Question

To catch speeders, a police radar gun detects the beat frequency between the signal it emits and that which reflects off a moving vehicle. What would be the beat frequency for an emitted signal of 900 Mhz reflected from a car moving at 30 m/s ?

Short answer

The beat frequency between the Doppler shifted frequency and original frequency is$180\mathrm{Hz}$ .

Step-by-step solution

Write the given data from the question

The speed of the car,$v_s=30m/s$

The frequency of the emitted signal, f=900Mhz

Beat frequency

The expression for the Doppler shift is given as follows.

$\frac{\mathbf{∆}\mathbf{f}}{\mathbf{f}}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{5}}}{\mathbf{c}}$

Here, ${\mathbf{v}}_{\mathbf{5}}$ is the speed of the source, c is the speed of the light and$\mathbf{∆}\mathbf{f}$ is the beat frequency.

Calculate the beat frequency for the emitted signal

To calculate the beat frequency, equation for beat frequency and Doppler shift required.

$\frac{∆f}{f}=\frac{2V_{target}}{c}$ …… (i)

Calculate the beat frequency,

Substitute $3\times {10}^8\mathrm{m}/\mathrm{s}\mathrm{for}\mathrm{c},900\mathrm{MHz}\mathrm{for}\mathrm{f}\mathrm{and}30\mathrm{m}/\mathrm{s}\mathrm{for}{\mathrm{V}}_{\mathrm{s}}$into equation (i).

$$\begin{gathered} \frac{∆f}{900\times {10}^6}=\frac{2\times 30}{3\times {10}^8} \\ ∆f=\frac{60}{3\times {10}^8}\times 900\times {10}^6 \\ ∆f=\frac{540}{3} \\ ∆f=180Hz \end{gathered}$$

Hence the beat frequency between the Doppler shifted frequency and original frequency is 180 Hz .