# Uin/Uout = H(w) where to ask

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Im as is to see, quite new to this forum, and i was wondering where to ask questions about transition equasions (dont know if ive got the right word, still have to get a scientific dictionary)

the reason im asking it here is because it is mathematics, but its just because most normal electronics formulas dont aply, and the fact that it sometimes involves complex numbers (j)

example of what i mean: ive got an analog butterworth filter to the first degree, I just need to know the amplification factor formula, which is voltage_in(Uin)/voltage_out(Uuit)=H(w) where w is the frequency of the voltage, (the variable so to say)

ive got to: Uin-Uuit((1-(1+jwrc))jw2rc+1=0

r and c are given, i can substract Uuit/Uin, i just want to know if there is any way left to simplyfy ((1-(1+jwrc))jw2rc+1

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Work out parenthesis !

(a + bi)(c + di) = ac - bd + (bc + ad) i

Mandrake

PS : Here a,b,c and d are real ofcourse

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First, let's check whether I understand the formula you mean by:

Uin-Uuit((1-(1+jwrc))jw2rc+1=0

Using the fantastic tool implemented here (check for details how to use it in the Quick Latex tutorial thread), you can compose this in real looking mathematics.

$U_{in}-U_{out} \frac{j\omega2RC+1}{1-(1+j\omega RC)}=0$

If you want to express H(w) Uin/Uout, that would be

$\frac{j\omega2RC+1}{1-(1+j\omega RC)}$

$= \frac{j\omega2RC+1}{-j\omega RC}$

$= \frac{j\omega2RC}{-j\omega RC}+\frac{1}{-j\omega RC}$

$= -2 + \frac {j}{\omega RC}$

Of course, the end result will only be as perfect as the formula which I started with...

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First, let's check whether I understand the formula you mean by:

Uin-Uuit((1-(1+jwrc))jw2rc+1=0

Using the fantastic tool implemented here (check for details how to use it in the Quick Latex tutorial thread), you can compose this in real looking mathematics. I'll just asume I understood correctly and start from:

$U_{in}-U_{out} \frac{j\omega2RC+1}{1-(1+j\omega RC)}=0$

If you want to express H(w) = Uin/Uout, that would be

$\frac{j\omega2RC+1}{1-(1+j\omega RC)}$

$= \frac{j\omega2RC+1}{-j\omega RC}$

$= \frac{j\omega2RC}{-j\omega RC}+\frac{1}{-j\omega RC}$

$= -2 + \frac {j}{\omega RC}$

Of course, the end result will only be as perfect as the formula which I started with... Give me a buzz if I got it all wrong in the first place.

Next to the general rule for multiplication given by Mandrakeroot, one for dividing a complex number:

$\frac{1}{a+bj}=\frac{a}{a^2-b^2}-\frac{b}{a^2-b^2}j$ just multiply top & bottom with $a-bj$

Greetz,

Leo

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