I have not got it to work. The inputs are mixed up as are the power rails and the output.
LTspice expects a certain order in the PINATTRIB
pinname and spiceorder and i changed it around but i made a flaw.
I am at the moment a bit short in time to analyze in a relaxed moment what is going on. But i will try to solve it.
I do noticed a neat trick and have a usefull tip for you :
First, you should add always an capacitor between 10uF and 100uF in parallel over your battery.
For audio that is good enough. When you place a capacitor in parallel over your battery, the capacitor will when large enough act as a very low impedance (close to a short circuit) to an ac signal with a given frequency.
This is important for to simplify the calculation of the high pass filter on the input.
For Omega , you can write 2*pi*f where f is the frequency.
Here is an example :
I do not know if LTspice actually takes this in account. I think that the voltage sources are ideal 0Ω impedances unless you specify it by adding an inductors, resistor and capacitor network that represents the power source you are going to use. But if you add in parallel a large capacitor, that capacitor will be the dominant reactance and you can within boundaries ignore the other effects.
Electronics has the advantage that you can easily simplify circuits as long as you design with a little bit of breathing space.
The calculation is simple and you can use a neat trick :
The resistor divider of 2*1MΩ is for an ac signal really two resistor in parallel. Thus you have a first order hpf that has a resistance of 500kΩ and a capacitor of 100nF.
If you add that in the formula you get 1/(1*10^6 * pi * 100*10^-9)
= 3.18Hz. This means every frequency below this cutoff frequency will be continued to be attenuated.
As long as your power source has a very low impedance( near 0) you can use this :
If you use always powers of ten for the resistor, the intermediate result is always a power of ten times pi . For example :
For a resistor divider with equal resistors :
1MΩ in parallel is 500kΩ, 100kΩ in parallel is 50kΩ.
10KΩ in parallel is 5kΩ.
2*pi times this replacement resistance makes a power of ten * pi.
If you always use a capacitor that has a value that is also a power of ten
and the same capacitance as well, you always get a result that is 1 divided by (pi * a power of ten). If you shi
ft the powers of ten around :
When always using a 100nF capacitor :
For 10MΩ you get 1/pi = 0.318 Hz.
For 1MΩ you get 10/pi = 3.18Hz.
For 100kΩ you get 100/ pi = 31,8Hz.
For 10kΩ you get 1000/ pi = 318Hz.
You can use this for low pass filters as well since the same formula applies.