What are Bode Plots?

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What are Bode Plots? Transfer Function (TF) H(s) H(jω) = M (ω)6 θ(ω) M (ω) is the magnitude of H(s) as a function of ω θ(ω) is the angle of H(s) as a function of ω We need to plot both M (ω) and θ(ω) with respect to ω

Often the range of ω is from very low frequency to a high frequency. Low frequency could be as low as 0.01 rad/sec and high frequency could be as high as 100 K rad/sec. How to accommodate such a large range is some thing we need to look into. It turns out linear scale for ω is not well suited. We need to resort to logarithmic scale. Magnitude M (ω) could be plotted by itself or could be converted into deci Bells (dB). What is dB? Magnitude of M in dB = 20 log10 M .

2 Properties of logarithms: log(ab) = log(a) + log(b) log10 (10a) = log10 (10) + log10 (a) = 1 + log10 (a) log ab = log(a) − log(b)

Magnitude M

Magnitude M in dB = 20 log10 M

We cannot plot this 0

−∞

M > 0 1 √

2

√1 2

0

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−3

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6

1 2

−6

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100

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1000

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0.1

−20

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−40

0.001

−60

For every two fold increase, 6 dB gets added.

For every ten fold increase, 20 dB gets added.

Motivation for non-linear scaling along the frequency axis: • First explain linear scale

1

2

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5

• Disadvantage of linear scale when wide-spread data exists: Let us take an example. Consider that we are interested in a frequency range from 1 to 10,000 Hz. Suppose we can afford to have 10 CM graph along the frequency axis. Then in a linear scale each cm corresponds to 1000 HZ. Draw the axis In this scale it is hard to distinguish low frequencies, say 1 Hz from 10 Hz although there is a tenfold increase from 1 to 10. • Linear scale disregards the sense of proportionality: To explain this, for the sake of argument, let us say we can afford a graph of 1000 CM. (I know it is not feasible.) We can perhaps distinguish all frequencies then. Draw the axis In this case, a change from 1 Hz to 10 Hz is a tenfold change in frequency and it is represented by 1 CM of graph. On the other hand, a change from 1000 Hz to 1010 Hz is only a 1 % change in frequency and it is still represented by 1 CM of graph. In 1 % change in frequency, the characteristic we are trying to draw would not have changed much (unless 1000 Hz happens to be a critical frequency). What this says is that a linear scale lacks the sense of proportionality along the frequency axis. • A logarithmic scale takes into account such a sense of proportionality and cures both the above disadvantages.

Logarithmic Scale We are interested in plotting the magnitude and angle characteristics of H(s) when s = jω for a very broad range of values of ω. Linear scale along the ω axis is not feasible. We need a non-linear scale. A common non-linear scale used is the logarithmic scale. The following describes the nature of logarithmic scale. Note that the representation of zero along the logarithmic axis is not feasible since log10 (0) = −∞. Linear Versus Logarithmic Scale

Linear Progression ω 0 1 2 3 4 5 6 7 8 9 10

Non-linear Logarithmic Progression log10 (ω) −∞ (not feasible) 0 0.30103 0.47712 0.60206 0.69897 0.77815 0.84510 0.90309 0.95424 1

log10 (αω) = log10 (ω) + log10 (α) • Multiplying ω by a number α implies adding log10 (α) to log10 (ω) in logarithmic domain. • Let α = 10. In order to plot log10 (10ω), we need to add one unit to log10 (ω) since log10 (10) = 1. A ten fold increase in ω adds only one unit of length along the logarithmic axis. This means that the frequency axis is compressed in logarithmic domain. Clearly, as seen in the graph paper on the other side, each unit along the logarithmic axis corresponds to one cycle. Say we start at some number n. In order to reach 10n, we need to move one unit along the logarithmic axis.

• A range of magnitude extending from some value to ten times that value is called a decade. Thus, a one unit length on the logarithmic axis corresponds to one decade. Note that a range of magnitude extending from some value to two times that value is called an octave. • We can start with any number ωstart other than zero at the beginning of first cycle. If the graph paper has six cycles, then the end of sixth cycle would be 106 ωstart (a multiple of 10 or a decade for each cycle). • In Bode plots, either the magnitude in dB or angle of H(jω) is plotted with respect to frequency. To do so, we use semi-log sheets. Frequency axis is the Logarithmic Axis.





ωstart = 0.1

An increase or decrease of frequency by a factor 10 is referred to as a DECADE. An increase or decrease of frequency by a factor 2 is referred to as an OCTAVE.



• 0.95

0.85

0.75

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0.45

0.32

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In actuality, octave refers to eight and not two. The reason that doubling a frequency is called an octave comes from the musical world. An octave is a doubling up frequency, but it's eight notes in the scale to go up an octave.

Logarithmic Axis





ωstart = 1 •

• 9.5

8.5

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Logarithmic Axis

This is a two cycle semi-log sheet































9.5 8.5 7.5 6.5 5.5

4.5

3.2

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1.1

1.2

1.5

9.5 8.5 7.5 6.5 5.5

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1.2 1.1

9 8 7 6 5 4

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9 8 7 6 5 4

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Logarithmic Axis

This is a six cycle semi-log sheet

This is a 7 cycle semi-log sheet

















































































































































































































Logarithmic Axis

20 dB per decade is equivalent to 6 dB per octave. In music, there are eight scalesto go up when the frequency is doubled.

d

At 0.01, a positive slope starts At 0.5, a negative slope starts The effect of the above two renders the net slope zero from 0.5 on wards until 1. At 1, the positive slope started at 0.01 ends. The net effect from 1 onwards until 10 is a negative slope. At 10, another negative slope starts. the net effect is then twice the negative slope until 50 at which point the negative slope started at 0.5 disappears bringing the net slope to only one negative value. At 1000, the negative slope started at 10 disappears rendering the net slope 0 from 1000 omwards.

HW due date will be announced in the class Student’s name in capital letters: This is a HW problem, collected and graded. Problem 1: Construct both the magnitude and phase angle plots for the transfer function H(s) =

2000s (s + 2)(s + 10)

Correct the asymptotic plots as needed. Draw the plots on semi-log sheets with appropriate scale so that the entire graph with all details fills the graph sheet. Problem 2: Construct both the magnitude and phase angle plots for the transfer function H(s) =

100(s + 10) (s + 1)(1 + s + s2 )

Correct the asymptotic plots as needed. Correct the asymptotic plots as needed. Draw the plots on semi-log sheets with appropriate scale so that the entire graph with all details fills the graph sheet.

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What are Bode Plots?

What are Bode Plots? Transfer Function (TF) H(s) H(jω) = M (ω)6 θ(ω) M (ω) is the magnitude of H(s) as a function of ω θ(ω) is the angle of H(s) as a ...

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