitances I would look I would like to use the same techniques that I have for resistances and voltages direct voltages to analyze the circuit. And therefore, this idea of generalizing the concept of resistance to impedance Laplace domain impedance is very useful in analyzing more general circuits than just resistive. In fact, this Laplace domain impedance together with the Laplace transform of the input and the output can help you handle a much wider class of circuit situations where you have resistances, inductances, capacitances and sources that could be sinusoidal starting from a point or exponential or a step or even ramp. Let us now proceed to an example. Let us now draw the circuit again and let us write down the input-output relationship as we did earlier. So, I have input voltage V in T and the output voltage across the capacitor V out T. And I know the relation between V in T and V out T. We have done that many times before. So, there we go. We say V in T is a sum of two voltages. It is the voltage across the resistance which is R times the current in the resistance. The current in the resistance is the same as the current in the capacitance which is C times dV out T dT. And then you have V out T being added to it. So, this gives us the simple relationship between the input and the output. And now we can take the Laplace transform on both sides. To give V in S, the Laplace transform of the input is R C S times V out S plus V out S. Essentially, here this is the Laplace transform of the input and this is the Laplace transform of the output. And we have a system here. The system is the R C circuit. And we can now write down the ratio V out S by V in S which is 1 by 1 plus R C S. Let us put R C equal to the time constant tau. So, we get 1 by 1 plus tau S. This is the system function of this system. A simple example of system function. And now, of course, we can find the impulse response very easily. We can find the impulse response by inverting the Laplace transform. Now, of course, here it is very easy to invert because you know which region of convergence to choose. You see, when you have a rational Laplace transform, all that you have to remember is that the poles of the Laplace transform cannot be a part of the region of convergence. So, here in this case, we have only one pole. So, in the S plane, the pole is at minus 1 by tau on the negative real axis. And therefore, there are only two possible regions of convergence. You see, as I was saying in the Laplace transform, if you have a rational Laplace transform, the poles need to be excluded from the regions of convergence. And therefore, you can have two possible regions of convergence here. Let us mark them. Draw a vertical line through this pole. And the right shaded region is one of the regions of convergence. I will mark that here. And the green left shaded region is the other possible region of convergence. Now, this would give us the right-sided impulse response. And this would give us the other one, would give us the left-sided impulse response. But we have to choose the right-sided impulse response. This has to be chosen on account of causality. You will remember that if you want the system to be causal, its impulse response needs to be 0 for all negative t. And that is only possible if the signal is right-sided. So, we have a very clear choice to make here. Now, having chosen that, let us give a unit step input to the system. So, we have this. This is the system function here. We have made this choice. This is the ROC now. And let us give it a unit step input. That means, we are giving it U t as the input. And the Laplace transform of U t is very simple. Now, of course, here we need to worry about the region of convergence. And the region of convergence is going to be real part of S greater than 0. And for real part of S greater than 0, this becomes essentially 1 by S. So, this is the region of convergence. And this is the expression of the Laplace transform. And if we give this input, that means, if we make V in S equal to 1 by S, then V out S is going to be V in S times the system function. And that is 1 by S times the system function. And I can make a partial fraction expansion of this. To make a partial fraction expansion, it is easier for us to write it in the form that we know the standard form. So, divide numerator and denominator by tau. 1 by tau into 1 divided by S into S plus 1 by tau. And this can be expanded as something divided by S plus something divided by 1 or something divided by S plus 1 by tau. And to get this, we multiply by S and put S equal to 0, whereupon we would have 1. And to get this, we multiply by 1 or rather just S plus 1 by tau and put it equal to 0. And that gives you minus 1. And you can of course verify that this indeed the right hand side indeed sums to the left hand side. That is not very difficult to do. So, one can check. It is always a good idea to check your partial fraction decomposition. What I mean by check is here, for example, let us see. You see, I can simply aggregate this. This can be aggregated. This is S into S plus 1 by tau, S plus 1 by tau minus S. And that is indeed equal to the left hand side. So, checked. It is always a good idea to do that if you are not sure. So, now we have this partial fraction expansion and we need to recognize the regions of convergence. What regions of convergence do we have to associate? So, essentially V out S now is 1 by S minus 1 by S plus 1 by tau. And this came with the ROC, real part of S greater than 0. This came with the ROC, real part of S is greater than minus 1 by tau. And the only way in which we can reconcile these ROCs is to take their intersection. The intersection which is real part of S greater than 0 must be in the ROC. And we can use that to invert this. And when we use that to invert, we get V out t, very simple. Is U t, it is the inverse of 1 by S minus e raised to the power minus t by tau times U t. Or 1 minus e raised to the power minus t by tau times U t, which is indeed the step response that we had got much earlier. Simple and elegant. Laplace transform gives you a simple neat algebraic way using algebra first, just plain addition, multiplication and finally, the experience of inverting a rational Laplace transform. Now, it would be interesting to see what happens if you gave the same circuit not a unit step, but a unit ramp. So, let me explain what I mean by that. Suppose we gave this a unit ramp input now, essentially an input that looks like this ever increasing. You will notice that the unit ramp is the running integral of the unit step. So, it is equal to the integral from minus infinity to t U lambda d lambda. And that amounts to multiplying the Laplace transform by 1 by S. So, its Laplace transform is 1 by S into 1 by S, which is 1 by S squared. And now, we could ask what happens at the output. Now, you can solve this by two ways. One approach is to make a partial fraction expansion as usual. The second approach is to invoke the property of the linear shift invariant system. This operation of taking a running integral is itself a linear shift invariant system. So, if you give it an input x of t and produce the output y of t. And if y of t is the integral from minus infinity to t x lambda d lambda, the system is linear shift invariant. And the system function is 1 by S. Now, what you have done is to take the input unit step. You have given it to a running integrator. Let us call it S i. Let us call the system S i. You have given it to S i. And you have given the output of this to the R C circuit. That is what you have done. Now, this is a cascade of two LSI systems. We can interchange the order. And therefore, you could put S i after the R C circuit, which means you could have the situation as follows. You could give a unit step input, find the output to the R C circuit, which we already know. And this could be given to the running integrator. Now, the exercise for you to do, with which we shall conclude this session is verify that both approaches give the same answer. And with that, you would enhance and fortify your understanding of the Laplace transform. We shall see more in the next session. Thank you.