# Sketching Bode Plots (Which one is correct?)

Discussion in 'Homework Help' started by jegues, Jan 22, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Hello all,

I've got a question with regards to sketching Bode plots.

I'm supposed to sketch the bode plot of my transfer function,

$T(s) = \frac{-390000 - 0.585s}{164368 + 0.019302s}$

by hand.

EDIT: Whoops! I think I mixed my zeroes and poles up!

I should have a zero at,

$\omega_{z} = \frac{390000}{0.585}$

and one pole at,

$\omega_{p} = \frac{164368}{0.019302}$

When we compute these, it seems as though the zero occurs at a high frequency than the pole, so the bode plot would resemble a low pass filter, but when I modeled the circuit in Multisim it gives me a graph like the following,

I attached the original schematic of the circuit I'm working with as well incase someone can tell by inspection whether the circuit will act as a low, or high pass filter.

There is a possibility I may have made an error in my calculator computing the values of the transfer function, but I'd like to be sure before I redo all my calculations. (There's alot, and it's really messy)

EDIT: I just redid my calculations again and I arrive at the same result. Am I misunderstanding how to plot the bode plot correctly?

Now that I've got my zeros and poles sorted out, my plot should match this one!

What do you guys think?

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Last edited: Jan 22, 2011
2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
I kept trying to scratch the specks off my screen before I realized it was your graphic.

Can you size them a bit smaller?

3. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
In a Bode plot, when a pole occurs, the graph goes up by 20dB/decade. When a zero occurs, the graph goes down by 20dB/decade.

As a result, the graph will first rise when it meets the pole, and then level down when it reaches the zero.

Your assumption that it would be a low-pass filter is wrong, in my opinion.

<edit> This post is flowed. Read below for amends.

Last edited: Jan 22, 2011
4. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
I think it is just the opposite. A pole sends the slope of the magnitude down while a zero sends the slope of the magnitude up. In the equations given above, the zero occurs before the pole, so the resultant graph appears correct.

5. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
You guys are correct. I don't know how I remembered this the wrong way.

I admit my mistake and confirm that a zero sends the plot up and a pole sends it down. Sorry for any misunderstanding.