# Bode Plot Examples

posted by

*sri vastav.*
Several examples of the construction of Bode Plots are included in this file. Click on the transfer function in the table below to jump to that example.

# Bode Plot: Example 1

Draw the Bode Diagram for the transfer function:

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 0 polynomial, the denominator is order 1.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 2 components:

- A constant of 3.3
- A pole at s=-30

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 3.3 is equal to 10.4 dB). The phase is constant at 0 degrees.
- The pole at 30 rad/sec is the blue line. It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (3 rad/sec) then drops linearly down to -90 degrees at 10 times the break frequency (300 rad/sec).

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The overall asymptotic plot is the translucent pink line, the exact response is the black line.

# Bode Plot: Example 2

Draw the Bode Diagram for the transfer function:

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 1 polynomial, the denominator is order 2.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 4 components:

- A constant of 0.1
- A pole at s=-10
- A pole at s=-100
- A zero at s=-1

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 0.1 is equal to -20 dB). The phase is constant at 0 degrees.
- The pole at 10 rad/sec is the green line. It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (1 rad/sec) then drops linearly down to -90 degrees at 10 times the break frequency (100 rad/sec).
- The pole at 100 rad/sec is the blue line. It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (10 rad/sec) then drops linearly down to -90 degrees at 10 times the break frequency (1000 rad/sec).
- The zero at 1 rad/sec is the red line. It is 0 dB up to the break frequency, then rises at 20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (0.1 rad/sec) then rises linearly to 90 degrees at 10 times the break frequency (10 rad/sec).

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The overall asymptotic plot is the translucent pink line, the exact response is the black line.

# Bode Plot: Example 3

Draw the Bode Diagram for the transfer function:

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 1 polynomial, the denominator is order 2.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 4 components:

- A constant of 33.3
- A pole at s=-3
- A pole at s=0
- A zero at s=-10

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 33.3 is equal to 30 dB). The phase is constant at 0 degrees.
- The pole at 3 rad/sec is the green line. It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (0.3 rad/sec) then drops linearly down to -90 degrees at 10 times the break frequency (30 rad/sec).
- The pole at the origin. It is a straight line with a slope of -20 dB/dec. It goes through 0 dB at 1 rad/sec. The phase is -90 degrees.
- The zero at 10 rad/sec is the red line. It is 0 dB up to the break frequency, then rises at 20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (1 rad/sec) then rises linearly to 90 degrees at 10 times the break frequency (100 rad/sec).

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The overall asymptotic plot is the translucent pink line, the exact response is the black line.

# Bode Plot: Example 4

Draw the Bode Diagram for the transfer function:

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 1 polynomial, the denominator is order 3.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 4 components:

- A constant of -10
- A pole at s=-10
- A doubly repeated pole at s=-1
- A zero at the origin

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 10 is equal to 20 dB). The phase is constant at -180 degrees (constant is negative).
- The pole at 10 rad/sec is the blue line. It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency then drops linearly down to -90 degrees at 10 times the break frequency.
- The repeated pole at 1 rad/sec is the green line. It is 0 dB up to the break frequency, then drops off with a slope of -40 dB/dec. The phase is 0 degrees up to 1/10 the break frequency then drops linearly down to -180 degrees at 10 times the break frequency. The magnitude and phase drop twice as steeply as those for a single pole.
- The zero at the origin is the red line. It has a slope of +20 dB/dec and goes through 0 dB at 1 rad/sec. The phase is 90 degrees.

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The overall asymptotic plot is the translucent pink line, the exact response is the black line.

# Bode Plot: Example 5

Draw the Bode Diagram for the transfer function:

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 1 polynomial, the denominator is order 2.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 4 components:

- A constant of 6
- A zero at s=-10
- Complex conjugate poles at the roots of s
^{2}+3s+50,

with

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 6 is equal to 15.5 dB). The phase is constant at 0 degrees.
- The zero at 10 rad/sec is the green line. It is 0 dB up to the break frequency, then rises with a slope of +20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency then rises linearly to +90 degrees at 10 times the break frequency.
- The plots for the complex conjugate poles are shown in blue. They cause a peak of:

at a frequency of

This is shown by the blue circle. The phase goes from the low frequency asymptote (0 degrees) at

to the high frequency asymptote at

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The exact response is the black line.

# Bode Plot: Example 6

Draw the Bode Diagram for the transfer function:

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 2 polynomial, the denominator is order 3.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 4 components:

- A constant of 1
- A pole at s=-100
- A repeated pole at the origin (s=0)
- Complex conjugate zeros at the roots of s
^{2}+s+25,

with

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 1 is equal to 0 dB). The phase is constant at 0 degrees.
- The pole at 100 rad/sec is the green line. It is 0 dB up to the break frequency, then falls with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency then falls linearly to -90 degrees at 10 times the break frequency.
- The repeated poles at the origin are shown with the blue line. The slope is -40 dB/decade (because pole is repeated), and goes through 0 dB at 1 rad/sec. The slope is -180 degrees (again because of double pole).
- The complex zero is shown by the red line. The zeros give a dip in the magnitude plot of

at a frequency of 5 rad/sec (because ζ is small, ω_{r}≈ω_{0}). This is shown by the red circle. The phase goes from the low frequency asymptote (0 degrees) at

to the high frequency asymptote at

Again, because ζ is so small, this line is close to vertical.

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The exact response is the black line.

# Bode Plot: Example 7

Draw the Bode Diagram for the transfer function:

This is the same as "Example 1," but has a 0.01 second time delay. We have not seen a time delay before this, but we can easily handle it as we would any other constituent part of the transfer function. The magnitude and phase of a time delay are described here.

## Step 1: Rewrite the transfer function in proper form.

Make both the lowest order term in the numerator and denominator unity. The numerator is an order 0 polynomial, the denominator is order 1.

## Step 2: Separate the transfer function into its constituent parts.

The transfer function has 3 components:

- A constant of 3.3
- A pole at s=-30
- A time delay of 0.01 seconds (magnitude and phase of time delay described here).

## Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

- The constant is the cyan line (A quantity of 3.3 is equal to 10.4 dB). The phase is constant at 0 degrees.
- The pole at 30 rad/sec is the blue line. It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (3 rad/sec) then drops linearly down to -90 degrees at 10 times the break frequency (300 rad/sec).
- The time delay is the red line. It is 0 dB at all frequencies. The phase of the time delayis given by -0.01·ω rad, or -0.01·ω·180/π° (at ω=100 rad/sec, the phase is -0.01·100·180/π≈-30°). There is no asymptotic approximation for the phase of a time delay. Though the equation for the phase is linear with frequency, it looks exponential on the graph because the horizontal axis is logarithmic.

## Step 4: Draw the overall Bode diagram by adding up the results from step 3.

The exact response is the black line.

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