4
\$\begingroup\$

I'm trying to brush up on some electronics theory and am working through this PDF.

At the bottom of page F-6 there is a problem, F.1, that I am currently trying to solve. The circuit is as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

In order to find the transfer function, $$T(s) = \frac{Vo(s)}{Vi(s)}$$

I first obtained the following equivalent circuit:

schematic

simulate this circuit

where $$V'i = \frac{Vi*C1}{C1+C2}$$

From this I get the transfer function using the Laplace transform: $$T(s) = \frac{s}{s + \frac{1}{R(C1 + C2)}}$$

At this point I'm not entirely sure if I have done everything correctly so I would like some verification. If this is not correct then please let me know so that I can edit the question showing my math. This should help us narrow down to where the mistake is. I just don't want to post all of the math unless I absolutely need to.

The second part of the question asks if this is a single time constant circuit, which I expect it is because the circuit can be reduced to a single capacitor and a single resistor. The type would be high-pass.

The third part of the question says that for the element values shown, find the poles and zeros. It is clear from the transfer function that there is a zero at s = 0 rad/sec and a pole at $$\frac{1}{R(C1+C2)} = \frac{1}{100k*2*(0.5*10^{-6})} = 10$$ rad/sec.

Are these answers all correct, and are they complete? Have I missed something? I have not done this for some time so I am in dire need of a refresher course.

EDIT:

The last part of the problem asks to sketch the magnitude and phase response Bode plots. I am having some trouble with this part. I have the following for the magnitude:

enter image description here

where plot A is the final gain. Did I do this part correctly? As for the phase, I'm not even sure where to begin. I have that $$Φ = -tan^{-1}(\frac{w}{10})$$ and I believe that the s = 0 term starts us off at +90°, so that would give us a straight line on the plot of degrees vs. rad/sec. I don't know where to go from there. Some assistance on this part would be appreciated. What is the next step (to plot the phase of the pole)? I know the phase would drop but where the drop begins/ends I do not know.

I am happy to provide further clarification wherever necessary.

\$\endgroup\$
9
  • \$\begingroup\$ Looks good to me. The time constant is t = R*(C1+C2) because when we are looking from resistor point of view into the capacitors, we see two capacitors in parallel. \$\endgroup\$
    – G36
    Feb 19, 2017 at 19:33
  • \$\begingroup\$ Exactly, that's what I was thinking \$\endgroup\$
    – DerStrom8
    Feb 19, 2017 at 19:39
  • \$\begingroup\$ But if I was the pedantic person I will point out that the pole should be negative. \$\endgroup\$
    – G36
    Feb 19, 2017 at 19:47
  • \$\begingroup\$ I was wondering about that, the actual root would be s = -10, but I guess I don't know how you can have a negative frequency. Never did that before. \$\endgroup\$
    – DerStrom8
    Feb 19, 2017 at 19:55
  • 1
    \$\begingroup\$ In the s-plane, we can have a negative frequency. \$\endgroup\$
    – G36
    Feb 19, 2017 at 20:00

3 Answers 3

4
\$\begingroup\$

Using impedances (forgive the lack of standard forms along the way) and going long-hand from scratch, I get:

$$\begin{align*} \frac{V_O}{V_I}&= \frac{R\:\vert\vert\: C_2}{C_1+R\:\vert\vert\: C_2}\\\\ &=\frac{\frac{R}{1+s R C_2}}{\frac{1}{s C_1}+\frac{R}{1+s R C_2}}\\\\&=\frac{\frac{R}{1+s R C_2}}{\frac{1+s R C_2}{s C_1 \left(1+s R C_2\right)}+\frac{s R C_1}{s C_1 \left(1+s R C_2\right)}}\\\\ &=\frac{\frac{R}{1+s R C_2}}{\frac{1+s R\cdot\left(C_1+C_2\right)}{s C_1 \left(1+s R C_2\right)}}=\frac{R}{1+s R C_2}\cdot\frac{s C_1 \left(1+s R C_2\right)}{1+s R\cdot\left(C_1+C_2\right)}\\\\&=\frac{s R C_1}{1+s R\cdot\left(C_1+C_2\right)}=\frac{\frac{s C_1}{C_1+C_2}}{s+\frac{1}{R\cdot\left(C_1+C_2\right)}}\\\\&=\left[\frac{s}{s+\frac{1}{R\cdot\left(C_1+C_2\right)}}\right]\cdot\left[\frac{C_1}{C_1+C_2}\right] \end{align*}$$

So I guess I agree with your results (in the first part.)

\$\endgroup\$
1
  • \$\begingroup\$ Ah yes, thank you for posting the math. That is similar to what I did, though I think I did things in a slightly different order. I did drop the C1/(C1+C2) from the transfer function, however, and I think that was an error. I need to keep it as a multiplication of the "s" in the numerator because that will affect the Bode plot. There will be a constant of 0.5 multiplied by 's' which means a straight line at around -6dB. Is this correct? \$\endgroup\$
    – DerStrom8
    Feb 19, 2017 at 21:20
1
\$\begingroup\$

A contribution: Regarding to the math, I believe the easiest way to get the transfer function \$V_o(s)/V_i(s)\$ is to apply the KCL to the top node:

Dirceu Rodrigues Jr.

\$\endgroup\$
0
\$\begingroup\$

You pole and zero are correct, but the transfer function should be T(s) = sRC/(s2RC+1). Rearranged, your transfer function would be T(s) = s2RC/(s2RC +1). You didn't show all your steps, so I can't see where you went wrong.

\$\endgroup\$
4
  • \$\begingroup\$ If you divide by RC in the numerator and the denominator you end up with s/(s+1/2RC), or if the capacitors are different, s/(s+1/R(C1+C2)), which is what I have. You really want T(s) to be in the form s/(s+a) \$\endgroup\$
    – DerStrom8
    Feb 19, 2017 at 19:54
  • \$\begingroup\$ Intuitive, or slipshod analysis: at high frequency can the output equal the input, or should it be 1/2 the input? \$\endgroup\$
    – user128351
    Feb 20, 2017 at 1:15
  • \$\begingroup\$ I don't see how your transfer function is any different. s2RC/(s2RC+1), at higher frequencies, the '+1' is swamped out and since the values for s, R, and C are the same in the numerator and the denominator, it approaches 1/1 (Vo~Vi) \$\endgroup\$
    – DerStrom8
    Feb 20, 2017 at 2:24
  • \$\begingroup\$ My expression has sRC rather than s2RC in the numerator. \$\endgroup\$
    – user128351
    Feb 21, 2017 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.