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Bode Plot Sketching

  • Thread starter Dafe
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  • #1
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Homework Statement


Sketch the Bode Plot for the following Transfer Function:

G(s)=10/s(1+ts)

where t=0.1sec

The Attempt at a Solution



G(jw)=10/jw(1+tjw) - frequency response...

Gain = 20log(10) - 20log(tw^2+w)

Does this mean that the Gain apporaches infinity as w approaches 0?

I really don't understand this, hope someone can give me a couple of hints...
 

Answers and Replies

  • #2
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You have a pole in the transfer function and an integrator, so to solve it if you are unsure split them up into two separate transfer functions and add the graphs.

First put the transfer function in familiar form:

[tex]G_{1}(s) = \frac{100}{(s+10)}[/tex]

and

[tex]G_{2}(s) = \frac{1}{s}[/tex]

For [tex]G_{1}(s)[/tex]

At low frequency the gain will be:

[tex]20log_{10}(100/10) = 20db[/tex]

At a value of around [tex]\omega = 10 [/tex] The pole will kick in and produce an asymptote of -20db/decade.

For the integrator:

The integrator will have a value of 0db at [tex]\omega = 1[/tex] so if you start your graph at [tex]\omega = .01[/tex] it will start with a value of 40db and slope downwards at 20db per decade.

Adding both of those graphs gives a magnitude plot that starts at +60db (for omega = .01) and goes down -20db/decade until 10db, where it goes down -40db/decade for all omega.

This is a very simple transfer function, how were you taught these? Perhaps there is some fundamental misunderstanding.
 
  • #3
145
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There sure is a fundamental misunderstanding :) I'm actually trying to learn them by my self with the aid of a book called Modern Control Systems (dorf).. I think I'm getting there, thanks alot!
 
  • #4
23
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The general rules for these asymptotic plots are:

-Poles cause -20db/decade slope at omega = a, where the the pole is [tex]\frac{1}{s+a}[/tex]

-Zeros do the opposite, +20db/decade at omega = a, the zero is [tex]s+a[/tex]

-Integrators and differentiators must have a value of zero db at omega = 1. Other than that they are pure slope (+/- 20db/decade)

-To find behavior at low frequency factor out the a, so for a pole you would get:

[tex]\frac{1}{a(\frac{s}{a}+1)}[/tex]

You can see here that as s goes to zero, the frequency becomes [tex]\frac{1}{a}[/tex]

It is the opposite for a zero, and this is all multiplied by the gain.

tl;dr go here:

http://lims.mech.northwestern.edu/~lynch/courses/ME391/2003/bodesketching.pdf
 
  • #5
145
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Thanks alot, I really appretiate it!
 

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