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

1. Apr 8, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

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

3. 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!

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

Last edited: Apr 8, 2007
2. Apr 8, 2007

### denverdoc

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

3. Apr 9, 2007

### VinnyCee

That gives:

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

What do I do with the omega cubed term now?

4. Apr 9, 2007

5. Apr 9, 2007

### AlephZero

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

6. Apr 9, 2007

### 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.

7. Apr 9, 2007

### 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.)

8. Apr 11, 2007

### 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?

9. Apr 11, 2007

### 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, thats a good one for practice.

Last edited: Apr 11, 2007
10. Apr 11, 2007

### 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.

11. Apr 11, 2007

### denverdoc

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

12. Apr 11, 2007

### 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?

13. Apr 11, 2007

### denverdoc

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

14. Apr 11, 2007

### 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?

15. Apr 11, 2007

### denverdoc

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

16. Apr 11, 2007

### VinnyCee

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

17. Apr 11, 2007

### denverdoc

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

18. Apr 11, 2007

### 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?

Last edited: Apr 11, 2007
19. Apr 11, 2007

### denverdoc

Count me out, those give me headaches,:yuck:

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.

20. Apr 11, 2007

### VinnyCee

Thanks a lot!