Let's talk now about Stability. Stability is important because it guarantees that the system will not behave unexpectedly if the input to the system is well-behaved. In signal processing, a well-behaved input is a bounded input. Namely, a signal, for which we know the maximum excursion. We want a system to behave well when the input is bounded. And the concept of bounded-input bounded-out of the stability, or BIBO stability for short, formalizes this by requiring that a system produces a bounded output when the input is bounded, so a well-behaved output for a well-behaved input. The fundamental stability theorem for their filters, states that a filter is BIBO stable if and only if its impulse response is absolutely summable. So we're going to prove this and we're going to prove both the necessary and sufficient condition. So we started with this sufficient condition and we say that our hypothesis are that our input is bounded and that they impulse response of the filter is absolutely summable. And what we want to prove is that in this case, the output will be bounded. Now the proof is very simple, we start by considering the magnitude of the output of the filter. Which is just the magnitude of the convolution sum, written here as the sum for k that goes to minus infinity to plus infinity of h[k]x[n- k]. Notice here that we use the commutative property of the convolution and we actually time reverse and delay the input rather than the impulse response. Now, the magnitude of the sum is maximized by the sum of the magnitude of its terms. And this in turn is maximized by the following expression where we have replaced the magnitude of the input by its upper bound M. And we're left therefore with M that multiplies the sum from minus infinity to plus infinity of the magnitude equals response. But we know that the equals response is absolute summable and therefore the out put would be bounded by the product ML which is finite as for the necessary part? We'll proceed as follows, the hypothesis now are that both input and output are bounded and we want to prove that in this case H of N must be absolutely summable. The proof will proceed by contradiction and it is a little bit tricky, but not difficult at all. So, assume, by contradiction, that the impulse response is not absolutely summable, therefore the sum of its magnitude is plus infinity. At which point we can build an artificial input as follows. x[n] will be equal to +1 if h[-n] is positive and -1 if h[-n] is negative. Well, clearly, x[n] built as such a bounded signal because it's between +1 and -1. However, if we try to compute the response of the system defined by h of n in 0. We have the output by 0 is the convolution between our artificial imput x of n and the inputs response computed in 0. So we write out the convolution sum explicitly, so it's a sum for k that goes from minus infinity to plus infinity of h of k times x of minus k, because remember n is equal to 0 here in this convolution. And, of course, by definition this will be the sum of the magnitude of h of k, because we built a signal like so, and because of our initial assumption, this will be plus infinity. So, we have proven that h[n] must be absolutely summable, because if it were not I could find a bounded input that would make the system explode. So, the good news is that FIR filters are always stable, because their impis response only contains a finite number of non-zero values, and therefore the sum of their absolute values will always be finite. When it comes to IIR filters on the other hand, we have to explicitly check for stability. And for the time being the only way we know how to do so is by going through the sum of the absolute values of the inputs response. And if we do the exercise with our friend the Leaky Integrator, we got the sum for n that goes from 0 to infinity. Of magnitude of lambda to the power of N multiply by a leading factor of 1 minus lambda and magnitude. This is just the geometric sum and we know that it's equal to the limit for N that goes to Infinity of 1 minus magnitude of lambda to the N plus 1 divided by 1 minus lambda. Always multiply by this factor that doesn't really bother us. This limit converges only if lambda is less than 1 in magnitude. And therefore, stability of the of the leaky integrator is guaranteed only for lambda less than 1 in magnitude. Later on in this module, we will study indirect methods for filter stability that will allow us to determine whether a filter is stable or not. Without going through the Able's response explicitly.
