Hamid Roozbahani

A Bode plot is a graph of the transfer function of a linear,

time-invariant system versus frequency, plotted with a log-

frequency axis, to show the system's frequency response. It

is usually a combination of a Bode magnitude plot,

expressing the magnitude of the frequency response gain,

and a Bode phase plot, expressing the frequency response

phase shift. [1]

Bode Plots

Bode diagram

Let us assume that transfer function can be divided into parts

Thus the frequency response of the whole G can be calculated as a sum of the

frequency responses of G1…Gn

Magnitude

Phase

The idea in Bode’s method is to plot magnitude curves using a logarithmic scale and

phase curves using a linear scale. This strategy allows us to plot a high order G (jw) by

simply adding the separate terms graphically

Advantages of Bode plots

1. Dynamic compensator design can be based entirely on bode plots

2. Bode plots can be determined experimentally

3. Bode plots of system in series simply add which is quite convenient

4. The use of a log scale permits a much wider range of frequencies to be displayed

on the single plot than is possible with linear scales

Bode diagram of constant G(s)=K ,G(s)=1/s, G(s)=1/s2 or s

Open-loop bode diagram

Km=gain margin

Φm=phase margin

If the both are positive the system is stable

Recommended values

Closed loop Bode-diagram

Mp=resonance peak < 2.3…3.5

dB (recommended for servo

systems)

The bandwidth of the system is a measure of speed of response. For control system it is

defined as the frequency corresponding to 0.707 (3db) in closed-loop magnitude Bode

plot.

Relation between resonance peak and overshoot

Calculation of responses by Matlab

The open-loop transfer function of a servo is

The bode diagram is calculated by typing

NUM= [30] (nominator polynomial coefficients)

DEN= [2.3e-4 1.2e-2 1 0] (denominator coefficients)

Sys=tf (NUM, DEN) (calculates transfer function)

Margin (sys) (calculates and plots Bode-diagram with gain and phase margins)

With direct feedback H(s)=1, the closed loop transfer function is

This in Matlab is

NUM=[30]

DEN=[2.3e-4 1.2e-2 1 30]

bode(NUM,DEN)

sys=tf(NUM,DEN)

(calculates Bode-diagram without margins)

Closed-loop frequency response

-If in the open-loop Bode-diagram the gain margin Km is bigger than 6dB and phase

margin Φm bigger than 45o increase the gain until one reaches the limit

-If Km or Φm or both are smaller than required the stability of servo is insufficient =>

increase the cylinder and valve or apply better control method

-If –3dB bandwidth in closed-loop Bode-diagram is smaller than required ω-3dB the gain

must be increased. If stability becomes insufficient => increase the cylinder and valve or

apply better control method [3]

Hand Sketching: Step-by-step approach

1- Put transfer function in ZPK form (factored zeros and poles, with a constant

multiplier K out front).

2- Identify breakpoints: distance of poles and zeros from the origin. Mark those on

the frequency axis of the plot. Remember to convert from rad/sec to Hz.

3- Determine low or high frequency constant asymptote of gain by taking the limit

of H(s) as s → 0 or infinity, respectively. Convert to dB.

4- Start at one of the asymptotes that is constant. (see below if neither is constant).

Move along in frequency until you get to a breakpoint. Each breakpoint is

associated with a change in slope of +/-20 dB/decade (+/-6 dB/octave). From left

to right, a zero produces an increase in slope (The increase could be from

negative to less negative, or from positive to more positive, etc.) Each pole

produces a decrease in slope. Work through all the breakpoints, and check that

the final asymptote is correct.

5- Sketch in a smoother curve, 3 dB below or above each breakpoint (unless it is a

double pole or zero in which case it is 6 dB below or above).

6- Now fill in the phase by the same procedure: Find the phase on the asymptotes

by looking at the limit of H(jω) as ω → 0 or ω → infinity. If the limit is real, the

phase is 180 degrees or zero. If the limit is imaginary, the phase is +/-90 degress

(-90 for a negative imaginary limit). If the limit is zero, you have to look at what

direction it came in to zero from. If it came from a positive imaginary number, the

phase is 90 degree; a negative imaginary number -90 degrees. If it came from a

positive real number, the phase is 0, and if it came from a negative real number,

the phase is -90° Then from left to right, each pole causes a –90° transition in

phase, and every zero causes a +90° transition in phase. You can use the rules

for the slope of the transition, but it’s usually not worth the trouble to get that

exact.

7- OK, what if neither asymptote is flat? For example H(s) = s/[(s+1)(s+100)]:

HF limit: s/100 -> 0

LF limit: s/(s2) -> 0

Solution: Consider the shape—it will look like:

With breakpoints at 1/(2π) and 100/(2π)

Instead of starting at HF or LF asymptotes, you can start with the flat center part.

Consider the approximate value there, where s>>1 and s<<100. Then we can write the

transfer function, putting the negligible stuff in a small font: H(s) = s/[(s+1)(s+100)] Then.

Thus, we can approximate H(s) as s/[100s] = 1/100 or –40 dB in the flat middle part.

Then we can start there and sketch the rest. (It slopes down with +/- 20 dB/decade

slope on either side of the plateau.) [4]

Bode Plots by hand

1. Bode Plots by Hand

Bode Phase Plots

[2]

Also you can find helpful information regarding Bode plots in below web address too:

http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/BodeHow.html

References:

[1] Wikipedia- Bode plot-http://en.wikipedia.org/wiki/Bode_plot

[2] Original EE 105 Discussion Notes from Meghdad Hajimorad (“Amin”)- Last Modified by: Bill Hung- Date: 5 August 2006

[3] Course material by Prof. Huapeng Wu

[4] Engineering Sciences 22 — Systems Summer 2004 -Bode Plots Page 1-BODE PLOTS