 A warm welcome to the 27th session of the fourth module in signals and systems, we are now pretty much into an understanding of the stability of rational systems at least in the continuous independent variable. So, we have shown a very simple and elegant check for the stability of a continuous independent variable rational system. Let us write it down and let us explain it a little more in detail. So, theorem. So, continuous independent variable linear shift invariant or LSI rational system, remember we are talking only about rational systems. So, what is a rational system? An LSI system whose impulse response has a rational Laplace transform is stable if and only if remember it is an if and only if. There are two things that are satisfied. It has no hanging or loose differentiator or integrator components because those inherently make it unstable. And given this the region of convergence includes the imaginary axis. So, the imaginary axis is the critical contour. You look at whether the imaginary axis is in the region of convergence and you can Now, you see we talked about poles to the left and poles the right of the imaginary axis. So, to the right of the imaginary axis is the right half plane, right half s plane if you like. And to the left of the imaginary axis is the left half s plane. So, what we are saying essentially is that if there are poles on the right half plane, then the ROC must be to the left of those poles. If there are poles in the left half plane, there must be an ROC to the right of those poles. What happens if the pole is on the imaginary axis? Then obviously, the imaginary axis cannot be included in the region of convergence and that means the system must be unstable. In fact, that is correct because for example, if you have an integrator, there is a pole at s equal to 0. The moment there is a pole at s equal to 0, the system is unstable because there are specific inputs for which the output is unbounded although the input is bounded. Now, you know this poles on the imaginary axis is a bit of a special case. Let us discuss it in some depth with a little bit of discussion. So, poles on the imaginary axis, there is a pole at s equal to 0, 1 or more. It is very clear that these cause instability because these are essentially integrators. But suppose the poles are not at s equal to 0. So, for example, suppose you have poles at a complex conjugate pair of points. Now, what kind of a term do they contribute in the impulse response? Let us see. So, if you have a simple pole, then you are essentially talking about something like a divided by s minus j omega c plus a conjugate divided by s plus j omega c in the partial fraction expansion. Assuming all coefficients are real and this essentially contributes something like e raised to the power j omega c t times a plus a bar in the same complex conjugate with u t u minus t does not matter. So, essentially something like you know it is some of two complex conjugate quantities. So, two times the real part of this. So, you can write a, a can be written as a 0 e raised to the power j phi 0. And then you can take the real power, you get two times a 0 cos omega c t plus phi 0. So, essentially you get a sinusoidal component. That is interesting. You get a sinusoidal component in the impulse response. Why does a sinusoidal component make the system unstable? Now, of course, you can always argue. The modulus of the sinusoidal component is not absolutely integrable. So, that itself makes it unstable. That is of course, what you might call a direct argument. But you know a skeptic may ask you. So, what if there is a sinusoidal component in the impulse response? It is not unbounded unlike an exponential which grows where the diversions of the absolute integral is so clear. Here it is a little more hidden, you know it is more subtle. So, in fact that is true. The instability of such systems is a little more subtle. Many bounded inputs go out with unbounded with in fact with bounded outputs. Do not go out with unbounded outputs. There are specific bounded inputs which create unbounded outputs. And in this case, when you have a sinusoidal impulse response, the trouble making input is a sinusoid of the same frequency. So, that is the trouble making input. So, in fact I give you an exercise now. Take a linear shift invariant system, rational system if you please with such a pole pair. And give to it the input x of t is equal to cos omega c t which is of course bounded. Obtain the output y t. Show that it is of the form some constant let us say b times t plus c perhaps into cos omega c t plus phi 1. So, essentially here you have a so called poly x term, but the x in the poly x term is a sinusoid here. And now the polynomial is what causes an unbounded character there. So, you know you are multiplying and x the x part is neither growing nor decaying, but the polynomial part makes it grow without bound. So, you know you have a bounded input. There is a sinusoid, it is a bounded input. And because that frequency is the same, the frequency of the sinusoid which you have input and the frequency of the sinusoid present in the impulse response because they are the same, you get an unbounded output. This is called resonance. So, resonance is what makes these systems with poles on the imaginary axis unstable. So, what is the resonance for an integrator? The resonance for an integrator is an input of 0 frequency. In fact, you know when I wrote this input down, I should have said cos omega c t either to the left or the right to be precise, you know. So, that should have been appended. And here also I could have taken the ROC to be either to the left or the right and either case you would have had instability. And if the pole lies at 0, then the resonant frequency is 0. That means that if you give a 0 frequency input, what is the 0 frequency input? A 0 frequency input is a constant, a constant on one side, the left or the right. What is the constant towards the right? A unit step to the right. What is the constant to the left? A unit step to the left. Even not the unit of course, a step to the right or step to the left. So, a step to the right or a step to the left creates instability for an integrator and that we have seen before. This is the idea of resonance. So, resonance is a very subtle indication of instability. And in fact, these poles on the imaginary axis actually are present in what is called an oscillator. So, you are an oscillator when we design oscillators in electronic circuits. So, when we design oscillators even in other branches of engineering, mechanical oscillators or piezoelectric oscillators, whatever. So, oscillators actually if they are linear oscillators, they generate a sinusoid. And this generation of a sinusoid is because the impulse response has a sinusoid in it. That means, moment I disturb the system in some way, there you have an impulse response term coming out and that impulse response sinusoidal term is persistent and that is what the oscillator has. So, it is interesting. The imaginary axis needs to be clearly included in the region of convergence. If you have poles on the imaginary axis, that actually disqualifies the system from being stable. But it could make the system an oscillator. Now, of course, I leave it to you to check what happens if the poles are repeated. Suppose these poles are repeated, there is a repeated pole on the imaginary axis. What do you expect? You expect a poly-x term instead of a sinusoid, except that the x part is replaced by the sinusoid. But the poly term will not be trivial now. It will be a polynomial of degree non-zero. And there of course, it is very clear because of the polynomial term, the impulse response has a growing term, a term which cannot be absolutely integrated. And therefore, the system becomes unstable. So, we have essentially established very firmly the fact that the imaginary axis needs to be in the region of convergence. If you want the system to be stable. Now, what do you require if a system is both causal and stable? Let us put it down. When would, well, causal means the extreme right contour, real part of s tending to plus infinity is included in the region of convergence. Stable means imaginary axis is included in the region of convergence. Now, what does that mean? What kinds of regions of convergence do we have? Strictly between vertical lines passing through the poles. That means inside the region of convergence, there cannot be any poles. Now, you are saying the region of convergence must include the imaginary axis. It must also include the extreme right contour, real part of s tending to plus infinity. That means there cannot be any poles between them. That means there cannot be any poles in the right half plane. It is right that term. It is as simple as that. A linear shift invariant system, if it is rational, is stable and causal if and only if all its poles are strictly in the left half plane. So, the imaginary axis is also included and real part of s tending to plus infinity is also included. So, the entire right half plane is free of poles. Now, this is the better known version of stability. A lot of times people tell you and maybe rightly so that you do not have to allow, you cannot allow poles in the right half plane if you want the system to be stable. Now, they forget that you are saying it should be both causal and stable. And you know I can probably understand that because most of the time we are talking about causal systems. So, for causal system another way of saying it if you know the system is causal, then stability requires that all the poles be in the left half plane. We shall see more in the next session by looking at what happens now in the case of discrete independent variables. Thank you.