Bode Plot By hand

  • Thread starter kwbake01
  • Start date
  • #1
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
 

Answers and Replies

  • #2
berkeman
Mentor
58,010
8,070
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
10
0
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
berkeman
Mentor
58,010
8,070
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
10
0
It is a block diagram with no feedback so the two blocks above are in series
 
  • #6
berkeman
Mentor
58,010
8,070
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
10
0
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
 
  • #9
10
0
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
berkeman
Mentor
58,010
8,070
  • #11
10
0
So for the numerator i get 4s(s+10) so then theres a zero at s=-10 and s=0?
 
  • #12
berkeman
Mentor
58,010
8,070
So for the numerator i get 4s(s+10) so then theres a zero at s=-10 and s=0?
Yes, if that's the numerator, then those are the zeros.
 
  • #13
10
0
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
10
0
Now those are the poles with 1 real and two imaginary
 

Related Threads on Bode Plot By hand

  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
646
  • Last Post
Replies
4
Views
671
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
15K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
815
  • Last Post
Replies
4
Views
3K
Top