Calculating Transfer Function off Bode Plot?

[jean28]

Hey guys. I was wondering if anyone could give me some hints or help me understand well how to get transfer functions off bode plots. Here is an example I need to solve:

http://i1226.photobucket.com/albums/ee410/jean28x/bodeplotcalc.jpg

 

[NascentOxygen]

If the slope changes by an amount ± 20dB/decade then you can conclude that a 1st order filter section has come into play at that corner frequency. (Had the slope changed by 40dB/decade then you would say that a 2nd order filter section had become significant at that frequency.)

 

[DragonPetter]

http://en.wikipedia.org/wiki/Bode_plot
they have examples, and there is also a controls wiki website with examples, I believe: http://en.wikibooks.org/wiki/Control_Systems/Bode_Plots.

Given a bode plot, you look for changes in slope. At any frequency that the slope changes, first look to see how much the slope is changing at that frequency. It will be in multiples of 20dB/decade. If you have just a 20db/decade change, you have 1 pole or zero depending if it was negative or positive slope change. Remember poles and zeroes cancel each other, and so you will never have both at the same frequency.

Say you find a slope change at a frequency. Examples:

-20dB/decade -> 1 pole @ frequency
-40dB/decade -> 2 poles @ frequency
+20dB/decade -> 1 zero @ frequency
+40dB/decade -> 2 zeroes @ frequencySo just make a list of your poles and zeroes: z1,z2,z3,...,zn and p1,p2,p3,...,pn as you look at the bode plot. Now you write your transfer function in the form [K*(s-z1)*(s-z2)*...*(s-zn)]/[(s-p1)*(s-p2)*...*(s-pn)].

Then with all of the poles and zeroes in place in the equation, you look back at the plot and find the DC gain (the magnitude at f = 0). Remember the bode plot is a log plot while your transfer function is not, and so you must take the inverse log to get the correct DC gain. If you now take your transfer function and set all the s variables equal to 0 and set the equation equal to the DC gain, you can solve for the K multiplying factor.

 

[alagez]

i m having problem with 1. Sketch Bode asymptotic approximation plots of magnitude and phase for the following systems

G(s)=(s+3)(s+5)/s(s+2)(s+4) please help me sir

 

[jean28]

i m having problem with 1. Sketch Bode asymptotic approximation plots of magnitude and phase for the following systems

G(s)=(s+3)(s+5)/s(s+2)(s+4) please help me sir

First thing you need to do is turn the G(s) in a general form i.e. make it 3 * 5 * (s/3 + 1)(s/5 + 1) etc...

After that, you take whichever constant you get (from the 3 * 5 divided by whatever you get in the denominator). Let's call that constant K. You evaluate 20log(K) and you will get a new constant.This new constant is where you will start in the y-axis (dB). Ex say the constant k is 5*3/(5) = 3. 20 log (3) = 9.54. You will start at the coordinate (0, 9.54) on the bode plot.

Now, starting from (0, 9.54), you will write a straight line until (3,9.54) and (5,9.54). In both of these coordinates you will have to INCREASE the line by 20 dB/decade. This is because they are in the numerator.

You do the same process for the ones in the denominator, except that now the line will DECREASE by 20 dB/decade (this is because they are poles, i.e. they are in the denominator).

The s in the denominator is a straight downwards line that starts at the origin as well, and this line will go downwards 20 dB/ decade immediately.

Explaining Bode plots is a bit difficult through chat so feel free to ask your doubts if you don't understand fully. I know how frustrating it is to work with these things and have nobody to help so I'm willing to help in any way I can.

Cheers.

 

[alagez]

its really helpful...thank you sir...i got the answer,..