Chapter 12, Problem 12.80P

Consider a feedback amplifier for which the open-loop gain is given by
$A(f)=\frac{2\times {10}^3}{\left(1+j\frac{f}{5\times {10}^3}\right){\left(1+j\frac{f}{{10}^5}\right)}^2}$

(a) Determine the frequency $f_{180}$ at which the phase of $A(f)$ is $-180$ degrees. (b) For $\beta =0.0045,$ determine the magnitude of the loop gain $T(f)$ at the frequency $f=f_{180}$ and determine the phase of $A(f)$ when $|T(f)|=1$ Determine the closed-loop, low-frequency gain. Is the system stable or unstable? (c) Repeat part (b) for $\beta =0.15$ .

To determine

The frequency $f_{180°}$ for the given specifications.

Answer to Problem 12.80P

The value of the voltage $f_{180°}$ is $100000.05\text{Hz}$

Explanation of Solution

Given:

The expression for the open loop gain of the amplifier is given by,

  $A\left(f\right)=\frac{2\times {10}^3}{\left(1+j\frac{f}{5\times {10}^3}\right){\left(1+j\frac{f}{{10}^5}\right)}^2}$

The phase of $A\left(f\right)=-180°$

Calculation:

The expression for the phase transfer function $A\left(f\right)$ is given by,

  $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{0}{2\times {10}^3}\right)-\left({\tan }^{-1}\left(\frac{\frac{f}{5\times {10}^3}}{1}\right)+{\tan }^{-1}\left(\frac{\frac{f}{{10}^5}}{1}\right)+{\tan }^{-1}\left(\frac{\frac{f}{{10}^5}}{1}\right)\right)\\ =-\left({\tan }^{-1}\left(\frac{f}{5\times {10}^3}\right)+{\tan }^{-1}\left(\frac{2f}{{10}^{10}-f^2}\right)\right)\end{array}$

Substitute $-180°$ for $\varphi$ in the above equation.

  $\begin{array}{l}-180°=-\left({\tan }^{-1}\left(\frac{f}{5\times {10}^3}\right)+{\tan }^{-1}\left(\frac{2f}{{10}^{10}-f^2}\right)\right)\\ 0=\left(\frac{1.00001\times {10}^{10}f-f^3}{5\times {10}^{13}f-5000f^2}\right)\\ f_{180}=100000.05\text{Hz}\end{array}$

The expression for the magnitude of the transfer

Conclusion:

Therefore, the value of the voltage $f_{180°}$ is $100000.05\text{Hz}$

To determine

The magnitude of the loop gain $T\left(f\right)$ at $f_{180°}$ and the phase gain for the given specifications.

To identify: Whether the system is stable or not.

Answer to Problem 12.80P

Thevalue of $A_f\left(0\right)$ for $100$ , $\left|T\left(f_{180°}\right)\right|$ is $0.225$ , $f$ is $38.793\text{kHz}$ and $\varphi$ is $-125.06°$ , the system is stable.

Explanation of Solution

Given:

The expression for the open loop gain of the amplifier is given by,

  $A\left(f\right)=\frac{2\times {10}^3}{\left(1+j\frac{f}{5\times {10}^3}\right){\left(1+j\frac{f}{{10}^5}\right)}^2}$

The value of $\left|T\left(f\right)\right|$ is $1$

Calculation:

The given expression for the transfer function is shown below,

  $T\left(f\right)=\frac{\left(\beta \right)\left(2\times {10}^3\right)}{\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{5\times {10}^5}\right)}^3\right)}}$ ……. (1)

Substitute $0.0045$ for $\beta$ and $100000.05\text{Hz}$ for $f_{180°}$ in the above equation.

  $\begin{array}{c}\left|T\left(f_{180°}\right)\right|=\frac{\left(0.0045\right)\left(2\times {10}^3\right)}{\sqrt{\left(1+{\left(\frac{100000.05\text{Hz}}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{100000.05\text{Hz}}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{100000.05\text{Hz}}{5\times {10}^3}\right)}^2\right)}}\\ =0.225\end{array}$

Substitute $1$ for $\left|T\right|$ and $0.0045$ for $\beta$ in equation (1).

  $\begin{array}{l}T\left(f\right)=\frac{\left(0.0045\right)\left(2\times {10}^3\right)}{\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{5\times {10}^5}\right)}^3\right)}}\\ f^6+2002510\times {10}^6{f}^4+1005\times {10}^{12}{f}^2=20\times {10}^{28}\\ f=38.793\text{kHz}\end{array}$

The expression for the phase transfer function is given by,

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

Substitute $38.793\text{kHz}$ for $f$ in the above equation.

  $\begin{array}{c}\varphi =\left(-{\tan }^{-1}\left(\frac{38.793\text{kHz}}{{10}^4}\right)+{\tan }^{-1}\left(\frac{38.793\text{kHz}}{{10}^5}\right)+{\tan }^{-1}\left(\frac{38.793\text{kHz}}{{10}^6}\right)\right)\\ =-125.06°\end{array}$

The expression for the closed loop low frequency gain is given by,

  $A_f\left(0\right)=\frac{A_O}{1+A_O\beta }$

Substitute $2\times {10}^3$ for $A_O$ and $0.045$ for $\beta$ in the above equation.

  $\begin{array}{c}{A}_f\left(0\right)=\frac{2\times {10}^3}{1+A_O\left(0.045\right)}\\ =100\end{array}$

The magnitude of the loop gain $T\left(f\right)$ at the frequency $f_{180}$

  $0.025$ and is less than unity therefore the system is stable.

Conclusion:

Therefore, the value of $A_f\left(0\right)$ for $100$ , $\left|T\left(f_{180°}\right)\right|$ is $0.225$ , $f$ is $38.793\text{kHz}$ and $\varphi$ is $-125.06°$ , the system is stable.

To determine

The magnitude of the loop gain $T\left(f\right)$ at $f_{180°}$ and the phase gain for the given specifications.

To identify: Whether the system is stable or not.

Answer to Problem 12.80P

Thevalue of $A_f\left(0\right)$ is $100$ , $\left|T\left(f_{180°}\right)\right|$ is $7.49$ , $f$ is $233.1\text{kHz}$ and $\varphi$ is $-222.33°$ , the system is unstable.

Explanation of Solution

Given:

The expression for the open loop gain of the amplifier is given by,

  $A\left(f\right)=\frac{2\times {10}^3}{\left(1+j\frac{f}{5\times {10}^3}\right){\left(1+j\frac{f}{{10}^5}\right)}^2}$

The $\left|T\left(f\right)\right|$ is $1$

Calculation:

The given expression for the transfer function is shown below,

  $T\left(f\right)=\frac{\left(\beta \right)\left(2\times {10}^3\right)}{\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{5\times {10}^5}\right)}^3\right)}}$ ……. (2)

Substitute $0.15$ for $\beta$ and $100000.05\text{Hz}$ for $f_{180°}$ in the above equation.

  $\begin{array}{c}\left|T\left(f_{180°}\right)\right|=\frac{\left(0.15\right)\left(2\times {10}^3\right)}{\sqrt{\left(1+{\left(\frac{100000.05\text{Hz}}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{100000.05\text{Hz}}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{100000.05\text{Hz}}{5\times {10}^3}\right)}^2\right)}}\\ =7.49\end{array}$

Substitute $1$ for $\left|T\right|$ and $0.15$ for $\beta$ in equation (1).

  $\begin{array}{l}T\left(f\right)=\frac{\left(0.15\right)\left(2\times {10}^3\right)}{\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{{10}^5}\right)}^2\right)}\sqrt{\left(1+{\left(\frac{f}{5\times {10}^5}\right)}^3\right)}}\\ f^6+2002510\times {10}^6{f}^4+1005\times {10}^{12}{f}^2=22499.75\times {10}^{28}\\ f=233.1\text{kHz}\end{array}$

The expression for the phase transfer function is given by,

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

Substitute $233.1\text{kHz}$ for $f$ in the above equation.

  $\begin{array}{c}\varphi =\left(-{\tan }^{-1}\left(\frac{233.1\text{kHz}}{{10}^4}\right)+{\tan }^{-1}\left(\frac{233.1\text{kHz}}{{10}^5}\right)+{\tan }^{-1}\left(\frac{233.1\text{kHz}}{{10}^6}\right)\right)\\ =-222.33°\end{array}$

The expression for the closed loop low frequency gain is given by,

  $A_f\left(0\right)=\frac{A_O}{1+A_O\beta }$

Substitute $2\times {10}^3$ for $A_O$ and $0.045$ for $\beta$ in the above equation.

  $\begin{array}{c}{A}_f\left(0\right)=\frac{2\times {10}^3}{1+A_O\left(0.045\right)}\\ =100\end{array}$

The magnitude of the loop gain $T\left(f\right)$ at the frequency $f_{180}$

  $7.49$ and is more than unity therefore the system is unstable.

Conclusion:

Therefore, the value of $A_f\left(0\right)$ is $100$ , $\left|T\left(f_{180°}\right)\right|$ is $7.49$ , $f$ is $233.1\text{kHz}$ and $\varphi$ is $-222.33°$ , the system is unstable.