Pole

The concept of a Pole of a Complex Function comes from Complex Analysis: Pole.

In Digital Signal Processing, when we say Pole, we are typically referrring to a complex analysis pole of the Transfer Function $H(z)$, i.e. a root $p_0\in \mathbb{C}$ where the denominator of $H(z)$ is zero.

Each pole corresponds to a fundamental mode of the Impulse Response, with its location in the complex plane determining that mode’s growth, decay, and oscillation. Concepts from Differential Equations involving complex exponential solutions may be relevant here. Note also that poles only exist for Feedback loops. So essentially, the poles of a system correspond to the gains of the equivalent parallel feedback loops that comprise it; see Equivalence to Parallel First-Order System.

Poles literally are Eigenvalues. Poles are the $z$ for which the Transfer Function experiences a singularity. Eigenvalues are the $z$ for which their Matrix becomes Singular in $(zI-A)$. Poles are when $H$ blows up. Eigenvalues are when $(zI-A)$ blows up. They come from different mathematical angles, but are precisely the same concept.

Example Systems

Canonical Single-Pole System

The pole is at $z=p_0$, directly visible as the feedback gain. The denominator of the Transfer Function $1-p_0\mathcal{R}$ has a single root at $p_0$.

Transfer Function

$$(1-p_0\mathcal{R})Y=X⟹H=\frac{Y}{X}=\frac{1}{1-p_0\mathcal{R}}$$

Impulse Response

$$y[n]=\{\begin{array}{ll}{p}_0^n& n\ge 0\\ 0& \text{otherwise}\end{array}$$

The Impulse Response for this system is geometric, where the pole is just the gain $p_0$.

Stability of the Impulse Response

The Stability of the system when disturbed by $\delta [n]$ is determined by the gain $p_0$

• $|p_0|<1$: converges to zero
• $|p_0|=1$: constant magnitude
• $|p_0|>1$: diverges

Monotonicity of the Impulse Response

The Monotonicity of the system when disturbed by $\delta [n]$ is determined by the sign of $p_0$

• $p_0\ge 0$: monotonic
• $p_0<0$: alternating sign each sample

Second Order System

$$\begin{array}{rl}Y& =X+1.6\mathcal{R}Y-0.63{\mathcal{R}}^{2}Y\\ Y-1.6\mathcal{R}Y+0.63{\mathcal{R}}^{2}Y& =X\\ (1-1.6\mathcal{R}+0.63{\mathcal{R}}^2)Y& =X\\ H=\frac{Y}{X}& =\frac{1}{1-1.6\mathcal{R}+0.63{\mathcal{R}}^2}\end{array}$$

The poles are the roots of $1-1.6\mathcal{R}+0.63{\mathcal{R}}^2$, which are $p_0=0.7$ and $p_1=0.9$. Both are real, positive, and inside the unit circle, so the system is stable with monotonic decay.

Notice how the Transfer Function contains the reciprocal of a second order polynomial. Because it is a reciprocal, that tells us that it is a Feedback system. Because it is a second order polynomial, that tells us that the system is second order.

Equivalence to Cascaded First-Order Systems

We can algebraically manipulate and factor the second-order system’s operator form into a Cascaded System.

$$\begin{array}{rl}(1-1.6\mathcal{R}+0.63{\mathcal{R}}^2)Y& =X\\ (1-0.7\mathcal{R})(1-0.9\mathcal{R})Y& =X\\ H=\frac{Y}{X}& =\frac{1}{(1-0.7\mathcal{R})(1-0.9\mathcal{R})}\end{array}$$

The first equivalent first order system:

$$(1-0.7\mathcal{R})Y_2=X,(1-0.9\mathcal{R})Y=Y_2$$

The second equivalent first order system. This is because of the commutative property.

$$(1-0.9\mathcal{R})Y_1=X,(1-0.7\mathcal{R})Y=Y_1$$

The factored form $(1-0.7\mathcal{R})(1-0.9\mathcal{R})$ makes the poles explicit. Each factor contributes one pole, and each subsystem’s feedback gain is its pole.

Based on the Transfer Function, we can tell that the system is either combination of cascading these feedback, first order systems.

Equivalence to Feed-Forward Infinite-Order System

No denominator polynomial, so formally no poles. The poles are encoded in the convergence of the geometric series: the base of each series ($0.7$, $0.9$) is the pole, and the series only converges when the pole is inside the unit circle.

$$\begin{array}{rl}H& =\frac{1}{(1-0.7\mathcal{R})(1-0.9\mathcal{R})}\\ & =\frac{1}{1-0.7\mathcal{R}}\cdot \frac{1}{1-0.9\mathcal{R}}\\ & =\left(\sum _{n=0}^{\infty }0.7^n{\mathcal{R}}^n\right)\cdot \left(\sum _{n=0}^{\infty }0.9^n{\mathcal{R}}^n\right)\\ & =(1+0.7\mathcal{R}+0.7^2{\mathcal{R}}^2+\cdots )\cdot (1+0.9\mathcal{R}+0.9^2{\mathcal{R}}^2+\cdots )\\ & =(1+0.9\mathcal{R}+0.9^2{\mathcal{R}}^2+\cdots )\cdot (1+0.7\mathcal{R}+0.7^2{\mathcal{R}}^2+\cdots )\end{array}$$

For the Transfer Function $H=(1+0.9\mathcal{R}+0.9^2{\mathcal{R}}^2+\cdots )\cdot (1+0.7\mathcal{R}+0.7^2{\mathcal{R}}^2+\cdots )$, we can tell that the signal first goes through an infinite-feed forward system with gains of the exponentials of 0.7, before going through the same thing but with 0.9.

For the Transfer Function $H=(1+0.7\mathcal{R}+0.7^2{\mathcal{R}}^2+\cdots )\cdot (1+0.9\mathcal{R}+0.9^2{\mathcal{R}}^2+\cdots )$, we can tell that the signal first goes through an infinite-feed forward system with gains of the exponentials of 0.9, before going through the same thing but with 0.7.

In practice, if you wanted to compute an IIR system quickly on parallel hardware (e.g. CUDA), you could truncate this feed-forward representation at $N$ terms to approximate it. Because the Feed-Forward representation has no data dependencies between samples, it is very parallelizable. The pole magnitude determines how many terms you need: poles close to the unit circle require large $N$, while poles deep inside converge with small $N$. If a pole is outside the unit circle, the series diverges and this approach does not work.

Equivalence to Parallel First-Order System

We can also algebraicly manipulate the starting Transfer Function to be a sum of two first order systems using Partial Fraction Decomposition. Whereas a product of systems is cascading, a sum of systems means that they are being summed together after running in parallel.

$$\begin{array}{rl}H& =\frac{1}{(1-0.7\mathcal{R})(1-0.9\mathcal{R})}\\ & =\frac{A}{1-0.9\mathcal{R}}+\frac{B}{1-0.7\mathcal{R}}\\ 1& =A(1-0.7\mathcal{R})+B(1-0.9\mathcal{R})\\ 0.9\mathcal{R}=1:1& =A\left(\frac{2}{9}\right)⟹A=\frac{9}{2}=4.5\\ 0.7\mathcal{R}=1:1& =B\left(\frac{-2}{7}\right)⟹B=\frac{-7}{2}=-3.5\\ H& =\frac{4.5}{1-0.9\mathcal{R}}-\frac{3.5}{1-0.7\mathcal{R}}\end{array}$$

Based on this new transfer function

• The summation of the expressions tells us that we have two parallel systems being summed together
• The polynomials in the denominator in both expressions tells us that both systems are Feedback
• The order of both polynomials being 1 tells us that the systems are first order systems

Inspecting the Block Diagram, these features are visually apparent.

Each term’s denominator contributes one pole: $0.9$ from $\frac{4.5}{1-0.9\mathcal{R}}$ and $0.7$ from $\frac{-3.5}{1-0.7\mathcal{R}}$. The coefficients $4.5$ and $-3.5$ are the residues at each pole.

You can also intuit the behavior of the overall system based on the behavior of the subsystems: $y[n]=4.5(0.9{)}^n-3.5(0.7{)}^n$; both systems are exponential decay, but one decays way faster than the other. The result is a system that briefly surges before falling off.

Finding Poles

The key to simplifing a higher-order system is by identifying its Poles, i.e. complex roots of the denominator of the Transfer Function. Substituting $z^{-1}$ in for $\mathcal{R}$ may or may not be helpful. Factoring the denominator is the most straightforward approach. See some of the math in the Example Systems section for example approaches.

Example 1

We are given the Difference Equation of a system. We want to understand the long-term behavior of the Impulse Response of the system, and its poles.

$$y[n]=-\frac{1}{4}y[n-1]+\frac{1}{8}y[n-2]+x[n-1]-\frac{1}{2}x[n-2]$$

Start by rexpressing the difference equation in operator notation, and finding the Transfer Function.

$$\begin{array}{rl}Y& =-\frac{1}{4}\mathcal{R}Y+\frac{1}{8}{\mathcal{R}}^{2}Y+\mathcal{R}X-\frac{1}{2}{\mathcal{R}}^{2}X\\ \left(1+\frac{1}{4}\mathcal{R}-\frac{1}{8}{\mathcal{R}}^2\right)Y& =\left(\mathcal{R}-\frac{1}{2}{\mathcal{R}}^2\right)X\\ H=\frac{Y}{X}& =\frac{\mathcal{R}-\frac{1}{2}{\mathcal{R}}^2}{1+\frac{1}{4}\mathcal{R}-\frac{1}{8}{\mathcal{R}}^2}\end{array}$$

We can factor both the numerator and denominator.

$$\begin{array}{rl}H& =\frac{\mathcal{R}-\frac{1}{2}{\mathcal{R}}^2}{1+\frac{1}{4}\mathcal{R}-\frac{1}{8}{\mathcal{R}}^2}\\ & =\frac{\mathcal{R}\left(1-\frac{1}{2}\mathcal{R}\right)}{\left(1-\frac{1}{4}\mathcal{R}\right)\left(1+\frac{1}{2}\mathcal{R}\right)}\end{array}$$

Inspecting the denominator, we can solve for the poles by finding the roots of each of the factors.

How to find the poles from the factored Transfer function.

Find the first pole:

$$\begin{array}{rl}\left(1-\frac{1}{4}\mathcal{R}\right)& =0\\ \left(1-\frac{1}{4}{z}^{-1}\right)& =0\\ z^{-1}& =4\\ z& =\frac{1}{4}\end{array}$$

Find the second pole:

$$\begin{array}{rl}\left(1+\frac{1}{2}\mathcal{R}\right)& =0\\ \left(1+\frac{1}{2}{z}^{-1}\right)& =0\\ z^{-1}& =-2\\ z& =-\frac{1}{2}\end{array}$$

You will find that we get $z=\frac{1}{4},-\frac{1}{2}$. Now, the shortcut is just to look at the poles, to determine the behavior, but we will do the long way to maximize context.

Factor out the $\mathcal{R}$ from the numerator and decompose the rest using Partial Fractions:

$$\begin{array}{rl}H& =\mathcal{R}\cdot \frac{1-\frac{1}{2}\mathcal{R}}{\left(1-\frac{1}{4}\mathcal{R}\right)\left(1+\frac{1}{2}\mathcal{R}\right)}\\ & =\mathcal{R}\left(\frac{A}{1-\frac{1}{4}\mathcal{R}}+\frac{B}{1+\frac{1}{2}\mathcal{R}}\right)\\ 1-\frac{1}{2}\mathcal{R}& =A\left(1+\frac{1}{2}\mathcal{R}\right)+B\left(1-\frac{1}{4}\mathcal{R}\right)\\ \frac{1}{4}\mathcal{R}=1:-1& =3A⟹A=-\frac{1}{3}\\ -\frac{1}{2}\mathcal{R}=1:2& =\frac{3}{2}B⟹B=\frac{4}{3}\\ H& =\mathcal{R}\left(-\frac{1}{3}\cdot \frac{1}{1-\frac{1}{4}\mathcal{R}}+\frac{4}{3}\cdot \frac{1}{1+\frac{1}{2}\mathcal{R}}\right)\end{array}$$

Each $\frac{1}{1-p\mathcal{R}}$ is a first-order feedback system with impulse response $p^n$. The $\mathcal{R}$ in front delays everything by one sample. So the impulse response is

$$h[n]=-\frac{1}{3}{\left(\frac{1}{4}\right)}^{n-1}+\frac{4}{3}{\left(-\frac{1}{2}\right)}^{n-1},n\ge 1$$

with $h[0]=0$. The two fundamental modes are ${\left(\frac{1}{4}\right)}^n$ and ${\left(-\frac{1}{2}\right)}^n$. All of this was already known when we found the poles, but it is helpful to find the impulse response equation to kind of prove why it is the way it is. The $-\frac{1}{2}$ is the dominant exponential term. Because of this, the impulse response will exhibit alternating behavior. Both poles’ magnitudes are less than 1, so the overall behavior is decay.

Complex Poles

So far, we have only considered scenarios in which $\mathrm{Im}(p)=0$. However, this ignores the infinitely large set of possible values that are solidly within the Complex Plane, instead of just walking the Real Number line.

• Similar to complex solutions in Differential Equations
  • Complex poles always come in conjugate pairs, corresponding to $z=r{e}^{\pm j\Omega }$
  • The angle $\Omega$ determines oscillation
    • $\Omega =0$: no oscillation (positive real pole, monotonic)
    • $\Omega =\pi$: half-turn each step (negative real pole, alternating sign)
    • $0<\Omega <\pi$: oscillation with period $\frac{2\pi }{\Omega }$ samples
  • The magnitude $r$ determines exponential behavior
    • $r<1$: exponential decay
    • $r=1$: constant magnitude
    • $r>1$: exponential growth
• You can still do Partial Fractions. It’s exactly the same.
• The Impulse Response of a conjugate pair of poles combines into $r^n\cos (n\Omega )$, a sinusoid with an exponential envelope

Example 1

Find the poles for the following Transfer Function

$$H=\frac{1}{1-\mathcal{R}+{\mathcal{R}}^2}$$

Substituting $z^{-1}$ for $\mathcal{R}$ and finding the roots of the denominator:

$$\begin{array}{rl}1-z^{-1}+z^{-2}& =0\\ z^2-z+1& =0\\ z& =\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm j\sqrt{3}}{2}\\ z& =e^{\pm j\pi /3}\end{array}$$

The poles are a conjugate pair at $r=1$, $\Omega =\pi /3$. Since $r=1$, the impulse response neither grows nor decays. Since $\Omega =\pi /3$, the oscillation has period $\frac{2\pi }{\pi /3}=6$ samples.

$$h[n]=\cos \left(\frac{n\pi }{3}\right),n\ge 0$$