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I have the following problem:

Consider the ciruit and problems below.

enter image description here

The head of the voltage arrow denotes the \$-\$-pole of the voltage drop.

Okay, for the first problem I just used the simple formula for DC-gain for an op-amp modelled like this. I arrived at:

\$A_v = \frac{R_2}{R_1}=\frac{220 \text{k} \Omega}{51 \text{k} \Omega}=-4.31 \frac{V}{V}\$

But for problem 4.2, I'm not sure what I'm supposed to do. If it asked for the time constant for the capacitor in the circuit I might be able to figure out an answer. But they are asking for the time constant for the pole/zero for the filter.

Does a pole/zero have a pole? And if so, how do you the time constant.

I hope someone can help me with this.

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    \$\begingroup\$ Poles and zeros don't inherently have time constants (unless I'm mistaking something). The circuit has a zero but no poles. The time constant is RC where R = 51 kohm. It doesn't seem like a well-worded problem to me. \$\endgroup\$ – Andy aka Apr 22 '20 at 10:33
  • \$\begingroup\$ @Andyaka Yeah, that's a common theme in my instructors problem formulation I'm afraid. If the time constant is what you say, then I get a result of \$ \tau = RC = 19.89 \text{ms}\$ \$\endgroup\$ – Carl Apr 22 '20 at 10:47
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The pole of a 1st-order system is the inverse of its natural time constant obtained when the excitation is removed. Here, if you temporarily remove the cap. and "look" through its terminals while \$V_{in}\$ is zeroed, then because of the virtual ground, the time constant is zero.

enter image description here

So \$D(s)=1+s\tau=1\$.

Regarding the numerator, if you bring the stimulus back in place, what condition would prevent the stimulus from generating an ac signal on the output? In other words, what condition nulls the output? Look at the below circuit, it is when the impedance made of \$C_1\$ paralleled with \$R_1\$ approaches infinity:

enter image description here

It becomes infinite at the pole frequency of this impedance. The time constant is \$R_1C_1\$. Therefore, \$N(s)=1+s\tau_N=1+\frac{s}{\omega_z}\$ with \$\omega_z=\frac{1}{R_1C_1}\$.

The final transfer function is thus defined as \$H(s)=H_0(1+s\tau_N)=H_0(1+\frac{s}{\omega_z})\$ with \$H_0=-\frac{R_2}{R_1}\$

In my opinion, describing transfer functions via the time constants concept represents the best way as you can later infer or sense where the poles and zeroes are located when inspecting an electrical diagram whether it is passive or active. I have described the fast analytical techniques or FACTs in a book I published in 2016 but you can have a look at the seminar I taught at APEC in 2016, it is a quick introduction to the subject. If your teacher has the FACTs in mind, he must be encouraged to pursue this path!

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  • \$\begingroup\$ Verbal Kint ...I had a look at your seminar slides...great work. Gratulation! \$\endgroup\$ – LvW Apr 25 '20 at 15:48
  • \$\begingroup\$ Danke viel mal LvW :-) I am finishing a new book on transfer functions of switching converters and the next one will target the FACTs explained to students in a crash course on transfer functions of linear circuits. \$\endgroup\$ – Verbal Kint Apr 25 '20 at 16:02
  • \$\begingroup\$ Very nice analysis. \$\endgroup\$ – Leoman12 Apr 25 '20 at 16:17
  • \$\begingroup\$ Thank you Leoman12! The analysis based on time constants is truly intuitive, that is what I liked when I acquired the skill. Also, if you look at some other contributions I did, you can combine the analysis with a SPICE simulator and it really helps solve complicated circuits step by step. If you made a mistake at some point, you can easily correct the intermediate step and don't need to restart from scratch. Really cool. Try the FACTs and you won't back to classical analysis : ) \$\endgroup\$ – Verbal Kint Apr 25 '20 at 16:28
  • \$\begingroup\$ @Verbal Kint...what about writing a book or paper on transfer functions of switched-capacitor circuits? I have done some work on this subject already....(however, it is in German) \$\endgroup\$ – LvW Apr 26 '20 at 9:17

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