Find the Zero of a simple circuit

Question

I would like to better understand how to find the zero of the transfer function of a circuit by inspection using fast analytical techniques.

^{simulate this circuit – Schematic created using CircuitLab}

For example if I have this simple circuit I know that in order to find the zero I have to fix 0V at the output and look for which\$\ s\$ the condition is verified. In this case to have \$\ V_{out}=0\$ I need that the impedance \$\ 1/sC + R_2 = 0\$ and this condition is verified only for \$\ s=-\frac{1}{CR_2}\$.

Now my problem is: if I have 0V at the output it means that the current through \$\ R_1\$ is \$\ \frac{V_{in}}{R_1}\$. Now this current must flow through the capacitor C and R2, and the voltage of node Vx will rise to \$\ V_x=V_{in}\frac{R_2}{R_1}\$. But in this case it means that I have a positive voltage drop across C from Vout to Vx and so the current through C must flow in the opposite direction with respect to our assumption. What is wrong with this reasoning?

Answer 1

Well, finding the zero by inspection requires some mathematical abstraction. When you perform harmonic analysis, you only consider the vertical axis and \$s=j \omega\$. When you inspect a circuit, you now consider the entire \$s\$-plane.

The definition of the zero in the transfer function is a value of \$s\$ for which the stimulus cannot propagate through the circuit to create a response. In other words, you excite your circuit at \$s=s_z\$ and the output is nulled, 0 V in ac. If I add a load to your circuit, you can see that to have an output null - which is not a short circuit but more like an op-amp virtual ground - you need to have \$i_{out}=0\$. If you have 0 A in a resistance, then the voltage across its terminal is also 0 V. How can you have 0 A flowing in \$R_3\$ despite a stimulus? Well, this happens if the series connection of \$R_2\$ and \$C_1\$ becomes a transformed short. In other words, solve:

\$Z_1(s)=R_2+\frac{1}{sC_1}=0\$ which leads to \$s_z=-\frac{1}{R_2C_1}\$ or \$\omega=|s_z|=\frac{1}{R_2C_1}\$.

If inspection does not work, you can apply a so-called null double injection or NDI. I have shown how to do it with SPICE in my book and it is very good to verify calculations. If you simulate the below circuit and calculate the resistance "seen" from the capacitor's terminals during the null, you will find the resistance value \$R_2\$:

SPICE confirms the 0-V output and the resistance computed by the B-source in the left returns 100 \$\Omega\$ which the resistance in series with \$C_1\$. Time constant in this mode is \$\tau=R_2C_1\$ and the inverse of it is the zero you want.

You can do a similar exercise with the below picture: where is the zero? Well, for what condition the stimulus won't propagate and form a response?

In case the impedance made of \$R_1\$ in parallel with \$C_1\$ approaches infinity: \$Z_1(s)=R_1||(\frac{1}{sC_1})\$ approaches infinity. This occurs when the denominator becomes zero. In other words, you find the pole of this particular expression (the root of the denominator) to obtain the zero of the whole transfer function. And this is \$\omega_z=\frac{1}{R_1C_1}\$. A quick simulation will confirm this value:

Answer 2

You are correct up to the point \$\ s=-\frac{1}{CR_2} \$, but you should have stopped there and expanded as follows; \$\ s=\omega j=-\frac{1}{CR_2}\implies\omega=\frac{j}{CR_2} \$ and then you should have realized that you were looking for a purely imaginary frequency, which has no real amplitude, ie. you were looking for \$\ V_{in}=0 \$ which happens to be the only valid solution to \$\ V_{out}=0 \$ in your problem.

This is easy to judge just by looking at the circuit, it is a first order low pass filter, and you have a resistor in series with the capacitor, you are never going to get a voltage lower than the input voltage \$\ V_{in} \$ multiplied by the voltage divider ratio between \$\ R_{1} \$ and \$\ R_{2} \$ as follows; \$\ V_{out}=\frac{R_2}{R_1+R_2}V_{in} \$ and \$\ R_2=0 \$ is not a valid solution.