4 Engineering Example 1

4.1 Feedback applied to an amplifier

Feedback is applied to an amplifier such that

$A^{\prime }=\frac{A}{1-\beta A}$

where $A^{\prime },A$ and $\beta$ are complex quantities. $A$ is the amplifier gain, $A^{\prime }$ is the gain with feedback and $\beta$ is the proportion of the output which has been fed back.

Figure 8 :

1. If at 30 Hz, $A=-500$ and $\beta =0.005{\text{e}}^{8\pi i∕9}$ , calculate $A^{\prime }$ in exponential form.
2. At a particular frequency it is desired to have $A^{\prime }=300{\text{e}}^{5\pi i∕9}$ where it is known that $A=400{\text{e}}^{11\pi i∕18}$ . Find the value of $\beta$ necessary to achieve this gain modification.

Mathematical statement of the problem

For (1): substitute $A=-500$ and $\beta =0.005{\text{e}}^{8\pi i∕9}$ into $A^{\prime }=\frac{A}{1-\beta A}$ in order to find $A^{\prime }$ .

For (2): we need to solve for $\beta$ when $A^{\prime }=300{\text{e}}^{5\pi i∕9}$ and $A=400{\text{e}}^{11\pi i∕18}$ .

Mathematical analysis

1. $A^{\prime }=\frac{A}{1-\beta A}=\frac{-500}{1-0.005{\text{e}}^{8\pi i∕9}\times \left(-500\right)}=\frac{-500}{1+2.5{\text{e}}^{8\pi i∕9}}$

   Expressing the bottom line of this expression in Cartesian form this becomes:

   $A^{\prime }=\frac{-500}{1+2.5cos\left(\frac{8\pi }{9}\right)+2.5isin\left(\frac{8\pi }{9}\right)}\approx \frac{-500}{-1.349+0.855i}$

   Expressing both the top and bottom lines in exponential form we get:

   $A^{\prime }\approx \frac{500{\text{e}}^{i\pi }}{1.597{\text{e}}^{i2.576}}\approx 313{\text{e}}^{0.566i}$

2. $A^{\prime }=\frac{A}{1-\beta A}\to A^{\prime }\left(1-\beta A\right)=A\to -\beta A{A}^{\prime }=A-A^{\prime }$

   i.e. $\beta =\frac{A^{\prime }-A}{A{A}^{\prime }}\to \beta =\frac{1}{A}-\frac{1}{A^{\prime }}$

   So

   $\beta =\frac{1}{A}-\frac{1}{A^{\prime }}=\frac{1}{400{\text{e}}^{11\pi i∕18}}-\frac{1}{300{\text{e}}^{5\pi i∕9}}\approx 0.0025{\text{e}}^{-11\pi i18}-0.00333{\text{e}}^{-5\pi i∕9}$

   Expressing both complex numbers in Cartesian form gives

   $\beta =0.0025cos\left(-\frac{11\pi }{18}\right)+0.0025isin\left(-\frac{11\pi }{18}\right)-0.00333cos\left(-\frac{5\pi }{9}\right)-0.00333isin\left(-\frac{5\pi }{9}\right)$

   $=-2.768\times 10^{-4}+9.3017\times 10^{-4}i=9.7048\times 10^{-4}{\text{e}}^{1.86i}$

   So to 3 significant figures $\beta =9.70\times 10^{-4}{\text{e}}^{1.86i}$

Exercises

1. Two standard identities in trigonometry are $sin2z\equiv 2sinzcosz$ and $cos2z\equiv {cos}^{2}z-{sin}^{2}z$ . Use Osborne’s rule to obtain the corresponding identities for hyperbolic functions.
2. Express $sinh\left(a+ib\right)$ in Cartesian form.
3. Express the following complex numbers in Cartesian form
   1. $3{\text{e}}^{i\pi ∕3}$
   2. ${\text{e}}^{-2\pi i}$
   3. ${\text{e}}^{i\pi ∕2}{\text{e}}^{i\pi ∕4}$ .
4. Express the following complex numbers in exponential form
   1. $z=2-i$
   2. $z=4-3i$
   3. $z^{-1}$ where $z=2-3i$ .
5. Obtain the real and imaginary parts of $sinh\left(1+\frac{i\pi }{6}\right)$ .

Answer