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?


2 Answers 2


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}\$.

enter image description here

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\$: enter image description here

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?

enter image description here

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: enter image description here

  • \$\begingroup\$ If \$\ \omega=|s_z|=\frac{1}{R_2C_1} \$ let \$\ R_2=1M\Omega \$ and \$\ C_1=1\mu F \$ this gives \$\ \omega=2\pi f_z=1\implies f_z=\frac{1}{2\pi} \$.. so if C1 is 1uF and R2 is 1Mohm I should get a zero (no output stimulus), from this low pass filter, at approx. 1/6 of a Herz? I don't understand your reasoning. \$\endgroup\$
    – user173292
    Commented Jan 17, 2020 at 20:01
  • \$\begingroup\$ @Vinzent thats right. You would get a zero at ~ 1/6 Hz. \$\endgroup\$
    – efox29
    Commented Jan 17, 2020 at 20:27
  • \$\begingroup\$ @efox29 that doesn't make any sense, I can easily experimentally disprove that, but I wan't bother to because I know it just isn't true. You are basically claiming, since R1's value is irelevant I can set it to 1ohm, and if I apply 10V at ~1/6 Hz across 1Mohm and 1uF, through a 1ohm resistor, I will get 0V output across the 1Mohm and 1uF. I don't understand how you can't see that this doesn't make any sense. I know that I would get almost the full 10V signal at the output. \$\endgroup\$
    – user173292
    Commented Jan 17, 2020 at 20:34
  • \$\begingroup\$ @Vinzent I just simulated it in LTSPICE. I took your circuit, R1 = 20meg, C1 = 1u, R2 = 1meg. I set R1 = 20meg so as to not affect the zero, because the pole and the zero would interact if they are close to each other. With it, I can clearly see a zero (the AC response has the expected form, the attentuation of the zero aligns with calculations and the phase at the zero is -45 deg. why do you think a zero does not exist there ? Don't confuse zero with Vout = 0V. They are not the same thing. \$\endgroup\$
    – efox29
    Commented Jan 17, 2020 at 20:39
  • \$\begingroup\$ @Vinzent, your circuit has a pole and a zero. The zero is at 1/R2C and a pole at 1/C(R1+R2). If wp and wz are near each other, they will interact with each other and it may harder to see them. Which is why in my comment above, I kept them seperate. If I were to increase R1 and kept R2 the same, I would only be shifting the pole closer to 0 Hz. \$\endgroup\$
    – efox29
    Commented Jan 17, 2020 at 20:43

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.

  • \$\begingroup\$ Nice explanation \$\endgroup\$
    – Huisman
    Commented Jan 17, 2020 at 18:46
  • \$\begingroup\$ Thx, I hope it was useful (: \$\endgroup\$
    – user173292
    Commented Jan 17, 2020 at 18:49
  • \$\begingroup\$ I think The derivation provided by VerbalKint is correct. It shows via the Spice simulation that it is possible to find a voltage on a circuit equal to zero for a particular input 1 and for a particular input 2. This refers to the Null double injection theory and the Extra theorem element for finding transfer function. \$\endgroup\$
    – Jess
    Commented Jan 14, 2021 at 10:25

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