Question

Consider two different radio station which broadcast signals whose angular frequency differs by f hertz. They are received by a receiver at origin. Determine the time for which receiver remains idle in each cycle if it can detect amplitude more that 1.5 A only.

Answer 1

Answer on Question #51292, Physics, Mechanics | Kinematics | Dynamics

Consider two different radio stations which broadcast signals whose angular frequency differs by $f$ hertz. They are received by a receiver at origin. Determine the time for which receiver remains idle in each cycle if it can detect amplitude more than 1.5 A only.

Solution:

Consider the superposition of two signals of different frequencies

$$A (t) = A _ {0} \cos (\omega_ {1} t) + A _ {0} \cos (\omega_ {2} t)$$

The addition of the two cosines now gives

$$A (t) = 2 A _ {0} \cos \left(\frac {\omega_ {1} t - \omega_ {2} t}{2}\right) \cos \left(\frac {\omega_ {1} t + \omega_ {2} t}{2}\right)$$

i.e.

$$A (t) = 2 A _ {0} \cos \left(\frac {(\omega_ {1} - \omega_ {2}) t}{2}\right) \cos \left(\frac {(\omega_ {1} + \omega_ {2}) t}{2}\right)$$

From given

$$\omega_ {1} - \omega_ {2} = f$$

Hence

$$A (t) = 2 A _ {0} \cos \left(\frac {f t}{2}\right) \cos \left(\frac {\left(\omega_ {1} + \omega_ {2}\right) t}{2}\right)$$

In this case $|\omega 1 - \omega 2| \ll (\omega 1 + \omega 2)$ the first cosine function is a slowly varying function of time, as compared with the rapidly varying second cosine function.

The displacement - time graph of this superposition of signals is shown in Figure for the case where the two equal amplitude original oscillations have frequencies in the ratio 13 : 11.

The slow periodic amplitude oscillations are referred to as beats. The beat frequency is equal to $|\omega_1 - \omega_2| / 2\pi$ .

The time for which receiver remains idle in each cycle if it can detect amplitude more than $1.5A_{0}$ only is in period where

$$\left| 2 A _ {0} \cos \left(\frac {f t}{2}\right) \right| \leq 1. 5 A _ {0}$$ $$\left| \cos \left(\frac {f t}{2}\right) \right| \leq 0. 7 5$$ $$- 0. 7 5 \leq \cos \left(\frac {f t}{2}\right) \leq 0. 7 5$$

From $\cos\left(\frac{ft}{2}\right) = -0.75$ we obtain

$$\frac{ft}{2} = \cos^{-1}(-0.75) = 2.419$$ $$t_{-0.75} = 2 * \frac{2.4189}{f} = \frac{4.8378}{f}$$

From $\cos\left(\frac{ft}{2}\right) = 0.75$ we obtain

$$t_{0.75} = 2 * \frac{0.7227}{f} = \frac{1.4454}{f}$$

Time is

$$t = t_{-0.75} - t_{0.75} = \frac{1}{f} (4.8378 - 1.4454) = \frac{3.3924}{f}$$

Answer: $t = \frac{3.3924}{f}$

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