Hi.. does anyone know how to do the transformation for this type 2 S-transfer function,

H(s) = 10s^2/(s + 2pi*500k)(s+2pi*10k).

Thanks.

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- Mar 27th 2008, 11:25 AMashesS-transform
Hi.. does anyone know how to do the transformation for this type 2 S-transfer function,

H(s) = 10s^2/(s + 2pi*500k)(s+2pi*10k).

Thanks. - Mar 27th 2008, 11:32 AMCaptainBlack
- Mar 28th 2008, 08:53 AMashesreply
well i know it is a transfer function of a high pass filter but how do i know how many poles and zeroes does it have..?I mean..how do i go about doing the transformation..S-transform is pretty new to me..

- Mar 28th 2008, 09:10 AMCaptainBlack
- Mar 28th 2008, 02:00 PMCoomast
In case you need the inverse transform, you must split it up using partial fractions. After this do the polynomial division because you have s^2 in the numerator and only a first degree polynomial in the denominator. Then you can obtain the inverse in the time domain. It is an odd one. You will require a delta function and two exponentials.

- Mar 30th 2008, 07:12 AMashesreply
I tried to expand the function cos i need to plot the amplitude and phase response bode plot anyway does anyone have any idea how?..this is my function...THanks

10*s/(s+2pi*10k)*s/(s+2pi*500k)*1/[(s/2pi*10k+1)(s/2pi*500k+1) - Mar 30th 2008, 07:53 AMCoomast
You're not being very clear on this function. You should use parenthesis were appropriate. Is the function the following:

$\displaystyle H(s)=\frac{10\cdot s^2}{(s+2\pi10k)(s+2\pi500k)}\cdot \frac{\displaystyle 1}{\displaystyle \left(\frac{s}{2\pi10k}+1\right)\cdot \left(\frac{s}{2\pi500k}+1\right)}$

If so, you can rewrite it as:

$\displaystyle H(s)=K\cdot \frac{s^2}{(s+a)^2(s+b)^2}$

With:

$\displaystyle a=2\pi10k$

$\displaystyle b=2\pi500k$

$\displaystyle K=10ab$

Now replace:

$\displaystyle s --> j\omega$

And obtain from this the absolute value and the phase. That should give you enough information to plot the bode diagram. It has been a long time since I've done this, so please confirm this with your textbook.