I have seen the expression for the transfer function of an RLC tank given as the following.
$$H(j\omega) = \frac{H_o}{1+j2Q\frac{\omega-\omega_o}{\omega_o}}$$
I am somewhat able to get this result, but I still have some minor confusions and am unsure if my analysis is correct. I believe you can inject a current (Norton model) into the RLC parallel branch. Where $$H(j\omega) = \frac{V_{o}}{I_{\text{in}}}$$
Then finding the admittance.
$$Y(j\omega) = \frac{1}{R}+\frac{1}{jwL}+jwC$$
$$Y(j\omega) = \frac{1}{R}+jwC+\frac{-j}{wL}$$
$$Y(j\omega) = \frac{1}{R}+j\left(wC-\frac{1}{wL}\right)$$
$$Y(j\omega) = \frac{1}{R}+\frac{j}{wL}(w^2CL-1)$$
$$\omega_o^2 = \frac{1}{LC}$$
$$Y(j\omega) = \frac{1}{R}+\frac{j}{wL}\left(\frac{w^2}{w_o^2}-1\right)$$
$$Y(j\omega) = \frac{1}{R}+\frac{j}{wL}\left(\frac{w^2-w_o^2}{w_o^2}\right)$$
Now getting the impedance.
$$Z(j\omega) = \frac{1}{\frac{1}{R}+\frac{j}{wL}\left(\frac{w^2-w_o^2}{w_o^2}\right)}$$
$$Z(j\omega) = \frac{R}{1+\frac{jR}{wL}\left(\frac{w^2-w_o^2}{w_o^2}\right)}$$
Given that
$$Q = \frac{R}{wL}$$
$$Z(j\omega) = \frac{R}{1+jQ\left(\frac{w^2-w_o^2}{w_o^2}\right)}$$
I believe $$\frac{w^2-w_o^2}{w_o^2} \approx 2\frac{w-w_o}{w_o}$$
Therefore,
$$H(j\omega) = \frac{V_o}{I_{in}} = Z(j\omega) = \frac{R}{1+j2Q\left(\frac{w-w_o}{w_o}\right)}$$
Where $$H_o = R$$
Is this correct? And if so I have a question regarding the phase shift. I have seen this result given for the phase shift.
$$\alpha = \frac{\pi}{2} – \tan^{-1}\frac{L\omega\times \omega_o^2}{R\times (\omega^2-\omega_o^2)}$$
I can see how this could be the result but where does the \$\pi/2\$ come from?
Best Answer
I believe the calculation for the first formula goes something like:
Using \$\omega_0^2 = (LC)^{-1}\$:
$$\begin{align} Z(j\omega) &= \frac{1}{\frac{1}{R} + \frac{1}{j\omega L} + j\omega C} \\ &= \frac{j\omega RL}{j\omega L + R + j\omega RLC} \\ &= \frac{j\left(\frac{\omega}{\omega_0}\right)\omega_0RL}{j\left(\frac{\omega}{\omega_0}\right)\omega_0L+ R\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)} \\ &= \frac{j\omega_0RL}{j\omega_0L + R\left(\frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\right)} \end{align}$$
We can linearize \$f(\omega) = \frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\$ around \$\omega_0\$ using a Taylor expansion:
$$f(\omega) \approx f(\omega_0) + \left.\frac{df}{d\omega}\right|_{\omega=\omega_0}(\omega - \omega_0) + ...$$
You can then find that
$$f(\omega) \approx -\frac{2}{\omega_0}(\omega - \omega_0)$$
Allowing us to continue with \$Z(j\omega)\$:
$$\begin{align} Z(j\omega) &= \frac{j\omega_0RL}{j\omega_0L + R\left(\frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\right)} \\ &\approx \frac{j\omega_0RL}{j\omega_0L - \frac{2R}{\omega_0}(\omega - \omega_0)} \\ &= \frac{R}{1 + j2\frac{R}{\omega_0L}\frac{\omega - \omega_0}{\omega_0}} \end{align}$$
Now we just use the definition for the Q-factor for parallel RLC circuits:
$$Q = R\sqrt{\frac{L}{C}} = \frac{R}{\omega_0L}$$
Note that this is not the same as the Q-factor of an individual inductor or capacitor, which is usually defined as \$Q_C = \frac{1}{\omega_0CR_S}\$ or \$Q_L = \frac{\omega_0L}{R_S}\$ where \$R_S\$ is the series resistance.
Plugging this into the previous equation will result in the equation you found.
The phase of \$Z(j\omega)\$ is found by using:
$$\begin{align} \angle Z(j\omega) &= \tan^{-1}\left(\frac{\mathcal{I}\{Z(j\omega)\}}{\mathcal{R}\{Z(j\omega)\}}\right) \\ &= \frac{\pi}{2} - \tan^{-1}\left(\frac{\mathcal{R}\{Z(j\omega)\}}{\mathcal{I}\{Z(j\omega)\}}\right) \end{align}$$
The second one is just using the trigonometric identity:
$$\tan\left(\frac{\pi}{2} - \alpha\right) = \cot(\alpha) = \frac{1}{\tan(\alpha)}$$
They probably used this version for their calculations.