Chapter 5.6, Problem 28E

FM Radio Signals The carrier wave for an FM radio signal is modeled by the function

$y=a\sin (2\pi (9.15\times {10}^7)t)$

where t is measured in seconds. Find the period and frequency of the carrier wave.

To determine

The value of Period, frequency for a carrier wave.

Answer to Problem 28E

The frequency and period of a carrier wave is $9.15\times {10}^7$ and $1.09\times {10}^{-8}\mathrm{s}$.

Explanation of Solution

Given:

The function for the carrier wave for an FM radio is,

$y=a\sin \left(2\pi \left(9.15\times \right){10}^7\right)t$ (1)

Where,

t is in seconds.

Calculation:

The function for the simple harmonic motion which describes the displacement y of an object at time t is,

$y=A\sin \omega t$ (2)

Compare the equation (1) and equation (2),

The values obtained are A is a and the values of $\omega$ is $2\pi \left(9.15\times \right){10}^7$.

Now, calculate the value of period,

The formula for period is,

$p=\frac{2\pi }{\omega }$ (3)

Substitute the values A for a and $2\pi \left(9.15\times \right){10}^7$ for $\omega$ in equation (3),

$\begin{array}{c}p=\frac{2\pi }{2\pi \left(9.15\times \right){10}^7}\\ =1.09\times {10}^{-8}\mathrm{s}\end{array}$

Thus, the period for the carrier wave of a FM radio is $1.09\times {10}^{-8}\mathrm{s}$.

Frequency is calculated by,

$f=\frac{1}{p}$ (4)

Where,

f is frequency of the object.

p is the period of the object.

Substitute $1.09\times {10}^{-8}\mathrm{s}$ for p in equation (4),

$\begin{array}{c}f=\frac{1}{1.09\times {10}^{-8}}\\ =9.15\times {10}^7\end{array}$

Thus the value of frequency is $9.15\times {10}^7$.