Complex Poles: Extra Notes

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In lecture this week, we talked at length about poles, numbers that we can use to determine features of the unit sample response of a system, but we did not get to talk about complex-valued poles. This page will present a small bit of review of what we learned in lecture about poles, before discussing what happens when a system's poles are complex-valued.

1) Single-Pole System

When introducing poles, we started by noticing that the simple system below has a unit sample response y[n] = p^n:

We were able to use this fact to reason about the behavior of the system as the parameter p changes (note that this system's behavior is completely characterized by the single number p):

For values of p that are greater than 1, the unit sample response will diverge monotonically (y[n]\to\infty as n\to\infty)
For values of p that are less than 1 but positive, the unit sample response will converge to 0 monotonically (y[n]\to0 as n\to\infty)
For values of p that are between -1 and 0, the unit sample response will converge to 0, flipping sign on each timestep (|y[n]|\to0 as n\to\infty)
For values of p that are less the -1, the unit sample response will diverge, flipping sign on each timestep (|y[n]|\to\infty as n\to\infty)

2) More Complicated System

The next important step came from realizing that arbitrary LTI systems can be factored into a form representable as a sum of scaled versions of the single-pole system above.

The example system we looked at in lecture was represented by the following block diagram:

In system functional form, we can represent this system by:

Y = X + 1.6\mathcal{R}Y - 0.63 \mathcal{R}^2 Y

which can be factored into the following form:

(1 - 1.6\mathcal{R} + 0.63\mathcal{R}^2)Y = X

(1-0.7\mathcal{R})(1-0.9\mathcal{R})Y = X

\frac{Y}{X} = \frac{1}{(1-0.9\mathcal{R})(1-0.7\mathcal{R})}

Futhermore, we can do a partial fraction decomposition to express this new system as a sum, rather than a cascade, of these simpler systems:

\frac{Y}{X} = \frac{1}{(1-0.9\mathcal{R})(1-0.7\mathcal{R})} = \frac{4.5}{1-0.9\mathcal{R}} - \frac{3.5}{1-0.7\mathcal{R}}

This form can be represented by the following block diagram:

Notice that we have here two systems of the form we discussed earlier (hooray!). So we have y_1[n] = 0.9^n and y_2[n] = 0.7^n, and combining these together, we have:

y[n] = 4.5(0.9)^n - 3.5(0.7)^n

This is really nice, because it allows us to reason about the behavior of this big complicated system by thinking about its poles.

3) General Method for Finding Poles

In fact, we can use this idea to reason about the behavior of arbitrary LTI systems. If we find the poles p_0, p_1, \ldots of a system, then the unit sample response of the system will go like a sum of scaled geometric sequences, the bases of which are the poles of the system:

y[n] \sim \sum_i c_ip_i^n

So if we can find the poles of a system, we can use what we know about geometric sequences to reason about that system's unit sample response.

Below, we will generalize the procedure from above and see how to find the poles of an arbitrary system.

The poles can be identified by factoring the denominator of the system functional into the form we would expect from the simple system we first looked at:

\frac{Y}{X} = \frac{b_0 + b_1\mathcal{R} + b_2\mathcal{R}^2 + \ldots}{1+a_1\mathcal{R} + a_2\mathcal{R}^2 + \ldots}

\frac{Y}{X} = \frac{b_0 + b_1\mathcal{R} + b_2\mathcal{R}^2 + \ldots}{(1-p_0\mathcal{R})(1-p_1\mathcal{R})(1-p_2\mathcal{R})\ldots}

The poles are the p_i values, and one geometric mode p_i^n arises from each pole.

Unfortunately, factoring things into that form is actually quite difficult. One thing we do know, however, is how to find the roots of polynomials. So we will use the "trick" of constructing a new Polynomial whose roots are the p_i values in the form above.

We will do this by substituting throughout {\cal R} = \frac{1}{z}. For example, consider our system from earlier:

\frac{Y}{X} = \frac{1}{1-1.6\mathcal{R}+0.63\mathcal{R}^2} = \frac{1}{1-\frac{1.6}{z} + \frac{0.63}{z^2}}

And multiplying through to form a rational polynomial in z:

\frac{Y}{X} = \frac{z^2}{z^2 - 1.6z + 0.63}

Now, the poles p_i are the roots of the denominator polynomial expressed in z:

p_0, p_1 = 0.7, 0.9

And we were able to get here without doing the partial fraction decomposition by hand!

This method is completely general, and you will get some practice with it in the design lab, and also in this week's exercises.

4) Long-term Behavior

When analyzing systems' poles, we are interested in long-term behavior (not specific samples; there are arguably better toold for that!).

When we say long-term, we mean: as n\rightarrow \infty, how does y[n] behave?

We have seen that a system's unit sample response can be written in the form:

y[n] \sim \sum_{k} c_k p_k^n

In the "large-n" case, all poles but the one with the largest magnitude die away, and so looking at the dominant pole alone tells us about the behavior of the system in that case.

5) Complex Poles

As you may already be aware, when we are finding the roots of a second-order (or higher) polynomial, as we do to find the poles, it is possible that one or more of those roots is complex-valued (as opposed to purely real).

It's worth noting that we'll use the name j to represent \sqrt{-1} throughout 6.01. This is common practice in engineering (whereas in other disciplines, i is more common). j is the better name for our purposes, because we will use i to denote electrical current when we start to talk about circuits, and re-using those names can be confusing.

So the question is, what happens to our analysis if some (or all) of the poles have non-zero imaginary parts?

The good news is that the unit sample response still goes like the poles raised to the power n! But in order to make sense of this, we will need to understand what happens when complex numbers are raised to integer powers.

Let's start by considering a system with poles at a_0 \pm b_0j, with b_0 \neq 0:

Remember that what we are interested in (since this number represents a pole) is what happens as we raise this value to integer powers of n.

In the form we have this above, it is really a pain to think about this:

(a_0+b_0j)^n

...I don't want to do that algebra. We can help things a little bit by doing some trig. Firstly, notice that this number not only has its real and imaginary parts, but it also has a magnitude and an angle. Let's call these r and \theta for now:

A little bit of trig shows us how we can relate these values:

\begin{aligned} a_0 + b_0j &= r\cos\theta + (r\sin\theta)j\\ &= r(\cos\theta + j\sin\theta) \end{aligned}

Check Yourself 1:

How can you determine r and \theta from a_0 and b_0?

Even when we have things in this form, thinking about raising to higher powers looks painful:

\left(r(\cos\theta + j\sin\theta)\right)^n

Luckily, we can use Euler's formula to make one final leap from here:

\begin{aligned} a_0 + b_0j &= r\cos\theta + (r\sin\theta)j\\ &= r(\cos\theta + j\sin\theta)\\ &= re^{j\theta} \end{aligned}

Note

We will call the original form the rectangular form, and this new form the polar form.

Once we have our number in polar form, it becomes easier to think about what happens when we raise this value to a higher power (because exponentiation distributes over multiplication). We have:

\left(re^{j\theta}\right)^n = \left(r^n\right)\left(e^{jn\theta}\right)

Importantly, note that r^ne^{jn\theta} is just another complex number, whose magnitude is r^n and whose angle is n\theta.

If it is helpful, we can also think about converting this back to our earlier form:

r^ne^{jn\theta} = r^n\left(\cos(n\theta) + j\sin(n\theta)\right)

For now, though, I will stick with the form of r^ne^{jn\theta}.

If we raise this value to power 2, we have r^2e^{j2\theta}, which is a number with magnitude r^2 and angle 2\theta (here plotted next to our pole's original value):

Raising to the third power gives us r^3e^{j3\theta}:

Notice a pattern:

the angle of p^n is changing linearly (every time n increases by 1, we add \theta to the angle)
the magnitude of p^n is changing geometrically (every time n increases by1, we multiply the previous magnitude by r)

So as we continue to increase n, p^n will spiral through the plane.

Consider the following two example videos showing this process (p^n for n=0 through n=99) for two different bases p.

In this first, we consider p = 0.98e^{0.2j}:

And then $p = 1.01e^{0.2j}$:

Notice that in these examples, the _angle_ of the base is the same, and so the rate at which $p^n$ moves around the circle is the same between them (as it depends only on the angle of $p$). What is different is the _magnitude_ of the poles, which determines whether the response decays to 0 or explodes to infinity.

Note

If you don't have a lot of experience working with complex numbers and the explanation above doesn't suffice, there is a really nice series of videos on Youtube called "Imaginary Numbers are Real" that talks through complex numbers more completely from the basics.

The whole series is quite nice, but is a bit long. The particularly relevant pieces for 6.01 are parts 5, 6, and 7.

You may also find this optional exercise helpful.

6) Real Response from Complex Poles?

We are almost home. The remaining piece we have to deal with is the fact that complex modes like those above tend to result in complex numbers on their own (as they spiral through the plane). But ultimately, the unit sample response of a system we want to characterize is purely real. So how does it come to be that these complex modes combine to form a purely-real response?

It is important to remember that we will not be considering each pole in isolation when thinking about the unit sample response of the system. Remember that the unit sample response will be a sum of scaled geometric sequences, one for each pole:

y[n] \sim \sum_i c_ip_i^n

It also happens (we won't prove this here) that complex poles always come in complex conjugate pairs. So if a_0+b_0j is a pole, then a_0-b_0j must also be a pole (or, in polar form, if re^{j\theta} is a pole, then re^{-j\theta} must also be a pole).

It will be useful to think about what happens when these values sum together, since that is ultimately what will happen in the expression for the unit sample response above.

Let's look at a few examples. In the videos below, we show p_0^n in blue, p_1^n in red, and their sum (scaled by a constant) in gold:

The first example is when p_0, p_1 = 0.98e^{\pm 0.2j}:

And the second, when $p_0, p_1 = 1.01e^{\pm 0.2j}$:

This is really neat! As one mode spirals in one direction, the other mode spirals in the other direction; so when we sum them, we are _exactly canceling out the imaginary parts_, and we are left with a purely-real result! Notice also that this result is _oscillatory_ (rather than monotonic, as we saw in the case of purely real-valued poles).

We saw earlier that the magnitude of the dominant pole tells us whether the system's unit sample response will diverge or converge, and how quickly. The angle of the pole tells us something about the oscillation of the system. If we want to measure the period of the unit sample response of a system, we want to count the number of timesteps it takes for the "gold" arrow in the examples above to move from one "peak" value to the next "peak" value; notice, graphically, that this is the same thing as the number of timesteps it takes for the blue (or red) arrow to make one full revolution. We can thus use the angle of the pole to predict the system's period of oscillation.

Check Yourself 2:

If a system has complex poles with magnitude r and angle \theta in the complex plane, what is an expression for the period of oscillation of that system's unit sample response?

Note also that in the examples above, the angle is the same, and so the period of oscillation is the same between them (as it depends only on the angle of the pole(s) in the complex plane). What is different between the two examples is the magnitude of the poles, which governs the system's convergence/divergence.

Check Yourself 3:

If we were to make a similar video for p_0, p_1 = 0.5e^{\pm 0.2j}, how would it differ from those above? What about p_0, p_1 = 0.98e^{\pm 0.4j}?

7) Takeaways

A few quick notes on the important points to take away from the above:

Poles are useful tools that let us reason about the long-term behavior of a system without simulating sample-by-sample.
We learned a method for solving for the poles that is worth practicing and committing to memory.
One geometric mode arises in the unit sample response for each pole.
In the "long term" (as n\to\infty), most of these geometric modes are relatively small; the one that "sticks around" is the one that arises from the dominant pole (the pole with the largest magnitude).
Poles with nonzero angle lead to oscillatory unit sample responses (with a period that can be determined based on the angle of the pole in the complex plane).
Regardless of whether the dominant pole is real or complex, its magnitude is what determines the convergence/divergence of the system (magnitude < 1 and the system's unit sample response will converge to 0; magnitude > 1 and it will diverge).