Chapter 11, Problem 38P

(a)

To determine

To Find: The wavelength of the AM signals.

Answer to Problem 38P

The wavelength of the AM signals of $550\text{kHz}$ is $545\text{ m}$ and wavelength of the AM signals of $1600\text{kHz}$ is $187.5\text{ m}$ .

Explanation of Solution

Given:

Lower frequency of AM signals, $f_1=550\text{ kHz}$

Higher value of frequency of AM signals, $f_2=1600\text{ kHz}$

Speed of wave, $v=3\times {10}^8\text{ m/s}$

Formula used:

The expression for wave speed.

  $v=f\times \lambda$

Here, $\lambda$ is the wavelength of the AM Signal.

  $v$ is speed.

  $f$ is frequency of signal.

Thus, wavelength can be written as:

  $\lambda =\frac{v}{f}$

Calculation:

Substitute the given values in the above equation,

For $550\text{kHz}$ , Wavelength

  ${\lambda }_1=\frac{3\times {10}^8}{550\times {10}^3}=545\text{ m}$

For $1600\text{kHz}$ , Wavelength

  ${\lambda }_2=\frac{3\times {10}^8}{1600\times {10}^3}=187.5\text{ m}$

Conclusion:

The wave length of the AM signals of $550\text{kHz}$ is $545\text{ m}$ and wave length of the AM signals of $1600\text{kHz}$ is $187.5\text{ m}$ .

(b)

To determine

To Find: The wavelength of the FM signals.

Answer to Problem 38P

The wave length of the FM signals of $88\text{ MHz}$ is $3.41\text{ m}$ and wave length of $108\text{ MHz}$ FM signals is $2.78\text{ m}$ .

Explanation of Solution

Given:

Lower frequency of AM signals, $f_1=550\text{ kHz}$

Higher value of frequency of AM signals, $f_2=1600\text{ kHz}$

Speed of wave, $v=3\times {10}^8\text{ m/s}$

Formula used:

The expression for wave speed.

  $v=f\times \lambda$

Here, $\lambda$ is the wavelength of the AM signals. Thus, wavelength can be written as:

  $\lambda =\frac{v}{f}$

Calculation:

Substitute the given values in the above equation,

For $88\text{ MHz}$ , Wave length ${\lambda }_1=\frac{3\times {10}^8}{88\times {10}^6}=3.41\text{ m}$

For $108\text{ MHz}$ , Wave length ${\lambda }_2=\frac{3\times {10}^8}{108\times {10}^6}=2.78\text{ m}$

Conclusion:

Thus, the wave length of the FM signals of $88\text{ MHz}$ is $3.41\text{ m}$ and wave length of $108\text{ MHz}$ FM signals is $2.78\text{ m}$ .