Chapter 12, Problem 24P

(a)

To determine

Prove ${\beta }_2={\beta }_1+10\text{dB}$.

Answer to Problem 24P

The value is $\underset{\_}{{\beta }_2={\beta }_1+10\text{dB}}$.

Explanation of Solution

Write the equation of intensity.

$\begin{array}{l}\ln \beta =\left(10\text{dB}\right)\log \left(I/I_o\right)\\ \frac{I}{I_o}={10}^{\left(\beta \right)/\left(10\text{dB}\right)}\\ I=I_o{10}^{\left(\beta \right)/\left(10\text{dB}\right)}\end{array}$ (I)

Here, $I$ is the intensity, $I_o$ is the initial intensity.

Write the given equation,

$I_2=\left(10.0\right)I_1$ (II)

Conclusion:

Substitute, $I_o{10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}$ for $I_2$, $I_o{10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}$ for $I_1$ in equation (II)

$\begin{array}{l}{I}_o{10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}=\left(10.0\right)I_o{10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}\\ {10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}=\left(10.0\right){10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}\\ log{10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}=\log \left(10.0\right)+\log {10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}\\ {\beta }_2={\beta }_1+\left(10.0\text{dB}\right)\end{array}$

Thus, the value is $\underset{\_}{{\beta }_2={\beta }_1+10\text{dB}}$.

(b)

To determine

Prove ${\beta }_2={\beta }_1+\left(3.0\right)\text{dB}$.

Answer to Problem 24P

The value is $\underset{\_}{{\beta }_2={\beta }_1+\left(3.0\right)\text{dB}}$.

Explanation of Solution

Write the equation of intensity.

$\begin{array}{l}\ln \beta =\left(10\text{dB}\right)\log \left(I/I_o\right)\\ \frac{I}{I_o}={10}^{\left(\beta \right)/\left(10\text{dB}\right)}\\ I=I_o{10}^{\left(\beta \right)/\left(10\text{dB}\right)}\end{array}$ (I)

Here, $I$ is the intensity, $I_o$ is the initial intensity.

Write the given equation,

$I_2=\left(2.0\right)I_1$ (II)

Conclusion:

Substitute, $I_o{10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}$ for $I_2$, $I_o{10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}$ for $I_1$ in equation (II)

$\begin{array}{l}{I}_o{10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}=\left(2.0\right)I_o{10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}\\ {10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}=\left(2.0\right){10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}\\ log{10}^{\left(\frac{{\beta }_2}{10\text{dB}}\right)}=\log \left(2.0\right)+\log {10}^{\left(\frac{{\beta }_1}{10\text{dB}}\right)}\\ {\beta }_2={\beta }_1+\left(3.0\text{dB}\right)\end{array}$

Thus, the value is $\underset{\_}{{\beta }_2={\beta }_1+\left(3.0\right)\text{dB}}$.