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.
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.
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.