Chapter 14, Problem 88A

(a)

To determine

The signals’ range of wavelengths.

Answer to Problem 88A

The required answer is for AM the wavelength range is $190\text{ m to }550\text{ m}$ and for FM the wavelength range is $\text{2}\text{.77 m to }3.4\text{ m}$ .

Explanation of Solution

Given:

The velocity of the signal, is $v=3\times {10}^8\text{ m/s}$ .

For AM waves,

Minimum frequencyis $f_1=550\text{ kHz = 550}\times {\text{10}}^3\text{ Hz}$ .

Maximum frequency $f_2=1600\text{ kHz =1600}\times {\text{10}}^3\text{ Hz}$

Formula used:

The relation between velocity and wavelength is,

  $v={\lambda }_1{f}_1$    .......(1)

  $v={\lambda }_2{f}_2$    .......(2)

Here, $v$ is the velocity of the signal, ${\lambda }_1$ is the wavelength of minimum frequency wave, ${\lambda }_2$ is the wavelength of maximum frequency wave, $f_1$ is the minimum frequency and $f_2$ is the maximum frequency.

Calculation:

Substituting $v=3\times {10}^8\text{ m/s}$ in $v$ , $\text{550}\times {\text{10}}^3\text{ Hz}$ in $f_1$ in equation (1) and $\text{1600}\times {\text{10}}^3\text{ Hz}$ in $f_2$ in equation (2),

  $\begin{array}{l}3\times {10}^8={\lambda }_1\times 550\times {10}^3\\ {\lambda }_1=550\text{ m}\end{array}$

And,

  $\begin{array}{l}3\times {10}^8={\lambda }_1\times 1600\times {10}^3\\ {\lambda }_1=190\text{ m}\end{array}$

The AM radio waves wavelengths ranges from 190 mto 550 m.

Conclusion:

Hence, the required answer is for AM the wavelength range is $190\text{ m to }550\text{ m}$ .

(b)

To determine

The range of FM wavelengths.

Answer to Problem 88A

The required answer is for AM the wavelength range is $190\text{ m to }550\text{ m}$ and for FM the wavelength range is $\text{2}\text{.77 m to }3.4\text{ m}$ .

Explanation of Solution

Given:

The velocity of the signal, $v=3\times {10}^8\text{ m/s}$ .

For FM waves,

Minimum frequency is $f_1=88\text{ MHz=88}\times {\text{10}}^6\text{ Hz}$ .

Maximum frequency is $f_2=108\text{ MHz=108}\times {\text{10}}^6\text{ Hz}$

Formula used:

The relation between velocity and wavelength is,

  $v={\lambda }_1{f}_1$    .......(1)

  $v={\lambda }_2{f}_2$    .......(2)

Here, $v$ is the velocity of the signal, ${\lambda }_1$ is the wavelength of minimum frequency wave, ${\lambda }_2$ is the wavelength of maximum frequency wave, $f_1$ is the minimum frequency and $f_2$ is the maximum frequency.

Calculation:

(b)Substituting $v=3\times {10}^8\text{ m/s}$ in $v$ , $\text{88}\times {\text{10}}^6\text{ Hz}$ in $f_1$ in equation (1) and $\text{108}\times {\text{10}}^6\text{ Hz}$ in $f_2$ in equation (2),

  $\begin{array}{l}3\times {10}^8={\lambda }_1\times 88\times {10}^6\\ {\lambda }_1=3.4\text{ m}\end{array}$

  $\begin{array}{l}3\times {10}^8={\lambda }_1\times 108\times {10}^6\\ {\lambda }_1=2.77\text{ m}\end{array}$

The FM radio waves wavelengths ranges from 2.77 m to 3.4 m.

Conclusion:

Hence, the required FM the wavelength range is $\text{2}\text{.77 m to }3.4\text{ m}$ .