Question: Use Eq. 9.19 to determine$A_3$and${\delta }_3$in terms ofrole="math" localid="1653473428327" $A_1$,$A_2$,${\delta }_1$, and${\delta }_2$.

Write the amplitude equation 9.19.

$A_3{e}^{i{\delta }_3}=A_1{e}^{i{\delta }_1}+A_2{e}^{i{\delta }_2}$

It is known that $e^{i\theta }=\cos \theta +i\sin \theta$. So, the equation (1) becomes,

$A_3\left(cos{\delta }_3+isin{\delta }_3\right)=A_1\left(cos{\delta }_1+isin{\delta }_1\right)+A_2\left(cos{\delta }_2+isin{\delta }_2\right)$

Solve for the real term.

$A_3\cos {\delta }_3=A_1\cos {\delta }_1+A_2\cos {\delta }_2$ ….. (2)

Solve for the imaginary term.

$A_3\sin {\delta }_3=A_1\sin {\delta }_1+A_2\sin {\delta }_2$ ….. (3)

Squaring equations (2) and (3) and add them.

$$\begin{gathered} A_3^2{\cos }^2{\delta }_3+A_3^2{\sin }^2{\delta }_3={\left(A_1\cos {\delta }_1+A_2\cos {\delta }_2\right)}^2+{\left(A_1\sin {\delta }_1+A_2\sin {\delta }_2\right)}^2 \\ {A}_3^2\left({\cos }^2{\delta }_3+{\sin }^2{\delta }_3\right)={\left(A_1\cos {\delta }_1+A_2\cos {\delta }_2\right)}^2+{\left(A_1\sin {\delta }_1+A_2\sin {\delta }_2\right)}^2 \\ {A}_3^2={\left(A_1\cos {\delta }_1+A_2\cos {\delta }_2\right)}^2+{\left(A_1\sin {\delta }_1+A_2\sin {\delta }_2\right)}^2 \\ {A}_3^2=\left[\left(A_1^2{\cos }^2{\delta }_1+A_2^2{\cos }^2{\delta }_2+2A_1{A}_2\cos {\delta }_1\cos {\delta }_2\right)+A_1^2{\sin }^2{\delta }_1+A_2^2{\sin }^2{\delta }_2+2A_1{A}_2\sin {\delta }_1\sin {\delta }_2\right] \end{gathered}$$

Simplify further.

$$\begin{gathered} A_3^2=A_1^2\left({\cos }^2{\delta }_1+{\sin }^2{\delta }_1\right)+A_2^2\left({\cos }^2{\delta }_2+{\sin }^2{\delta }_2\right)+2A_1{A}_2\left(\cos {\delta }_1\cos {\delta }_2+\sin {\delta }_1\sin {\delta }_2\right) \\ {A}_3^2=A_1^2\left(1\right)+A_2^2\left(1\right)+2A_1{A}_2\cos \left({\delta }_1-{\delta }_2\right) \\ {A}_3^2=A_1^2+A_2^2+2A_1{A}_2\cos \left({\delta }_1-{\delta }_2\right) \\ {A}_3=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \left({\delta }_1-{\delta }_2\right)} \end{gathered}$$

Therefore, the value of $A_3$is $A_3=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \left({\delta }_1-{\delta }_2\right)}$.

Divide equation (3) by (2).

$$\begin{gathered} \frac{A_3\sin {\delta }_3}{A_3\cos {\delta }_3}=\frac{A_1\sin {\delta }_1+A_1\sin {\delta }_2}{A_1\cos {\delta }_1+A_1\cos {\delta }_2} \\ \tan {\delta }_3=\frac{A_1\sin {\delta }_1+A_1\sin {\delta }_2}{A_1\cos {\delta }_1+A_1\cos {\delta }_2} \\ {\delta }_3={\tan }^{-1}\left(\frac{A_1\sin {\delta }_1+A_1\sin {\delta }_2}{A_1\cos {\delta }_1+A_1\cos {\delta }_2}\right) \end{gathered}$$

Therefore, the value of ${\delta }_3$is ${\delta }_3={\tan }^{-1}\left(\frac{A_1\sin {\delta }_1+A_1\sin {\delta }_2}{A_1\cos {\delta }_1+A_1\cos {\delta }_2}\right)$.