Revision as of 13:03, 2 February 2011

Lecture 10, ECE301 Spring 2011, by Prof. Boutin

Wednesday February 2, 2011 (Week 4) - See Course Schedule.

Today Purdue is closed due to the current extreme weather conditions. Thus Lecture 10 will be presented online on this page. Note that we will not cover this material again in class, as time does not permit it (unless an extra day is added to the semester, which as far as I know, is out of question).

It is assumed that you already read the relevant sections of the book before reading this page. If you have questions, please feel free to ask them directly on this page, and I will try to answer them as soon as I can.

Lecture Plan

1. Remaining properties of LTI systems:
   1. Causality for LTI systems (Section 2.3.6)
   2. Stability for LTI systems (Section 2.3.7)
2. Causal CT systems described by differential equations (Sections 2.4.1 and 2.4.3)
3. Causal DT systems described by difference equations (Sections 2.4.2 and 2.4.3)

1. Remaining Properties of LTI systems

1.1 Causality for LTI systems

Previously, we have seen the definition of a "causal" sytem. If you recall, a "causal system" is a system whose response at time t only depends on the input at previous times, i.e. x(t') for t'<t.

In the case of an LTI system, one can determine whether or not it is causal by looking at its unit impulse response (h(t) for a CT system/ h[n] for a DT system). The trick is based on the following fact.

The same holds for DT systems:

Where do these facts come from? Actually, it is very easy to prove them if we remember that the output of an LTI system is the convolution between the input and the unit impulse response of the system.

More specifically, for a DT LTI system we have

$ \begin{align} y[n] &= x[n] * h[n] ,\\ & = \sum_{k=-\infty}^\infty x[k] h[n-k]. \end{align} $

Now if you look at the last equality, you see that the output at time n is a linear combination of the values of the input at all times (the x[k]'s). The coefficients for each x[k] is given by the values of h[n-k]. So, if the output at n does not depend on x[k] for k>n, this means that the coefficient of all x[k] with k>n must be zero. Thus h[n-k] must be zero for all k>n. Observe that saying "k>n" is equivalent to saying "k-n<0". So, if we think of k-n as a new variable u, what we are really saying is that h[u ] must be zero whenever u<0. But of course, u is just a place holder. We can replace it by n, and our statement becomes h[n]=0 for all n<0.

Similarly, for a CT LTI system we have

$ \begin{align} y(t) &= x(t) * h(t),\\ & = \int_{-\infty}^\infty x(t') h(t-t') dt'. \end{align} $

Again, if you look at the last equality, you see that the output at time t is a combination of the values of the input at all times (i.e. x(t') for t' between minus infinity and infinity). So all the values of x(t') (for all the different times t') influences the output, unless their coefficient h(t-t') happens to be zero. So if the system is to be causal, the coefficients h(t-t') must be zero whenever t'>t. This way, the future values of the input signal are not influencing the sum. But when we say h(t-t') must be zero whenever t'>t, this is the same as saying h(u) must be zero whenever u<0. Replacing the variable u by the more commonly used variable t, we get that h(t) must be zero whenever t<0.

You may notice that the arguments above are phrased in such a way to prove that "if a system is causal, then the unit impulse response satisfies "h(t)=0 for t<0". Actually, the converse is also true, and it would not be difficult to change the language of the above arguments slightly to prove the "if and only if". You may try to do this at home if you feel like it. It is always good to practice your logic!

So from now on, you have another way to check for the causality of a system, provided that you know that the system is LTI. So in an exam question, I could state that a system is LTI and give you its unit impulse response. I could then ask you whether or not the system is causal. To answer the question, you should simply check whether "h(t)=0 for t<0" or "h[n]=0 for n<0", depending on whether it is a CT system or a DT system.

Are there any questions????

1.2 Stability for LTI systems

Previously, we have seen the definition of a (BIBO) "stable" sytem. If you recall, a "stable system" is a system whose response is bounded for any bounded input. You may also recall that proving this property can be a bit tricky. Fortunately, for LTI systems, there is a very easy trick to determine whether the system is stable or not. The trick is based on the following fact.

So if you are given the unit impulse response of a system and want to determine if the system is stable, all you have to do is compute the complex magnitude of h(t) and integrate it over all values of t. If the result is finite, then the system is stable; if the result is infinite, then the system is unstable. That should be much easier than trying to find an upper bound for y(t) for a bounded x(t)!

The same holds for DT systems:

Again, we should ask ourselves: where do these facts come from? Again, an easy way to obtain them is to remember that the output of an LTI system is the convolution between the input and the unit impulse response of the system. I will only give the proof for the DT statement. The proof for the CT statement is very similar.

Proof. Suppose the input x[n] of a DT LTI system is bounded. Then there exists a upper bound, say b, for |x[n]|. In other words, |x[n]|<b for all n. Now because the system is assumed to be LTI, we have

$ \begin{align} \left| y[n] \right| &= \left| x[n] * h[n] \right| ,\\ & = \left| \sum_{k=-\infty}^\infty x[k] h[n-k] \right| ,\\ & \leq \sum_{k=-\infty}^\infty \left| x[k] h[n-k] \right| ,\text{ by the triangle inequality},\\ & = \sum_{k=-\infty}^\infty \left| x[k] \right| \left| h[n-k] \right| ,\\ & \leq \sum_{k=-\infty}^\infty b \left| h[n-k] \right| ,\text{ since }, |x[n]|<b, \\ &= b \sum_{k=-\infty}^\infty\left| h[n-k] \right|,\\ &= b \sum_{k'=-\infty}^\infty\left| h[k'] \right| \text{ letting }k'=n-k. \end{align} $

So as soon as the last sum $ \sum_{k'=-\infty}^\infty\left| h[k'] \right| $ is finite, then |y[n] is bounded. (Of course, this proves the "if" but not the "only if". However, I think this is enough "serious" math for today.)

Are there any questions????

2. Causal CT systems described by differential equations

Note: we will cover this subject very quickly now, but we will get back to it later once we have seen the continuous-time Fourier transform.

3. Causal DT systems described by difference equations

Note: we will cover this subject very quickly now, but we will get back to it later once we have seen the continuous-time Fourier transform.

Action items before the next lecture:

• Read Sections 3.0, 3.1, 3.2, 3.3 in the book.
• Keep working on the third homework. Try to finish it by Friday if possible.

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