WHAT ARE TRANSFER FUNCTIONS, POLES, AND ZEROES?

# What Are Transfer Functions, Poles, And Zeroes?

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## 1. Overview

Transfer functions are a way of describing the frequency response of a system. The system can be anything with a measurable input and output, e.g. mechanical spring/mass/dampers, electronic RLC circuits, e.t.c. This page will put an emphasis on electrical transfer functions.

Transfer functions are usually written in the Laplace domain using the variable $$s$$, where $$s$$ is the complex angular frequency, as shown in $$Eq.\ \ref{eq:s-eq-j-w}$$.

\begin{align} \label{eq:s-eq-j-w} s = j\omega \end{align}

where:
$$j$$ is the imaginary number ($$j = \sqrt{-1}$$), also seen as $$i$$ in maths ($$j$$ is used in electronics as to not get $$i$$ confused with current)
$$\omega = 2\pi f$$

 Angular frequency $$\omega$$ is used here rather than Hertz $$f$$ just to keep the equations tidier, but it is trivial to convert from one to the other!

Transfer functions describe the relationship between output and input (typically of voltage, but it doesn’t have to be). $$Eq.\ \ref{eq:xfer-fn-vout-vin}$$ shows this relationship. We will use the symbol $$H(s)$$ is represent the transfer function.

\begin{align} \label{eq:xfer-fn-vout-vin} H(s) = \frac{v_{out}(s)}{v_{in}(s)} \end{align}

Note that because we are using $$s$$, $$Eq.\ \ref{eq:xfer-fn-vout-vin}$$ encodes both magnitude and phase relationships between the output and input. On it’s own, it’s complex in nature and not really useful for deducing anything. But, using two clever tricks, you can separately extract the magnitude and phase response equations from $$H(s)$$. The magnitude response is found by taking the magnitude of $$H(s)$$, which is $$|H(s)|$$ as shown in $$Eq.\ \ref{eq:xfer-fn-magnitude}$$.

\begin{align} \label{eq:xfer-fn-magnitude} | H(s) | = \left| \frac{v_{out}(s)}{v_{in}(s)} \right| \end{align}

The phase response $$\angle H(s)$$ is found by finding the angle of the complex number from the positive x-axis, as shown in $$Eq.\ \ref{eq:xfer-fn-phase}$$.

\begin{align} \label{eq:xfer-fn-phase} \angle H(s) = Arg{\left(\frac{\Im \{H(s)\}}{\Re \{H(s)\}}\right)} \end{align}

where
$$Arg$$ is the argument (implemented with $$atan2(y, x)$$ in many software packages) $$\Im\{H(s)\}$$ is the imaginary part of $$H(s)$$
$$\Re\{H(s)\}$$ is the real part of $$H(s)$$ +1

 This rule of finding the phase response can sometimes be wrong, as the $$\arctan$$ function throws away information about sign. You have to be aware of what quadrant of the Argand diagram you are in and compensate appropriately.

## 2. Zeroes And Poles

A value of $$s$$ that causes the a transfer function to be 0 is called a zero, and a value of $$s$$ that causes the transfer function to be infinite is called a pole. Zeroes generally occur when a factor in the numerator is 0 (one notable exception is that a zero can also occur if the denominator tends towards infinity, usually as $$s \rightarrow \infty$$), poles generally occur when a factor in the denominator is 0. Poles that have an imaginary component always come in pairs (conjugate pairs).

Intuitively, you can think of zeroes as places in where the system completely blocks a certain frequency. A poles is a place where the system has infinite response (at least mathematically).

The zeroes are the roots of the numerator polynomial, and the poles are the roots of the denominator polynomial. For this reason they are also referred to generally as roots.

The poles and zeros of a system can tell you much about how the system performs — it can tell you if the system is stable, how fast it responds.

For example, the transfer function in $$Eq. \ref{eq:xfer-fn-1-over-s}$$ has a pole at the origin and a zero at infinity. This simple transfer function represents an integrator. A constant voltage applied to it will result in an output climbs without any limit. However, at high frequencies, the output is essentially zero as the positive and negative parts of the waveform are averaged out over time.

\begin{align} \label{eq:xfer-fn-1-over-s} H(s) = \frac{1}{s} \end{align}

## 3. The Transfer Function Of A Low-Pass RC Filter

A first-order low-pass RC filter has the transfer function shown in $$Eq.\ \ref{eq:rc-xfer-fn}$$.

\begin{align} \label{eq:rc-xfer-fn} H(s) &= \frac{1}{1 + \b{s}RC} \end{align}

Using $$Eq.\ \ref{eq:s-eq-j-w}$$, we can replace $$s$$ with $$j\omega$$ to get $$Eq.\ \ref{eq:rc-xfer-fn-jw}$$.

\begin{align} \label{eq:rc-xfer-fn-jw} H(\omega) &= \frac{1}{1 + j\omega RC} \end{align}

This system has a pole at $$f = \frac{1}{2\pi R C}$$ and a zero at $$f = \infty$$.

We can find the magnitude response of this low-pass RC filter by taking the magnitude of $$H(f)$$, remembering that the magnitude of a complex number is defined as in $$Eq. \ref{eq:magnitude-of-complex-num}$$.

\begin{align} \label{eq:magnitude-of-complex-num} |a + jb| = \sqrt{(a + jb)(a - jb)} \end{align}
\begin{align} \label{eq:abc} | H(\omega) | &= \left| \frac{1}{1 + j\omega RC} \right| \nonumber \\ &= \frac{1}{\sqrt{(1 + j\omega RC)(1 - j\omega RC)}} \nonumber \\ &= \frac{1}{\sqrt{1 - (j\omega RC)^2}} \nonumber \\ &= \frac{1}{\sqrt{1 - (-1)(\omega RC)^2}} \nonumber \\ \label{eq:mag-response-lp-rc-filter} &= \frac{1}{\sqrt{1 + (\omega RC)^2}} \\ \end{align}

$$Eq. \ref{eq:mag-response-lp-rc-filter}$$ shows the final result. Notice that by finding the magnitude, the imaginary components are gone! We can plot this on a graph.

Figure 1. The magnitude response of the the low-pass RC filter, found by plotting $$Eq.\ \ref{eq:mag-response-lp-rc-filter}$$. Note that the magnitude has been converted into decibels with $$dB = 20log10(mag)$$.

We can find the phase response of the low-pass RC filter by using rule in $$Eq.\ \ref{eq:xfer-fn-phase}$$.

\begin{align} \angle H(j\omega) &= Arg\left(H(j\omega)\right) \nonumber \\ &= Arg\left(\frac{1}{1 + j\omega RC}\right) \nonumber \\ &= Arg(1) - Arg(1 + j\omega RC) \nonumber \\ &= 0 - arctan\left(\frac{j\omega RC}{1}\right) \nonumber \\ \label{eq:phase-response-lp-rc-filter} &= -arctan\left(j\omega RC\right) \\ \end{align}
 We can safely reduce $$Arg$$ to $$arctan$$ because we know that $$1 + j\omega RC$$ will be in the 1st quadrant of the Argand diagram for all positive real values of $$\omega$$.
Figure 2. The phase response of the the low-pass RC filter, found by plotting $$Eq.\ \ref{eq:phase-response-lp-rc-filter}$$

## 4. Pole Zero Plots

Poles and zeroes are plotted in a Argand diagram in what is called a pole-zero plot to give the reader an understanding on how the circuit responds.

• Zeroes contribute +90 of phase and increase the magnitude, above the zero frequency.

• Poles contribute -90 of phase and decrease the magnitude, above the pole frequency.

Poles are normally drawn as X’s on the graph, and zeroes as O’s. Unless you are building an oscillator, poles in the right-hand half of the plane (having a positive real component) are a bad thing, as they represent an instability.

Figure 3. Argand diagram showing how the location of poles (no zeroes shown) on a pole zero plots shows how components of the system respond to transients (i.e. impulses).