I am trying to do the transfer function of this circuit and also trying to understand the frequency domain thing of the Laplace transform.
I want to do a transfer function of VA/V1, considering B to be ground.
After doing the node analysis and if my equations are correct, I have ended with this transfer function on the s domain.
A calculator online gave these answers for the cubic equation on the denominator
x1 = -915.52763
x2 = -42.23618 + i * 3304.67537
x3 = -42.23618 - i * 3304.67537
These are my problems:
- what do these numbers represent?
- the equation is on the denominator, what about the s100 on the nominator?
- how do I find the frequency performance of this circuit as the frequency changes?
- what does this function say about the circuit?
Best Answer
May I suggest that you start by simplifying the 2 resistors into one resistor. I'm suggesting this because it makes the math a lot easier and I think you have a mistake in your final formula. So, the voltage source reduces from 10 volts to 9.09 volts in series with a resistance of 90.90 ohms due to standard circuit theorums.
This then has a standard formula: -
$$H(s) = \dfrac{\omega_n^2}{s^2 +2\zeta\omega_n s+ \omega_n^2}$$ Where \$\omega_n = \dfrac{1}{\sqrt{LC}}\$
And \$\zeta= \dfrac{R}{2}\sqrt{\dfrac{C}{L}}\$
The value for R is the newly recalculated value of 90.9 ohms. To offer a little more help, \$\omega_n\$ is the natural resonant frequency of the filter and \$\zeta\$ is called the damping ratio (also equal to the reciprocal of 2Q).
Now if you want to see what this looks like you can use this on-line calculator and plug-in the values: -
Resonance occurs at about 503 Hz and there is a peak in the response of about 11 dB. Circuit Q is about 3.5.
But don't forget that there is an overall attenuation (due to me simplying the two resistors into one resistor) so the formulas are multiplied by 0.909.
I haven't used your equations because I can see that there is an error in the final formula you derived.