3
\$\begingroup\$

I'm trying to understand what the book "Basic Engineering Circuit Analysis" by Irwin is saying at the chapter of variable frequency...

It's trying to amplify a signal from 50 Hz to 20000 Hz. Where it uses this circuit to do that:

enter image description here

enter image description here

So the transfer function is this:

$$G_v(s)=\dfrac{R_{\text{in}}}{R_{\text{in}}+\frac{1}{sC_{\text{in}}}}\times 1000\times \dfrac{1/sC_{0}}{R_0+1/sC_{0}}$$

Which I understand.

Then it says that reorganizing and substituting this equation it yields:

$$G_v(s)=\dfrac{s}{s+100\pi}\times 1000\times \dfrac{40000\pi}{s+40000\pi}$$.

I'm lost here. I understand that it uses the domain frequency \$j\omega\$ instead of the Laplacian domain, and it should be replaced with \$\omega=2\pi f\$. But I don't understand the following:

  1. How does the first equation turn into the second one?

  2. Where did the complex \$j\$ go?

  3. Why does it replace the low frequency in the first term and high \$f\$ on the second?

And my last question within the same example: How is the transfer function approximated as $$G_v(s) \approx \left[\frac{s}{s}\right]1000\left(\frac{1}{1+0}\right) $$

when the frequency is between the low and high frequencies?

\$\endgroup\$
10
  • \$\begingroup\$ Sorry if someone is already typing but I just understood that the values of R and C are selected specifically to yield that result. That makes my questions: (1) and (3) solved. \$\endgroup\$ May 29 '19 at 21:21
  • \$\begingroup\$ I also understood the approximation. I'm not used to approximate in that way, it is still far from the "approximated" function written. \$\endgroup\$ May 29 '19 at 21:26
  • \$\begingroup\$ What are the remaining questions? I'm not sure. \$\endgroup\$
    – jonk
    May 29 '19 at 22:07
  • \$\begingroup\$ The (2). What happened to the complex \$"j"\$? \$\endgroup\$ May 29 '19 at 22:08
  • \$\begingroup\$ In what equation, exactly? I don't see j specifically located in any of them right now. Implied, perhaps. But not showing explicitly. And you've got several equations in the question. I could guess, but I'd rather just ask to be sure. (Note: \$s=\sigma+j\,\omega\$.) \$\endgroup\$
    – jonk
    May 29 '19 at 22:10
1
\$\begingroup\$

The standard form for a two-pole bandpass filter is:

$$G_s=\frac{K\,2\,\zeta\,\omega_{_0}\,s}{s^2+2\,\zeta\,\omega_{_0}\,s+\omega_{_0}^2}=\frac{K\,\frac{\omega_{_0}}{Q}\,s}{s^2+\frac{\omega_{_0}}{Q}\,s+\omega_{_0}^2}$$

\$K\$ is the gain. \$\zeta\$ is the damping factor (with \$Q=\frac{1}{2\,\zeta}\$, being the ratio of the center frequency to the \$-3\:\text{dB}\$ frequency.)

Your equation starts out as:

$$\begin{align*} G_s&=\frac{R_\text{IN}}{R_\text{IN}+\frac{1}{s\,C_\text{IN}}}\cdot 1000\cdot\frac{\frac{1}{s\,C_0}}{R_0+\frac{1}{s\,C_0}}\\\\ &=\frac{1000\cdot R_\text{IN}\,C_\text{IN}\,s}{R_\text{IN}\,C_\text{IN}\,R_0\,C_0\,s^2+\left(R_\text{IN}\,C_\text{IN}+R_0\,C_0\right)s+1}\tag{1}\\\\ &=\frac{1000\cdot\frac{1}{R_0\,C_0}\,s}{s^2+\frac{R_\text{IN}\,C_\text{IN}+R_0\,C_0}{R_\text{IN}\,C_\text{IN}\,R_0\,C_0}\,s+\frac{1}{R_\text{IN}\,C_\text{IN}\,R_0\,C_0}}\tag{2} \end{align*}$$

This isn't normalized. But it is getting closer.

Looking at either equation (1) or equation (2), the denominator can be expressed as \$b_2\,s^2+b_1\,s+b_0\$. We can always compute \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$2\,\zeta=\frac{b_1}{\sqrt{b_0\,b_2}}\$. (I'll leave it as an algebraic exercize to verify what I just wrote. But it's true.) In your case, \$\omega_{_0}=\frac{1}{\sqrt{R_\text{IN}\,C_\text{IN}\,R_0\,C_0}}\approx 6.286\:\text{k}\frac{\text{rad}}{\text{s}}\$ (or \$f_\text{C}\approx 1\:\text{kHz}\$) and \$\zeta=\frac{R_\text{IN}\,C_\text{IN}+R_0\,C_0}{2\sqrt{R_\text{IN}\,C_\text{IN}\,R_0\,C_0}}\approx 10\$. (Using your values of \$R_\text{IN}=1\:\text{M}\Omega\$, \$C_\text{IN}=3.18\:\text{nF}\$, \$R_0=100\:\Omega\$, and \$C_0=79.58\:\text{nF}\$.)

Given that \$Q\$ is the ratio relative to \$\omega_{_0}\$ for the \$-3\:\text{dB}\$ points, this means that the two corner frequencies are at \$f_\text{L}\approx 50\:\text{Hz}\$ and at \$f_\text{H}\approx 20\:\text{kHz}\$. (Obviously, the gain rises to \$K\approx 1000\$ in between these. Not quite, because \$K=\frac{1000}{1+\frac{R_0\,C_0}{R_\text{IN}\,C_\text{IN}}}\approx 997.5\$.)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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