Bode plots with a pole at zero

Thread Starter

applefat

Joined Oct 12, 2007
10
Ok, so, say I have a transfer function of the form:

H(s) = 20 (0.1jw +1 ) / (jw)(jw+1 )(0.01 jw +1)

where w is my omega.

We can rewrite this as:

H(s) = 200 (s +10) / (s)(s+1)(s+100)

There are break points at 0, 1, 10, and 100, and I understand how figuring the slope works. But what confuses me, is finding the initial gain value at w = 0 !

If my understanding is correct, normally one can simply plug in w = 0, or, a w value smaller than the smallest break point, to find the initial value (then put that into 20(log(w)) for decibels). But if the plot starts at zero, how can we find the initial value? If one plugs in zero, the denominator becomes zero, and the result is undefined, or infinity, basically...

Edit: This is just a linear plotting btw, or I suppose, uncorrected Bode plot, it's called
 

The Electrician

Joined Oct 9, 2007
2,751
The usual thing when plotting Bode plots is for the frequency axis to be logarithmic, so the plot can't begin at zero, anyway.

But, yes, if there is a pole at zero, then the behavior is like an integrator, unbounded gain as the frequency decreases toward zero.

The gain is infinite wherever there is a pole; that's what a pole is, a point of infinite gain.
 

Thread Starter

applefat

Joined Oct 12, 2007
10
The usual thing when plotting Bode plots is for the frequency axis to be logarithmic, so the plot can't begin at zero, anyway.

But, yes, if there is a pole at zero, then the behavior is like an integrator, unbounded gain as the frequency decreases toward zero.

The gain is infinite wherever there is a pole; that's what a pole is, a point of infinite gain.
So... to find the initial gain for the purposes of plotting the first line segment, do I plug in some arbitrary value close to zero, such as 0.001?
 

The Electrician

Joined Oct 9, 2007
2,751
If by initial gain, you mean the gain at zero, then the initial gain is infinite.

Since as I said, Bode plots usually have a logarithmic frequency axis, they can't start at zero. So what you do is pick some frequency a little greater than zero, such as your example of .001, calculate the gain there and start your plot there.

If by the phrase "initial gain" you mean the gain where you choose the Bode plot to begin, then I guess the initial gain will be the gain at .001 for the example you gave.

You could certainly SAY that the initial gain (at zero) is infinite in the problem write-up. You just can't show it on a traditional Bode plot.
 

Thread Starter

applefat

Joined Oct 12, 2007
10
If by initial gain, you mean the gain at zero, then the initial gain is infinite.

Since as I said, Bode plots usually have a logarithmic frequency axis, they can't start at zero. So what you do is pick some frequency a little greater than zero, such as your example of .001, calculate the gain there and start your plot there.

If by the phrase "initial gain" you mean the gain where you choose the Bode plot to begin, then I guess the initial gain will be the gain at .001 for the example you gave.

You could certainly SAY that the initial gain (at zero) is infinite in the problem write-up. You just can't show it on a traditional Bode plot.
Ok cool, thank you very much.
 
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