1. ## S-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.

2. Originally Posted by ashes
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.
To what with it, its a system transfer function in Laplace transform form

RonL

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..

4. Originally Posted by ashes
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..
The zeros are points at which it is zero, that is zeros of the numerator. In this case it has two both at $0$.

Poles are zeros of the denominator, again this has two at $-2 \pi ~500k$ and $-2 \pi ~10k$.

RonL

5. 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.

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)

7. Originally Posted by ashes
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)
You're not being very clear on this function. You should use parenthesis were appropriate. Is the function the following:

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

If so, you can rewrite it as:

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

With:

$a=2\pi10k$
$b=2\pi500k$
$K=10ab$

Now replace:

$s --> j\omega$

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.