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Just started on the topic Circuit Analysis, and am supposed to find the transfer function of the following circuit:

enter image description here

So I know I am supposed to use KVL to solve this problem. I can first transform the circuit to: enter image description here

From the circuit, I can see that because u*Vc is in parallel with the output:

$$V_o(s) = uV_c(s)$$ $$V_c(s)= \frac{V_o(s)}{u}$$

Using KVL,

$$V_i(s) = (I_1*R)+(I_3*R)+V_c(s)$$

where: $$V_c(s) = \frac{V_o(s)}{u}$$

Now the problem is, how do I exactly get I1 and I2 from the circuit?

New to this, thanks :)

EDIT: I crafted another KVL equation, as well as being able to determine I3. $$V_i(s) = (I_1*R)+(I_2*1/sc)+V_0(s)$$ $$I_3(s) = \frac {V_c(s)}{1/sc}$$

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  • \$\begingroup\$ I would recommend to start by properly labeling all these elements like \$R_1\$, \$C_1\$, \$C_2\$ and so on. \$\endgroup\$ Jun 1, 2020 at 9:04

2 Answers 2

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enter image description here

First thing I did was redraw the circuit. I added a node 'Vx' to help me solve.

(1) $$I_1 = I_2 + I_3$$

(2) $$I_1 = \frac{V_i - V_x}{R}$$

(3) $$I_2 = (V_x - V_o) * sC$$

(4) $$I_3 = \frac{V_x - V_o/u}{R}$$

Plugging equations 2,3,4 into 1 gives:

(5) $$\frac{V_i - V_x}{R} = (V_x - V_o) sC +\frac{V_x - V_o/u}{R}$$

(6) $$V_i = V_x (sCR + 2) - V_o(sCR + 1/u)$$

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Now let's get Vx in terms of Vo. I do this using I_3

$$I_{3 (R)} = I_{3 (C)}$$

$$\frac{V_x - V_o/u}{R} = (V_o/u) * sC$$

$$V_x = V_o * \frac{sCR}{u} + \frac{V_o}{u}$$

$$V_x = V_o * \frac{sCR+1}{u}$$

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Now, I have the rest done, but believe that you are capable of taking it from here. I think your roadblock was using Vx, which helped me get I1, I2, and I3.

Check your work by setting 'u' equal to 3, you should see some major magnitude peaking. For picture below I used R = 10kΩ, C = 1uF, u = 3.

enter image description here

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  • \$\begingroup\$ Hi there :) Thanks for the detailed solution :) Can I ask when will I need to use nodal analysis (as per your solution), instead of other methods? Is nodal analysis always the preferred way to go? \$\endgroup\$
    – Jisbon
    Jun 1, 2020 at 13:36
  • \$\begingroup\$ @Jisbon Some circuits (especially cascaded filters) you don't need to start with nodal analysis. For example, an inverting op-amp configuration, I know the gain is -Zf / Zi (which comes from nodal analysis), so if I find the complex impedances I can find the transfer function. There is normally also some intuition involved. With your circuit I was expecting some sort of lowpass response based on the two resistors and capacitor to ground. From this I knew the DC gain was 'u' and high frequency gain -> 0. The feedback cap made me have to bust out node equations, but I'm sure someone here \$\endgroup\$ Jun 1, 2020 at 15:34
  • \$\begingroup\$ has some fast analytical techniques for this problem. For me at least, using nodal methods was the way to go here. See the answer to this question, where a user breakdown a fairly complicated circuit into building blocks, no nodal analysis needed. electronics.stackexchange.com/questions/421871/… \$\endgroup\$ Jun 1, 2020 at 15:36
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You can indeed apply the fast analytical circuits techniques or FACTs as described in my book. For a simple circuit like this, three individual sketches are enough to obtain the three time constants you need:

enter image description here

For the dc gain, open all capacitors and you see that \$H_0=A_{OL}\$ the open-loop gain of your voltage-controlled source. Then, reduce the input voltage to 0 V (replace it by a short circuit) and "look" through each capacitor to determine the resistance \$R\$ offered by their connecting terminals. That resistance multiplied by the capacitor will for the time constants we need, \$\tau_1=R_aC_1\$ and \$\tau_2=R_bC_2\$. In the last sketch, you "look" through \$C_2\$'s terminals while \$C_1\$ is a short circuit. When done, you immediately form the denominator as \$D(s)=1+s(\tau_1+\tau_2)+s^2(\tau_1\tau_{12})\$ and a transfer function of the form: \$H(s)=H_0\frac{1}{D(s)}\$. The below Mathcad sheet shows the resulting values with the arbitrary values taken from the previous answer:

enter image description here

As you can see, your configuration leads to two right-half-plane poles which are indicative of an unstable open-loop time-domain response (you can observe a phase lead for a two-pole denominator which is indicative of RHPPs presence). If the gain of the controlled source increases up to a high value, the lower capacitor \$C_1\$ creates a pole located at a very high frequency (because of the virtual ground) and the system turns into an integrator with a pole located close to the origin but in the RHP, again quite unusual. They would return in the left-half plane if the gain of the controlled source becomes negative though.

The below SPICE sim confirms the equation-based graphs:

enter image description here

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