Chapter 12, Problem 12.77P

A feedback system has an amplifier with a low-frequency open-loop gain of $5\times {10}^4$ and has poles at ${10}^3\text{Hz},{10}^5\text{Hz},$ and ${10}^7\text{Hz}$ (a) Determine the frequency $f_{180}$ at which the phase of the loop gain is 180 degrees. (b) Determine the feedback transfer function $\beta$ for which the phase margin of the system is 45 ${}^{°}.$ (c) Using the results of part (b), determine the low-frequency closed-loop gain.

To determine

Value of frequency at the given values.

Answer to Problem 12.77P

Value of frequency is 1.005×10⁶Hz.

Explanation of Solution

Given information:

Given values are −

  $\begin{array}{l}\text{Open}\text{loop}\text{gain}\text{=}{\text{5×10}}^{\text{4}}\\ \text{Poles}\text{are}\text{at}{\text{10}}^3\text{Hz,}{\text{10}}^5\text{Hz}\text{and}{\text{10}}^7\text{Hz}\text{.}\end{array}$

The phase of loop gain is 180 degrees

Calculation:

Loop gain can be written as −

  $\begin{array}{l}{A}_v=\frac{5\times {10}^4}{\left(1+j\frac{f}{{10}^3}\right)\left(1+j\frac{f}{{10}^5}\right)\left(1+j\frac{f}{{10}^7}\right)}\\ \end{array}$

Phase of loop gain can be written as −

  $\begin{array}{l}{A}_v=\frac{5\times {10}^4}{\left(1+j\frac{f}{{10}^3}\right)\left(1+j\frac{f}{{10}^5}\right)\left(1+j\frac{f}{{10}^7}\right)}\\ \varphi \left(f\right)=-{\tan }^{-1}\left(\frac{f}{{10}^3}\right)-{\tan }^{-1}\left(\frac{f}{{10}^5}\right)-{\tan }^{-1}\left(\frac{f}{{10}^7}\right)\end{array}$

Value of frequency is −

  $-{\tan }^{-1}\left(\frac{f}{{10}^3}\right)-{\tan }^{-1}\left(\frac{f}{{10}^5}\right)-{\tan }^{-1}\left(\frac{f}{{10}^7}\right)=-{180}^o$

  $\text{We}\text{know}\text{that,}$

  ${\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

  $\text{Using}\text{these}\text{equations,}$

  ${\tan }^{-1}\left(\frac{f}{{10}^3}\right)+{\tan }^{-1}\left(\frac{f}{{10}^5}\right)+{\tan }^{-1}\left(\frac{f}{{10}^7}\right)={180}^o$

  ${\tan }^{-1}\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+{\tan }^{-1}\left(\frac{f}{{10}^7}\right)={180}^o$

  ${\tan }^{-1}\left(\frac{\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)}{1-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)\left(\frac{f}{{10}^7}\right)}\right)={180}^o$

  $\left(\frac{\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)}{1-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)\left(\frac{f}{{10}^7}\right)}\right)=\tan \left({180}^o\right)$

  $\left(\frac{\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)}{1-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)\left(\frac{f}{{10}^7}\right)}\right)=0$

  $\begin{array}{l}\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)=0\\ \left(\frac{f}{{10}^3}+\frac{f}{{10}^5}\right)+\left(\frac{f}{{10}^7}\right)\left(1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)\right)=0\\ \frac{10100f}{{10}^7}+\left(\frac{f}{{10}^7}\right)\left(1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)\right)=0\\ \left(\frac{f}{{10}^7}\right)\left[\left(10100\right)-\frac{f^2}{{10}^8}\right]=0\\ f^2=10100\times {10}^8\\ f=1.005\times {10}^6\text{Hz}\end{array}$

To determine

Value of ß at the given values.

Answer to Problem 12.77P

Value of ß is 2.83×10-3.

Explanation of Solution

Given:

Given values are −

  $\begin{array}{l}{A}_v=\frac{5\times {10}^4\beta }{\left(1+j\frac{f}{{10}^3}\right)\left(1+j\frac{f}{{10}^5}\right)\left(1+j\frac{f}{{10}^7}\right)}\\ \text{Phase}\text{margin}=45°\end{array}$

Calculation:

Value of phase angle of loop gain is −

  $\begin{array}{l}\text{Phase}\text{margin}=45°\\ \varphi \left(f\right)=\text{Phase}\text{margin}-180°\\ \varphi \left(f\right)=-135°\end{array}$

Magnitude of frequency is −

  $\begin{array}{l}-{\tan }^{-1}\left(\frac{f}{{10}^3}\right)-{\tan }^{-1}\left(\frac{f}{{10}^5}\right)-{\tan }^{-1}\left(\frac{f}{{10}^7}\right)=-135°\\ \text{It}\text{is}\text{known}\text{that}\\ {\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)\\ \text{Using}\text{these}\text{equations,}\\ {\tan }^{-1}\left(\frac{f}{{10}^3}\right)+{\tan }^{-1}\left(\frac{f}{{10}^5}\right)+{\tan }^{-1}\left(\frac{f}{{10}^7}\right)=135°\end{array}$

  $\begin{array}{l}\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)=-1\\ f^3+10.101\times {10}^6-1.0101\times {10}^{12}f-{10}^{15}=0\\ \text{After}\text{solving}\text{this}\text{equation,}\text{value}\text{of}\text{frequency}\text{is}\text{=}\\ f=100\text{kHz}\end{array}$

  ${\tan }^{-1}\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+{\tan }^{-1}\left(\frac{f}{{10}^7}\right)=135°$

  ${\tan }^{-1}\left(\frac{\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)}{1-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)\left(\frac{f}{{10}^7}\right)}\right)=135°$

  $\left(\frac{\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)}{1-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)\left(\frac{f}{{10}^7}\right)}\right)=\tan \left(135°\right)$

  $\left(\frac{\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)+\left(\frac{f}{{10}^7}\right)}{1-\left(\frac{\frac{f}{{10}^3}+\frac{f}{{10}^5}}{1-\left(\frac{f}{{10}^3}\right)\left(\frac{f}{{10}^5}\right)}\right)\left(\frac{f}{{10}^7}\right)}\right)=-1$

Value of ß is −

  $\begin{array}{l}{A}_v=\frac{5\times {10}^4}{\left(1+j\frac{f}{{10}^3}\right)\left(1+j\frac{f}{{10}^5}\right)\left(1+j\frac{f}{{10}^7}\right)}\\ \left|A_v\right|=\frac{5\times {10}^4}{\sqrt{\left(1+{\left(\frac{f}{{10}^3}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{{10}^7}\right)}^2\right)}}\\ 1=\frac{5\times {10}^4\beta }{\sqrt{\left(1+{\left(\frac{100\times {10}^3}{{10}^3}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{100\times {10}^3}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{100\times {10}^3}{{10}^7}\right)}^2\right)}}\\ 1=\frac{5\times {10}^4\beta }{100\times 1.414\times 1}\\ \beta =2.83\times {10}^{-3}\end{array}$

To determine

Value of low frequency closed loop gain.

Answer to Problem 12.77P

Value of low frequency closed loop gain is 350.87.

Explanation of Solution

Given:

Given values are −

  $\beta =2.83\times {10}^{-3}$

Calculation:

Value of closed loop gain is given by −

  $A_{vf}=\frac{5\times {10}^4}{1+5\times {10}^4\beta }$

Value of low frequency gain is −

  $\begin{array}{l}{A}_{vf}=\frac{5\times {10}^4}{1+5\times {10}^4\beta }\\ A_{vf}=\frac{5\times {10}^4}{1+5\times {10}^4\times 2.83\times {10}^{-3}}\\ A_{vf}=350.87\end{array}$