Bode Plot By hand

1. Jun 29, 2011

kwbake01

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

2. Jun 29, 2011

Staff: Mentor

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. Jun 29, 2011

kwbake01

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. Jun 29, 2011

Staff: Mentor

Could you please show the block diagram? Do you mean it is a traditional feedback block diagram?

5. Jun 30, 2011

kwbake01

It is a block diagram with no feedback so the two blocks above are in series

6. Jun 30, 2011

Staff: Mentor

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. Jul 1, 2011

kwbake01

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. Jul 1, 2011

Staff: Mentor

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

9. Jul 1, 2011

kwbake01

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. Jul 1, 2011

Staff: Mentor

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).

.

11. Jul 1, 2011

kwbake01

So for the numerator i get 4s(s+10) so then theres a zero at s=-10 and s=0?

12. Jul 1, 2011

Staff: Mentor

Yes, if that's the numerator, then those are the zeros.

13. Jul 1, 2011

kwbake01

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. Jul 1, 2011

kwbake01

Now those are the poles with 1 real and two imaginary