# Stumped on a hand drawn bode plot.

Discussion in 'Homework Help' started by farmosh203, Nov 26, 2006.

1. ### farmosh203 Thread Starter Member

Nov 26, 2006
20
0
Hello all, I have been browsing these forums for a while, but decided to register today. I was wondering if someone could help me with a homework problem that I had about bode plots.

I have a transfer function, H(s) = (s-10)/(s+10). I have to graph the bode plot by hand but I was a little confused on how to do this. For the magnitude plot, the poles and zeros cancel so it is just a straight line at 0 dB. What I could not figure out is the Phase angle plot.

When graphing the Phase angle plot by a computer, it looks like the graph has a pole at omega equals 10 and drops down with a slope of -90 degrees. The horizontal asymptotes are at -180 degrees and -360 degrees. To get a graph like this, I think that the transfer function would have to be in the form of something like this: a*/(b*s^2*(s/10+1)^2) where a and b are any positive number.

I can't seem to get this problem in that form... and I'm a little stumped. Does anyone have any ideas or suggestions for this problem? Thanks for any help.

Feb 24, 2006
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3. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
Ok, here is my solution to this:

H(s) = (s-10)/(s+10)

Sub jw for s and normalise the real part of the zero factor to zero:

H(jw) = (jw-10)/(jw+10)

H(jw) = (j(w/10)-1)/(j(w/10)+1)

So represented logarithmically:

H(jw) = 20log(j(w/10)-1) - 20log(j(w/10)+1)

From this you can get your magnitude. You can check your answer for this against my expression.

For the phase:

For w<<1:

arg{(j(w/10)-1)} = 0
-arg{(j(w/10)+1)} = 0

Total phase at (w<<1) = 0

For w~=10:

arg{(j(w/10)-1)} ~= -45
-arg{(j(w/10)+1)} ~= -45

Total phase at (w~=10) = -90

For w>>10:

arg{(j(w/10)-1)} = -90
-arg{(j(w/10)+1)} = -90

Total phase at (w>>10) = -180

From the above information you should be able to draw the phase diagram of the transfer function.

Apologies of this is difficult to understand, any further questions post back and I will try to explain better.

Dave

4. ### farmosh203 Thread Starter Member

Nov 26, 2006
20
0
Thanks for the responses, I had a little trouble following what you were saying but I understand the general idea. Our teacher taught us how to do Bode plots a little differently, and the solutions were posted today.

Thanks again.

5. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
Would you care to share with us the way you/your teacher approached this? The method I used is an old method employed by control engineers who are used to doing stability analysis using Root Locus plots. It is based on fragmenting the transfer function into a series of components from which you can easily deduce the magnitude and phase characteristics (as you can see). It is useful for arbitrarily complex transfer functions.

Dave

6. ### farmosh203 Thread Starter Member

Nov 26, 2006
20
0
I just checked my solution, it turns out he used the same method as you Dave. I was a little confused on this problem because the way he taught us couldn't be used for that problem.

Our teacher told us to get each equation into the form on slide 5 and use slide 7 to determine the asymptotes.

Here is the link to the slide show he used to teach us the material:

http://www.eas.asu.edu/~holbert/eee202/eee202.html and click on the "Bode Plots (9-4 to 9-5)" lecture or the direct link is:

http://www.eas.asu.edu/~holbert/eee202/EEE202_Lect21_BodePlots.ppt

7. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
Although good practice, it seems to me that your teacher has taken a more difficult approach to this question than is required. That said my method is more a technique and is not necessarily the best way of learn the details of such a subject.

Slide 4 is the one that is of most advantage to the technique I have described above so I would advise that you learn that table of-by-heart. A final point to stress about using that table - be aware of whether you zero or pole is positive or negative - for the magnitude it will differentiate whether the change is +20dB/dec or -20dB/dec, and for the phase it will translate to a difference or 180°.

Any further questions or queries from your studies, post back and I'll try and give some guidance - although I haven't formally done Control Systems analysis for a good few years, I still use may of the techniques they teach you.

Dave

8. ### farmosh203 Thread Starter Member

Nov 26, 2006
20
0
Thanks for all the help Dave.