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Bode Plots in Control System

Last Updated : 21 Aug, 2023
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Bode plots describe linear time-invariant systems’ frequency response (change in magnitude and phase as a function of frequency). It helps in analyzing the stability of the control system. It applies to the minimum phase transfer function i.e. (poles and zeros should be in the left half of the s-plane).

Types of Bode Plot

1. Gain plot:

It represents the magnitude response of the system as a function of frequency. It is plotted on a logarithmic scale.

mag

2. Phase plot:

It depicts the frequency-dependent phase shift of the system’s output signal compared to its input signal. It’s also drawn on a logarithmic scale.

phase


Bode plot representation for the open loop system is:

20 log|G(jω)|

How to draw Bode Plot?

Step 1: Write the given transfer function in the standard form.

Transfer function:

G(s)= \frac{(s+a)(s+b)}{(s+p)(s+q)}          —– (1)

Standard form of equation 1:

G(s) = \frac{ab(1+ \frac{s}{a})(1+\frac{s}{b})}{pq (1+\frac{s}{p})(1+\frac{s}{q})}

 Take \frac{ab}{pq}          as a constant k.

Step 2: Identify the slope of the first line for the bode plot. The slope of the first line is based on poles and zeros at the origin. Refer to the following table.

Poles at origin

Slope of 1st line

1

-20 dB/decay

2

-40 dB/decay

Zeros at origin

The slope of 1st line

1

+20 dB/ decay

2

+40 dB/ decay

 Step 3: Find the gain of 1st line at ω=1 rad/sec

Gain|ω=1 = 20 log k

Where k = \frac{ab}{pq}

Step 4: Write all the corner frequencies in ascending order and define the slope of each line

Step 5: Write the phase equation and make a table of phase and frequency.

Φ = tan-1(\frac{w}{a}           ) + tan-1(\frac{w}{b}           ) – tan-1(\frac{w}{p}           ) – tan-1(\frac{w}{q}           )


Parameters of Bode Plot

Bode-Plots-in-Control-System-1


Figure 1 shows the gain and phase plot. The gain cross over frequency (wpc) and phase crossover frequency (wpc) can be calculated using gain plot and phase plot respectively.

Wgc is the value at 0dB whereas Wpc is the value at -180o.

Phase margin and gain can be calculated by extending the graph as shown in the figure.

Stability by bode plot:

 Ï‰pc > ωgc ->System is stable
 Ï‰pc < ωgc ->System is unstable
ωpc = ωgc ->System is marginally stable

What is Phase and Gain Margin?

Phase Margin

The phase margin indicates how much more phase shift we may put in the open loop transfer function before our system becomes unstable.

It can be calculated from the phase at the gain cross-over frequency.

Phase Margin (PM) = 180∘+∠G(jω)H(jω)∣ω=ωgc
= -180∘

Gain crossover frequency: 

It is the frequency at which the magnitude of G(s) H(s) is unity as seen in the figure 1.


|G(jω)H(jω)|ω=ωgc = 1

Gain Margin

The gain margin is the amount of open loop gain that can be increased before our system becomes unstable.

It can be calculated from the gain at the phase cross-over frequency.

Gain Margin (GM): \frac{1}{|G(jw)H(jw)|}_{w=w_{pc}}

Phase crossover frequency:

It is the frequency where the phase angle of G(s) H(s) is -180 degrees as seen in the figure 1.


∠G(jω)H(jω)∣ω=ωpc= -180∘

Advantages

  1. It helps in identifying the stability of the system.
  2. It helps in identifying phases and gaining margins with minimum calculation.
  3. It can be used to calculate the system’s transfer function.
  4. It can show the amplification and attenuation in the gain plot which is helpful in designing the filters.

Disadvantages

  1. It is only applicable to LTI (linear time-invariant) system.
  2. It is not suitable for the system having extremely high or low frequencies.
  3. It focuses on the frequency response without considering the transient time effect.

Solved examples

 Example: Find the transfer function from the bode plot given in the figure.

Bode-Plots-in-Control-System-2


Solution:

Step 1: Corner frequency: ω= 1, 10, 100

Step 2: Calculation of the slope

Slope

Increasing or decreasing

0 dB/dec

Initial value

+20 dB/dec

Increasing

0 dB/dec

Decreasing

-20 dB/dec

Decreasing

Therefore if the slope is increasing then it is a zero else it is a pole.

ω = 1 (zeros)
ω = 1 (pole)
ω = 1 (pole)

Step 3:   Calculating the gain of the first line at ω = 1

Gain = 20 log(k) + (slope of 1st line)log(ω)
-20    = 20log(k) + 0
k       = 0.1

Step 4: Writing the transfer function

T(s) = \frac{k(1+\frac{s}{1})}{(1+\frac{s}{10})(1+\frac{s}{100} )}

k= 0.1

T(s) = \frac{0.1(1+\frac{s}{1})}{(1+\frac{s}{10})(1+\frac{s}{100} )}

T(s) = \frac{100(s+1)}{(s+10)(s+100)}


FAQs: Bode Plots in Control System

1. Which graph paper is used for Bode plots?

Semi-Logarithmic graph paper.

2. What is the Bode plot’s stability?

According to the bode plot, system is said to be stable when gain margin is positive and the phase margin should be greater than the gain margin. 

3. What is the unit of Bode Plot?

The bode magnitude plot measures the system Input/Output ratio in special units called decibels. The Bode phase plot measures the phase shift in degrees (typically, but radians are also used).

4. What is decibels (dB)?

It is a unit of measurement which is used to calculate the logarithmic ratio of one value of a power or field quantity to another. The logarithmic quantity is known as the power level or field level, respectively.

 



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