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 $$\frac{w^2-w_o^2}{w_o^2} \approx 2\frac{w-w_o}{w_o}$$


$$H(j\omega) = \frac{V_o}{I_{in}} = Z(j\omega) = \frac{R}{1+j2Q\left(\frac{w-w_o}{w_o}\right)}$$

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?

  • \$\begingroup\$ Do you realize it is a "90 degree phase shift" (I'm fairly certain, but been a minute since I did anything analog), and your question remains as to where the phase shift comes from? \$\endgroup\$ – CapnJJ Jun 12 '18 at 22:31
  • \$\begingroup\$ Yea I know pi/2 is 90 degrees. But what is producing it? My question is, is my analysis correct? And how to arrive with that phase shift result. \$\endgroup\$ – user367640 Jun 12 '18 at 23:03
  • \$\begingroup\$ \$Q=\frac {\omega_0 L}{R}\$ \$\endgroup\$ – Chu Jun 13 '18 at 7:41
  • \$\begingroup\$ This is the parallel resistance model, assuming the circuit operates at resonance. If you assume Q is greater than 3, Q = Rp/woL. Yours is the definition for series resistance. \$\endgroup\$ – user367640 Jun 13 '18 at 18:01

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.


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