Drawing Bode plots: I need to convert to s-domain - but how?

[VinnyCee]

Homework Statement

The question asks the reader to make a Bode plot for the following equation:

$$H(\omega)\,=\,\frac{50\left(j\omega\,+\,1\right)}{j\omega\left(-\omega^2\,+\,10j\omega\,+\,25\right)}$$

Homework Equations

$$j\omega\,=\,s$$

The Attempt at a Solution

$$H(\omega)\,=\,\frac{50j\omega\,+\,50}{-j\omega^3\,+\,10\left(j\omega\right)^2\,+\,25j\omega}$$

But how do I deal with the omega squared term in the denominator? If I can figure that, I can plug the num. and denom. into MATlab and it will plot the Bode plots for me!

$$H(\omega)\,=\,\frac{50s\,+\,50}{-s\omega^2\,+\,10s^2\,+\,25s}$$

 

[denverdoc]

j is imaginary root of -1, jw^2=-w^2

 

[VinnyCee]

That gives:

$$H(\omega)\,=\,\frac{50s\,+\,50}{\omega^3\,+\,10s^2\,+2s}$$

What do I do with the omega cubed term now?

 

[denverdoc]

Normally these are factored to find zeros and poles, the numerator is trivial problem, the denominator is not so obvious to me. Sorry if I misled you, earlier. Here is at least a link for usual means of plotting without benefit of Matlab.

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

 

[AlephZero]

Your cube term should be $$-j\omega^3 = +j^3\omega^3 = s^3.$$ (because $$j^2 = -1$$)

 

[denverdoc]

thanks aleph, I should have noticed that.
so to the op, the denominator would then become s(s+5)^2, that looks much better.

 

[maverick280857]

Although converting to s-domain is the standard practice for doing this, we had a first semester (1st year) compulsory course on electrical science where we were taught how to draw Asymptotic Bode plots for such Transfer functions without using (explicitly) s-domain analysis methods. We would then write every factor as $(1 + j\omega/\alpha)$ and note that

1. $(1 + j\omega/\alpha)$ is a straight line which goes at 20db/decade starting from $\omega = \alpha$ (and for $\omega < \alpha$ is zero).

2. $1/(1 + j\omega/\alpha)$ is a straight line which goes at -20db/decade starting from $\omega = \alpha$ (and for $\omega < \alpha$ is zero).

(Thought I should post this here anyway--might be of some use to beginners learning to draw asymptotic Bode plots this way.)

 

[VinnyCee]

OK, I put the transfer function into the following form:

$$H(\omega)\,=\,\frac{10\left(1\,+\,j\omega\right)}{-j\omega\left(1\,+\,\frac{j\omega}{5}\right)^2}$$

I divided the equation by $\frac{-5}{-1}$ because it is the pole divided by the zero.

And I do something (I really don't know what it is I am doing though!) with the constant coefficient 10:

20 log 10 = 20

Now what?

 

[denverdoc]

The constant is just gain and on its own would be straight line with log amplitude of 1. So you have one, zero, a double pole, and an integrator (jw). (that should be positive BTW). So you are now ready to apply all rules.

This is a good site for summary of what to do with each:
http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/BodeHow.html

PS: I added a post to your lowpass circuit problem post suggesting if you're just getting started with these, that's a good one for practice.

 

[VinnyCee]

OK, I have this now:

$$20\,log\,|H(\omega)|\,=\,20\,log\,10\,+\,20\,log\left(\left|1\,+\,\frac{j\omega}{1}\right|\right)\,-\,20\,log\,\left(\left|j\omega\right|\right)\,-\,20\,log\,\left(\left|1\,+\,\frac{j\omega}{5}\right|\right)\,-\,20\,log\,\left(\left|1\,+\,\frac{j\omega}{5}\right|\right)$$

But I still don't understand why K = 1 if the 20 log 10 = 20.

 

[denverdoc]

sorry, for confusion, that was just the log, not expressed in decibels.

 

[VinnyCee]

Cool, so the Magnitude plot starts at 20dB.

Now to draw dotted lines! The first term after the constant is a zero at frequency 1? How is that drawn?

The next term is an "integrator"? How do I draw that?

The last two terms are the double pole at frequency 5? I know that would be a -40dB/decade change. Right?

 

[denverdoc]

thats a pole at the origin. so slope -20 from the get go.

 

[VinnyCee]

The lone jw term is a pole at the origin? So the plot starts out at 20dB and goes -20dB slope until frequency is 1? Then it goes to -60dB slope?

 

[denverdoc]

I see constant gain offset, pole at zero, than the zero (so slopes counteract for a ways) then dbl pole.

 

[VinnyCee]

So, the resulting magnitude approximation would look like this?:

http://img135.imageshack.us/img135/9598/magnitudewp4.jpg

 

[denverdoc]

I believe so, what does your Matlab plot look like?

 

[VinnyCee]

Looks kind of like that. Now what about the phase plot?

Start at -90 degrees because of the pole at origin, then what for the zero and then what for the double pole at 5?

 

[denverdoc]

Count me out, those give me headaches,

Seriously, the rules are all on the link I posted, that is really a wonderful site. If you spend a few minutes exploring it, you'll see all kinds of examples that cover all the elements in your transfer function, be happy to look it over once complete.

 

[VinnyCee]

Thanks a lot!

 

[denverdoc]

hey no problem, it was good for me as well :!)

ie, review of material I haven't thought about in ages that I will need if I elect to go back to grad school.