Plotting Bode Magnitude & Phase Plots By Hand

In summary, to plot the magnitude and phase Bode plots for the given system transfer function, first calculate the values of the poles and zeros by factoring the transfer function. Then, using those values, sketch the gain and phase versus frequency. If using MATLAB, input the polynomial into the program to find the roots, and use those values as the poles. The zeros can be found from the factored polynomial in the numerator.
  • #1
kwbake01
10
0
How would you plot the magnitude and phase Bode plots for the system transfer function of (4s^3+40s^2)/(4s^4+405s^3+504s^2+400s)
I plugged it into MATLAB and got a result but my professor wants hand calculations.
Thanks
 
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  • #2
kwbake01 said:
How would you plot the magnitude and phase Bode plots for the system transfer function of (4s^3+40s^2)/(4s^4+405s^3+504s^2+400s)
I plugged it into MATLAB and got a result but my professor wants hand calculations.
Thanks

Start by calculating the values of the poles and zeros of the transfer function...

Then once you have the poles and zeros, do you have an idea of how to start plotting the gain and phase versus frequency...?
 
  • #3
I gave you the wrong transfer function its from a block diagram that looks like this
[4+(40/s)]---->[s^2/(s+100)(4s^2+5s+4)] if i could get the transfer function from that i think i know how to get the poles and zeros
 
  • #4
kwbake01 said:
I gave you the wrong transfer function its from a block diagram that looks like this
[4+(40/s)]---->[s^2/(s+100)(4s^2+5s+4)] if i could get the transfer function from that i think i know how to get the poles and zeros

Could you please show the block diagram? Do you mean it is a traditional feedback block diagram?
 
  • #5
It is a block diagram with no feedback so the two blocks above are in series
 
  • #6
kwbake01 said:
It is a block diagram with no feedback so the two blocks above are in series

So as was pointed out to you in the other thread, how do you get the overall transfer function if the two transfer functions are in series?
 
  • #7
ok so the transfer function is (4s^2+40s)/(4s^3+409s^2+540s+4) now how do i sketch the bode plot? I know i have to convert the transfer function into a different form but how do i do that
 
  • #8
kwbake01 said:
ok so the transfer function is (4s^2+40s)/(4s^3+409s^2+540s+4) now how do i sketch the bode plot? I know i have to convert the transfer function into a different form but how do i do that

You need to factor the transfer function to find the poles and zeros. Those are what you use to sketch the transfer function.

http://i.cmpnet.com/powermanagementdesignline/2008/05/BassoFig319a.jpg

BassoFig319a.jpg
 
  • #9
instead of the 4 in the transfer function its 400 sorry about that
Berkeman, how do you factor the transfer function?
Thanks for your help so far
 
  • #10
kwbake01 said:
instead of the 4 in the transfer function its 400 sorry about that
Berkeman, how do you factor the transfer function?
Thanks for your help so far

You factor it like you do any polynomial. You will end up with a factored polynomial in the numerator (which gives you the zeros), and a factored polynomial in the denominator (which gives you the poles).

http://www.google.com/search?source...1T4GGLL_enUS301US302&q=factoring+a+polynomial

.
 
  • #11
So for the numerator i get 4s(s+10) so then there's a zero at s=-10 and s=0?
 
  • #12
kwbake01 said:
So for the numerator i get 4s(s+10) so then there's a zero at s=-10 and s=0?

Yes, if that's the numerator, then those are the zeros.
 
  • #13
for the denominator i but it into MATLAB and this is what i got

>> p = [4 409 540 400]; % p(x) =4s^3+409s^2+540s+400
format long; % print double-precision
roots(p)

ans =

1.0e+002 *

-1.009221534098646
-0.006639232950677 + 0.007416660864940i
-0.006639232950677 - 0.007416660864940i
 
  • #14
Now those are the poles with 1 real and two imaginary
 

Related to Plotting Bode Magnitude & Phase Plots By Hand

1. What is the purpose of plotting Bode magnitude and phase plots by hand?

The purpose of plotting Bode magnitude and phase plots by hand is to visually analyze the frequency response of a system and determine important characteristics such as gain, bandwidth, and phase shift.

2. What are the steps involved in plotting Bode magnitude and phase plots by hand?

The steps involved in plotting Bode magnitude and phase plots by hand include determining the transfer function of the system, converting it to a logarithmic form, calculating the magnitude and phase at different frequencies, and plotting the values on a graph using logarithmic scales.

3. What is the difference between Bode magnitude and phase plots?

Bode magnitude plots show the gain of a system at different frequencies, while Bode phase plots show the phase shift of the system at different frequencies.

4. How do you interpret Bode magnitude and phase plots?

In Bode magnitude plots, the slope of the line indicates the order of the system, and the intersection of the line with the frequency axis indicates the cut-off frequency. In Bode phase plots, the slope of the line represents the phase shift of the system, and the intersection of the line with the frequency axis represents the phase margin.

5. What are the limitations of plotting Bode magnitude and phase plots by hand?

Plotting Bode magnitude and phase plots by hand can be time-consuming and prone to human error. It also requires a good understanding of logarithmic scales and transfer functions. Additionally, it may not be suitable for complex systems with multiple poles and zeros.

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