# $V_o/V_s$ of given circuit

I'm trying to calculate $$\frac{V_o}{V_s}$$ of the given circuit, but don't really know what to do with the voltage dependent voltage source. As I'm not an engineering student, this stuff is very foreign to me and I don't know how to approach a problem like this.

Also, what type of filter would this be? To me it looks like a high pass filter, but am not sure.

Any help is much appreciated!

• I...did not know you could use MathJax in the question title. – DKNguyen Nov 23 '19 at 20:14
• @DKNguyen aye. @mrMoonpenguin Best to use the inline version though ($zzz$ rather than $$zzz$$ ) – Tom Carpenter Nov 23 '19 at 20:20
• Oh right, my bad, didn't know that was a thing – user236803 Nov 23 '19 at 20:22
• This is a simple mesh analysis problem. You can look up what that process is. This circuit has two loops/meshes so you will have two equations with 2 unknowns. This should be kind of like a low pass filter since caps have low impedance for high frequencies so will have small voltage drop across the capacitor which means less is V is read by the dependent V source in the 2nd loop. Inductors also block high frequency changes – DKNguyen Nov 23 '19 at 20:54
• What does $V_1$ do with varying $V_s$? Is that a low-pass or a high-pass behavior? Then, whatever this is at $V_1$, another voltage source applies gain $A$ to it and this attempts to vary the voltage at one end of $L$, right? But $L$ has a current in it (doesn't matter the exact value, right now) which must be generating a voltage drop across $R_2$. When the dependent voltage source varies, and let's say it varies rapidly for now, how does the current in $L$ vary in response? Quickly? Or slowly? And how does that affect the drop across $R_2$? – jonk Nov 23 '19 at 21:22

Well, we have the following circuit:

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write:

$$\begin{cases} \text{I}_1+\text{I}_3=\text{I}_2\\ \\ \text{I}_3+\text{I}_4=\text{I}_0\\ \\ \text{I}_0+\text{I}_5=\text{I}_4\\ \\ \text{I}_2+\text{I}_5=\text{I}_1 \end{cases}\tag1$$

Using KVL, we can write:

$$\begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_0-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_0-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_4=\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{V}_0=\text{n}\text{V}_2 \end{cases}\tag2$$

Now, we can solve for:

$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_1}=\frac{\text{n}\text{R}_2\text{R}_3\text{R}_5}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_4+\text{R}_5\right)-\text{n}\text{R}_1\text{R}_2\text{R}_4}\tag3$$

Now, we transform to the s-domain (Laplace transform). We know that:

• $$\text{R}_2\to\frac{1}{\text{sC}}\tag4$$
• $$\text{R}_4\to\text{sL}\tag5$$

So, in the end we get:

$$\mathcal{h}\left(\text{s}\right):=\frac{\text{v}_3\left(\text{s}\right)}{\text{v}_1\left(\text{s}\right)}=\frac{1}{\text{sC}}\cdot\frac{\text{n}\text{R}_3\text{R}_5}{\text{R}_3\left(\text{R}_1+\frac{1}{\text{sC}}\right)\left(\text{sL}+\text{R}_5\right)-\frac{\text{n}\text{R}_1\text{L}}{\text{C}}}\tag6$$

Now, in order to see what kind of filter this is we need to look at the amplitude of the function when $$\\text{s}=\text{j}\omega\$$. So, first we have to expand:

$$\text{R}_3\left(\text{R}_1+\frac{1}{\text{j}\omega\text{C}}\right)\left(\text{j}\omega\text{L}+\text{R}_5\right)=\left(\text{R}_1\text{R}_3-\frac{\text{R}_3}{\omega\text{C}}\cdot\text{j}\right)\left(\text{j}\omega\text{L}+\text{R}_5\right)=$$ $$\text{R}_1\text{R}_3\text{j}\omega\text{L}+\text{R}_1\text{R}_3\text{R}_5-\frac{\text{R}_3\text{j}\omega\text{L}}{\omega\text{C}}\cdot\text{j}-\frac{\text{R}_3\text{R}_5}{\omega\text{C}}\cdot\text{j}=$$ $$\text{R}_3\left(\text{R}_1\text{R}_5+\frac{\text{L}}{\text{C}}\right)+\text{R}_3\left(\text{R}_1\omega\text{L}-\frac{\text{R}_5}{\omega\text{C}}\right)\text{j}\tag7$$

So, we get:

$$\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{1}{\omega\text{C}}\cdot\frac{\text{n}\text{R}_3\text{R}_5}{\sqrt{\left(\text{R}_3\left(\text{R}_1\text{R}_5+\frac{\text{L}}{\text{C}}\right)-\frac{\text{n}\text{R}_1\text{L}}{\text{C}}\right)^2+\left(\text{R}_3\left(\text{R}_1\omega\text{L}-\frac{\text{R}_5}{\omega\text{C}}\right)\right)^2}}\tag8$$

In your application, we know that $$\\text{R}_3\to\infty\$$. So we get:

$$\mathcal{X}\left(\omega\right):=\lim_{\text{R}_3\to\infty}\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{\text{n}\text{R}_5}{\sqrt{\left(\text{R}_5^2+\text{L}^2\omega^2\right)\left(1+\text{C}^2\text{R}_1^2\omega^2\right)}}\tag9$$

Now we look at two cases:

1. When $$\\omega\to0\$$: $$\lim_{\omega\to0}\mathcal{X}\left(\omega\right)=\text{n}\tag{10}$$
2. When $$\\omega\to\infty\$$: $$\lim_{\omega\to\infty}\mathcal{X}\left(\omega\right)=0\tag{11}$$

So we have a lowpass filter.

And we want to know the roll-off of this filter:

• $$\lim_{\omega\to\infty}\left(20\log_{10}\left(\mathcal{X}\left(10\omega\right)\right)-20\log_{10}\left(\mathcal{X}\left(\omega\right)\right)\right)=-40\space\text{dB/decade}\tag{12}$$
• $$\lim_{\omega\to\infty}\left(20\log_{10}\left(\mathcal{X}\left(2\omega\right)\right)-20\log_{10}\left(\mathcal{X}\left(\omega\right)\right)\right)=-40\log_{10}\left(2\right)\space\text{dB/octave}\tag{13}$$
• Wow that is so interesting, solving it like this would never have occurred to me, thanks a bunch! – user236803 Nov 24 '19 at 10:39
• @mrMoonpenguin You're welcome, I'm glad that I could help – Jan Nov 24 '19 at 10:45
• @mrMoonpenguin I edited my answer a bit. – Jan Nov 25 '19 at 22:36

Most of EE are too lazy to solve this circuit the way the @jan did (too much math).

This circuit can be solved just using the voltage divider equation. Because we have two voltages dividers.

One at the input (R1 and C) and the second one at the output (L and R2).

$$\frac{V_O}{V_S} = \frac{\frac{1}{s C}}{R_1 + \frac{1}{s C}}\cdot A \cdot \frac{R_2}{s L + R_2} =\frac{1}{1 +s C R_1} \cdot A \cdot \frac{1}{1 + s\frac{L}{R_2}}$$