Obtain the output of basic circuits using Bode plots

**What are Bode Plots? ** Bode plots sketch the AC amplitude and phase at a chosen node in a circuit or control loop, as a function of frequency. They show, for example, the high or low frequency cutoffs of a filter circuit, or the positions of resonance peaks.

**Why Bode Plots?** Getting the response of simple networks is very easy and intuitive using this engineering technique. They provide a fast way to visualize the frequency response of basic networks with a minimum of complex algebra.

This post shows a simple progressive explanation of reactance and impedance. **Step 1** is to plot reactances for basic circuit elements: resistors, capacitors, and inductors. **Step 2** is to define and use the impedance concept. The final step in most textbooks is to rotate the impedance vector as a *phasor*. The discussion here departs from that approach and switches to s-space (as used in Laplace transforms). It is possible to do inverse transforms on s-space functions to obtain circuit response to transients, but that is not the purpose here. Rather **Step 3** in this ‘lab experiment’ is to determine the poles and zeroes of the transfer function of the circuit. The result is an immediate sketch of the output amplitude and phase of any chosen node in the circuit as a function of frequency.

Simple Example

This simple example is a high pass filter consisting of an inductor and a resistor.

**Step 1**: The resistance is R and the reactance is just ωL, where ω = 2π*frequency. These two components of impedance can be plotted as perpendicular vectors on the complex plane:

In the complex plane, R is a real quantity, and the reactance of the inductor is a positive imaginary (often written jωL, where j =√(-1) ). If the impedance were purely real, as in the case of a resistance, then the current and voltage have the same phase. If the impedance were purely inductive – a positive imaginary – then the current would be 90 degrees out of phase with the voltage. In fact considering the voltage as a real V_{0 }cos (ωt), and dividing by a positive imaginary, jωL, the resulting current is a negative imaginary. Thus the current vector lags 90º behind the voltage vector as these two phasors rotate counterclockwise on the complex axes. Thus the voltage ‘leads’ current in a pure inductor. The current is –(V_{0} /ωL) sin (ωt).

**Step 2**: In this simple example, the net impedance Z = (R ^{2} + (ωL)^{2} )^{½} . The net impedance is indicated in the above graph by the dashed arrow. The phase angle between voltage and current is the angle between Z and the real axis in the figure, thus φ = tan^{-1} (ωL/R). The voltage leads the current by the angle φ.

Power is not dissipated in the reactance L because the current and voltage are 90º out of phase in the inductor. Therefore the entire power dissipation in the circuit is in the resistance R. Since the current in the circuit is I_{rms} = V_{rms}/|Z|, the power in the circuit, equal to the power in the resistor, is I_{rms}^{2} R = (V_{rms }/ |Z| )^{2} R = V_{rms} I_{rms} cosφ.

**Step 3**: The above circuit is a simple voltage divider, where the voltage V_{in} is divided by the network to give V_{out }= V_{in } jωL/Z . These are complex numbers, dependent on frequency. The Bode plot allows a quick graphical analysis of the high pass filtering effect. Divide by V_{in } and take the Laplace transform of both sides, substituting the variable ‘s’ for jω:

H(s) must be put in the form of a quotient of two polynomials in s, and preferably the polynomials are factored. Then directly determine the zeroes in the numerator: s = 0 in this case; and the zeroes in the denominator: s = –1/τ. Use this information to generate the Bode plots. Go to the Bode plotting website: Bode website

Suppose for example that R = 10 and L = 25 mH. Then τ = .0025 sec. The zero of H(s) is at s = 0, and the pole is at s = -400 radians/sec. Entering this info in the Bode website obtain the following result:

The Bode plot is the Asymptotic plot. It is a quick approximation to the Real plot, and there are simple rules for drawing it. It shows that the transmission of the filter is very low at small frequencies, and becomes flat for frequencies above ∼400 rad/sec. The phase plot is also available:

The phase plot shows there is a 90 degree phase shift at low frequencies due to the inductor. But at high frequency, where the impedance of the inductor becomes much greater than that of the resistor, the phase tends toward zero degrees as if the inductor were an open circuit.

#### Worksheets for Students

We provided a few editable worksheets to download for students in PHY131 (Physics Science and Engineering students).

- Bode plot of a resonant circuit response – Resonant Circuit
- Bode plot of RC filter circuits – RC Circuits
- Bode plot of an RC bandpass filter, which has both low frequency and high frequency cutoffs: only intermediate frequencies are transmitted – Bandpass Circuit