- #1

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I plugged it into MATLAB and got a result but my professor wants hand calculations.

Thanks

- Thread starter kwbake01
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- #1

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I plugged it into MATLAB and got a result but my professor wants hand calculations.

Thanks

- #2

berkeman

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

I plugged it into MATLAB and got a result but my professor wants hand calculations.

Thanks

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

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[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

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

[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

- #5

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

- #6

berkeman

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

- #7

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- #8

berkeman

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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

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Berkeman, how do you factor the transfer function?

Thanks for your help so far

- #10

berkeman

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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).

Berkeman, how do you factor the transfer function?

Thanks for your help so far

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

.

- #11

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

- #12

berkeman

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Yes, if that's the numerator, then those are the zeros.So for the numerator i get 4s(s+10) so then theres a zero at s=-10 and s=0?

- #13

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>> 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

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Now those are the poles with 1 real and two imaginary

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