EE3300/EE5300 Electronics Applications

Recap of transfer functions and Bode plots

It is assumed that you are familiar with transfer functions from previous study of circuit theory, signals and systems, or similar courses. This section is a quick recap.

The definition of a transfer function

We previously defined the DC gain of an amplifier as the ratio of the output voltage to the input voltage,

We would like to generalise this definition so that it can be used for any arbitrary input signal (such as a sinusoid or some other waveform). You should recall that this type of “gain” equation can be written in the Laplace domain for any linear time-invariant system. Therefore, we define the transfer function as

Notice the direct analogy to Eqs. and . The transfer function is the “gain” of the circuit in the Laplace domain.

Given a Laplace domain transfer function , one may substitute to obtain the DC gain (Eq. ). In the case of sinusoidal inputs, we can substitute to obtain the phasor domain (complex number) frequency response of the circuit.

We can also perform circuit analysis in the Laplace domain using familiar circuit theory techniques (e.g. Ohm’s law, KCL, KVL, node analysis, etc). Laplace domain circuit impedances are shown in Figure 1.

Example 1

To refresh our memories about transfer functions, find the transfer function of the low-pass RC filter shown in Figure 2.

Solution

We recognise that is given by a voltage divider formula, and therefore write

It is convenient to write this as

where is the location of the pole. Recall that a pole is any value of where the denominator of . You should recall that stable systems have all poles in the left-half of the complex plane (), therefore in intuitive analysis the negative sign is often implied.

Some remarks about this transfer function:

• From Eq. , we can quickly identify the DC gain by substituting . In this case, .
• Equation has a simple interpretation: the pole shows the frequency at which the circuit response starts to decrease. Here, the response starts to decrease at a frequency of . This is the -3 dB cutoff frequency that should be familiar from circuit theory.

Bode plots

To analyse a circuit’s frequency response, we often use a graphical tool called a Bode plot. The Bode plot is obtained by substituting in the transfer function and then plotting the magnitude and phase of the resulting complex function.

There are simple rules to enable Bode plots to be sketched by hand. It is important to master these rules so that you can intuitively “see” the impact of moving a pole or adding a pole.

We start drawing at low frequencies and progress towards higher frequencies. Each time our sketch passes the frequency of a pole or zero, we change the slope of the magnitude plot. The changes are:

• At each pole: the magnitude plot slope changes by dB/decade.
• At each zero: the magnitude plot slope changes by dB/decade.

For the phase plot:

• At each pole: a total phase shift of degrees occurs, starting one decade before the pole and finishing one decade after the pole.
• At each zero: a total phase shift of degrees occurs, starting one decade before the zero and finishing one decade after the zero.

Example 2

Sketch the Bode plot of the RC low-pass filter in Figure 2 when and .

Solution

The DC gain is , therefore the magnitude plot starts at 0 dB and the phase plot starts at 0.

There is one pole at

The magnitude plots starts to decrease by dB/decade at 1 kHz. The phase plot starts to decrease one decade earlier (0.1 kHz) and finishes one decade later (10 kHz) for a total phase shift of degrees. Therefore, we can sketch the Bode plot as shown in Figure 3.

For comparison, an accurate plot produced using Matlab’s bode function is shown in Figure 4.

Show Matlab source code to generate the Bode plot

R = 1e3;
C = 160e-9;
s = tf('s'); % Define the Laplace variable
H = 1/(1 + s*R*C); % Define the transfer function
opts = bodeoptions();
opts.FreqUnits = 'Hz';
opts.XLim = [1e1 1e5];
bodeplot(H, opts);
grid on;