# Sketch a Bode Plot for the Loop Gain of the circuit

Below I have the transfer function, but I am still uncertain about Bode plots. How do I know the slope, how do I know where the zeros of the function are? • Your equations are not correct from the start. This is a non inverting amplifier so use the correct 1 + Zf/Zc equation where Zf is the series impedance of R+sL. – Mike Oct 22 at 3:29
• To sketch, rather than plot, you just need the Laplace TF. There's a simple set of rules for sketching the asymptotic (log) gain, that's easily googled. – Chu Oct 22 at 6:07
• As a first approach, you can count the energy-storing elements with independent state variables. There are two, it is a second-order circuit. You can think of your circuit when these energy-storing elements are set in their dc ($s=0$) or in high-frequency states ($s$ approaches infinity): the inductor is replaced by a short circuit while the capacitor is open-circuited. Then the inductor is open-circuited while the capacitor is a short. What is the gain in both cases? This already gives you two asymptotes. But as pointed out by Mike, have a look at the non-inverting configuration first. – Verbal Kint Oct 22 at 6:48

Well, with a bit of reasoning we can find:

$$\text{V}_-=\text{V}_+\space\Longrightarrow\space\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=1+\frac{\text{R}+\text{sL}}{\left(\frac{1}{\text{sC}}\right)}=\text{CLs}^2+\text{CRs}+1\tag1$$

In order to draw the bode plots, we need to look at:

$$\underline{\mathcal{H}}\left(\omega\text{j}\right)=1-\text{CL}\omega^2+\text{CR}\omega\text{j}\tag2$$

Where $$\\text{j}^2=-1\$$.

So, for the bode plot we get:

• Amplitude: $$\left|\underline{\mathcal{H}}\left(\omega\text{j}\right)\right|=\sqrt{\left(1-\text{CL}\omega\right)^2+\left(\text{CR}\omega\right)^2}\tag3$$
• Argument: $$\arg\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=\begin{cases} 0,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space1-\text{CL}\omega=0\\ \\ \arctan\left(\frac{\text{CR}\omega}{1-\text{CL}\omega}\right),\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space1-\text{CL}\omega>0\\ \\ \frac{\pi}{2}+\arctan\left(\frac{\left|1-\text{CL}\omega\right|}{\text{CR}\omega}\right),\space\space\space\space\space\text{when}\space\space1-\text{CL}\omega<0 \end{cases}\tag4$$

And the slope you can find, using:

1. In terms of dB/octave: $$\lim_{\omega\to\infty}\left(20\log_{10}\left(\left|\underline{\mathcal{H}}\left(2\omega\text{j}\right)\right|\right)-20\log_{10}\left(\left|\underline{\mathcal{H}}\left(\omega\text{j}\right)\right|\right)\right)=$$ $$40\log_{10}\left(2\right)\approx12.0412\space\text{dB/octave}\tag5$$
2. In terms of dB/decade: $$\lim_{\omega\to\infty}\left(20\log_{10}\left(\left|\underline{\mathcal{H}}\left(10\omega\text{j}\right)\right|\right)-20\log_{10}\left(\left|\underline{\mathcal{H}}\left(\omega\text{j}\right)\right|\right)\right)=40\space\text{dB/decade}\tag6$$

What important is to notice, is that the slope is independent of the values of the components.