# Variable Frequency-Response analysis (amplifier)

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:

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?

• 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. May 29 '19 at 21:21
• I also understood the approximation. I'm not used to approximate in that way, it is still far from the "approximated" function written. May 29 '19 at 21:26
• What are the remaining questions? I'm not sure.
– jonk
May 29 '19 at 22:07
• The (2). What happened to the complex $"j"$? May 29 '19 at 22:08
• 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: $s=\sigma+j\,\omega$.)
– jonk
May 29 '19 at 22:10

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.)

\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*}
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\$$.)