[guzzler]

There's always a lot of discussion about input high pass filters on things like CMoys, and I think it's to do with a lack of appreciation of the theory of it.

I present my little introduction to this topic. I hope it will serve as a general introduction to complex impedance, and also be readable + useful.

An introduction to complex impedance

I would be most grateful of any suggestions you may have. I don't have a PDF maker, so I make no guarentees it will run on browsers other than IE (I don't have any here). I'll flesh out the two stated formulae at the end when I have more time.

 

[guzzler]

Anybody read it?

 

[DaKi][er]

Quote:

Originally Posted by guzzler Anybody read it?

Yea, you seemed to explain it ok, but I only really know the basics of the maths used (high school level) and skimmed over most of it

I'm pretty happy to just use F=1/2piRC and dont really care too much where it came from for now

 

[Tril]

I did. It's good but I already learned that before so I can't say how good it is for a beginner. You could add that Gain = Vout/Vin. It's obvious but maybe not for everyone.

In the end, what's important to know is that the impedance of
a resistor is Z = R [Ohm]
a capacitor is Z = 1/(jwC) [Ohm]
an inductor is Z = jwL [Ohm]

Where w = 2*Pi*f [rad/s].
Where f is the frequency [in Hz] of the (usually sinusoidal) oscillation.

I used w for omega.

The impedance law (don't know if this name exists) is like the Ohm law.
V = RI (Ohm law)
E = ZI (impedance law)

You simply use the loop and currents laws and voltage dividers like you do with the ohm law in resistor circuits and you can easily analyse any passive circuit.

 

[grasshpr]

Quote:

Originally Posted by guzzler Anybody read it?

I think you have a mistake in your derivation of impedance with a resistor and capacitor. Look at the second and third line starting with Vout = ZR/(Zr+ZC)*Vin.

If you square both sides of the second line, you don't get the third line. Also, looking at the last equation of that subsection, i.e., where you replacing omega = 2*pi*f, you didn't square that quantity.

Hope this helps...

 

[dsavitsk]

Quote:

Originally Posted by DaKi][er Yea, you seemed to explain it ok, but I only really know the basics of the maths used (high school level) and skimmed over most of it.

I have a university math degree and I skimmed over it too

 

[JahJahBinks]

Another way to think about it is this:
Basically the characteristic of a filter is related to its transfer function (which is often expressed as the ratio of two polynomials). If you factor the transfer function, numerator roots are called "zeros" and denominator roots are called "poles". A zero or a pole can be either a real or complex number and it's represented on a pole-zero plane (very similar to complex number plane), that's another reason why complex numbers are involved here.

In college, this kind of theory is seen more in control courses. In circuits courses, more emphasis is put on the implementation of filter, especially in the area of active filters.

Another thing I want to mention is that I am glad you introduced Euler's equation. As someone who has a basic background in communication theory, I can't say enough that the Euler's equation is one of the most important and widely used mathematical discoveries ever existed to electrical engineers, if not THE most.

 

[guzzler]

Quote:

Originally Posted by grasshpr I think you have a mistake in your derivation of impedance with a resistor and capacitor. Look at the second and third line starting with Vout = ZR/(Zr+ZC)*Vin. If you square both sides of the second line, you don't get the third line. Also, looking at the last equation of that subsection, i.e., where you replacing omega = 2*pi*f, you didn't square that quantity.

The squaring is correct as we're dealing in complex numbers; to square a complex number means you multiply your number (a+ib) by its complex conjugate (a-ib)

You are correct in the second point, thanks for that!

Ah well, seems like it was a little unecessary. Still, the best way to understand things is to try and teach them

 

[grasshpr]

Quote:

Originally Posted by guzzler The squaring is correct as we're dealing in complex numbers; to square a complex number means you multiply your number (a+ib) by it's complex conjugate (a-ib) You are correct in the second point, thanks for that! Ah well, seems like it was a little unecessary. Still, the best way to understand things is to try and teach them

Your right, I thought you were simply squaring the signal and squarerooting it, but your really computing magnitude of V, sorry for the confusion.

 

[digitalmind]

I only skimmed it quickly, really need to go to bed now. Seems explained well, should be a good help to alot of people.

And remember, if you want to use a Low pass filter to create a bass boost, just swap the location of the capacitor and resistor. Calculation will be the same.