Any pointers to something which explains the math used to describe filters?
Three semesters of engineering school calculus were enough for me to learn math is all about the notation, although as a software guy I avoided the EE department beyond a few digital classes and picked up nothing about the notation needed to describe transfer functions.
Cursory web searches turn up documents which assume I remember that little omega is angular velocity and that I know EE guys write their i with a tail as j.
Three semesters of engineering school calculus were enough for me to learn math is all about the notation, although as a software guy I avoided the EE department beyond a few digital classes and picked up nothing about the notation needed to describe transfer functions.
Cursory web searches turn up documents which assume I remember that little omega is angular velocity and that I know EE guys write their i with a tail as j.
Look for info on Laplace transforms.
Impedance of a resistor is Xr=R
Impedance of a capacitor is Xc=1/(s*C)
Impedance of an inductor is Xl=s*L
From there it is simply Kirchoffs voltage and current laws. If I get my internet set up at home tonight, I'll do a quick example for ya.
What are you trying to do?
Impedance of a resistor is Xr=R
Impedance of a capacitor is Xc=1/(s*C)
Impedance of an inductor is Xl=s*L
From there it is simply Kirchoffs voltage and current laws. If I get my internet set up at home tonight, I'll do a quick example for ya.
What are you trying to do?
OK - "simple" example 😉
This probably isn't going to make much sense if you haven't done a few circuit analysis problems and have an awareness thereof., but here goes... focusing on "hows" rather than "whys"
Think of a driver (modeled as a resistor) in series with an inductor. The inductor ind resistor are connected together and the input voltage is across both of them.
Impedance of a resistor is Xr=R
Impedance of an inductor is Xl=s*L
This being a simple voltage divider, the voltage we are interested in is the voltage across the driver. Due to the nature of these circuits (hand-waving for brevity) the transfer function (simply Vout/Vin) of any circuit is simply the impedance of the driver (and any components in parallel with it) divided by the impedance of the entire circuit. Remember the rules of resistance, parallel/series values, etc?
Vout/Vin = R/(sL+R)
For a more complicated second order lowpass, with a capacitor in parallel with the driver and an inductor in series, you get:
Vout/Vin = (R/(s*C))/(R+1/(s*C))/(s*L+(R/(s*C))/(R+1/(s*C)))
Here, "(R/(s*C))/(R+1/(s*C))" is the impedance of the Driver (resistor) and the capacitor in parallel. You can see the math quickly gets very tedious.
Draw the circuits and look carefully and I think you will begin to see the pattern.
Now to plug in some numbers - to draw a frequency response curve or do any analysis, you need some real values.
The quantity "s" in the impedance equations above is the complex frequency variable - without explaining it, you simply substitute s=jw (where w = 2*pi*f and j=sqrt(-1)) into the equations above.
To calculate the phase, you need to separate the numerator and denominator into real and imaginary (those with j) parts and take the arctangent of each to get an angle. The angle of the numerator minus the angle of the denominator is the phase. With a computer, you can simply evaluate the complex quantity and take the "argument".
All this is pretty easy to do in excel if you have the analysis toolpak installed.
Ask questions. It'll help me understand where you are at with this.
This probably isn't going to make much sense if you haven't done a few circuit analysis problems and have an awareness thereof., but here goes... focusing on "hows" rather than "whys"
Think of a driver (modeled as a resistor) in series with an inductor. The inductor ind resistor are connected together and the input voltage is across both of them.
Impedance of a resistor is Xr=R
Impedance of an inductor is Xl=s*L
This being a simple voltage divider, the voltage we are interested in is the voltage across the driver. Due to the nature of these circuits (hand-waving for brevity) the transfer function (simply Vout/Vin) of any circuit is simply the impedance of the driver (and any components in parallel with it) divided by the impedance of the entire circuit. Remember the rules of resistance, parallel/series values, etc?
Vout/Vin = R/(sL+R)
For a more complicated second order lowpass, with a capacitor in parallel with the driver and an inductor in series, you get:
Vout/Vin = (R/(s*C))/(R+1/(s*C))/(s*L+(R/(s*C))/(R+1/(s*C)))
Here, "(R/(s*C))/(R+1/(s*C))" is the impedance of the Driver (resistor) and the capacitor in parallel. You can see the math quickly gets very tedious.
Draw the circuits and look carefully and I think you will begin to see the pattern.
Now to plug in some numbers - to draw a frequency response curve or do any analysis, you need some real values.
The quantity "s" in the impedance equations above is the complex frequency variable - without explaining it, you simply substitute s=jw (where w = 2*pi*f and j=sqrt(-1)) into the equations above.
To calculate the phase, you need to separate the numerator and denominator into real and imaginary (those with j) parts and take the arctangent of each to get an angle. The angle of the numerator minus the angle of the denominator is the phase. With a computer, you can simply evaluate the complex quantity and take the "argument".
All this is pretty easy to do in excel if you have the analysis toolpak installed.
Ask questions. It'll help me understand where you are at with this.
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