How Does the Doppler Effect Alter Radar Frequencies?

In summary, the frequency at the target vehicle (f.tar) is equal to ((c-u)/c)*f.sou, the frequency of the reflected radiation (f.ref) at the radar gun is equal to (c/(c+u))*f.tar, and the simplified expression relating the initial frequency (f.sou) to the reflected frequency (f.ref) is f.ref = (c/(c+u))*((c-u)/c)*f.sou. However, this expression is incorrect as it does not account for the case when u=0, which would result in f.ref = f.sou.
  • #1
Lengalicious
163
0

Homework Statement


Radar is used to measure the speed of passing vehicles. A stationary radar gun emits EM radiation, a proportion of which is reflected from the target vehicle which is moving away from the gun at 'u' ms^-1.

a)State the frequency at the target vehicle (f.tar) in terms of the frequency at the radar gun (f.sou).

The target vehicle then acts as a moving source emitting at frequency (f.tar), moving away from the radar gun.

b) State the frequency of the reflected radiation (f.ref) at the radar gun in terms of the frequency reflected by the target vehicle (f.tar).

c) Combine these expressions to relate the initial frequency (f.sou) to the reflected frequency (f.ref), simplifying the expression by assuming u<<c.


Homework Equations





The Attempt at a Solution



I say f.tar = ((c-u)/c)*f.sou
and f.ref = (c/(c+u))*f.tar

so, f.ref = (c/(c+u))*((c-u)/c)*f.sou

where f.ref = f.sou when u<<c. This surely can't be correct? What am i doing wrong?? Apparently there should be a beat frequency =S
 
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  • #2
Lengalicious said:
where f.ref = f.sou when u<<c. This surely can't be correct? What am i doing wrong?? Apparently there should be a beat frequency =S


Why is this incorrect? In the case that u=0, you would expect f.ref = f.sou. If there's no relative velocity there is no doppler shift.
 

Related to How Does the Doppler Effect Alter Radar Frequencies?

1. What is the Doppler effect for radar guns?

The Doppler effect for radar guns is a phenomenon in which the frequency of a radar signal changes when it is reflected off of a moving object. This change in frequency is caused by the relative motion between the radar gun and the object, and can be used to determine the speed of the object.

2. How does the Doppler effect work in radar guns?

The radar gun emits a high-frequency signal, which bounces off of the moving object and returns to the gun. As the object moves towards or away from the gun, the frequency of the returning signal changes due to the Doppler effect. The radar gun measures this change in frequency and uses it to calculate the speed of the object.

3. What factors can affect the accuracy of a radar gun's speed measurement?

There are several factors that can affect the accuracy of a radar gun's speed measurement. These include the angle at which the radar beam hits the object, the weather conditions (such as rain or fog), and the size and shape of the object being measured. Additionally, any interference from other objects or signals can also impact the accuracy of the measurement.

4. How is the Doppler effect used in other applications besides radar guns?

The Doppler effect is used in a variety of other applications besides radar guns. It is commonly used in weather forecasting to track the movement of storms and measure wind speeds. It is also used in medical imaging, such as ultrasound, to measure blood flow and detect abnormalities. Additionally, the Doppler effect is used in astronomy to measure the speed and distance of stars and other celestial objects.

5. How do radar guns compensate for the Doppler effect when measuring a moving object?

Radar guns have built-in algorithms that can compensate for the Doppler effect when measuring a moving object. These algorithms take into account the speed and direction of the radar gun and the object, as well as any other factors that may affect the accuracy of the measurement. By using these calculations, the radar gun is able to provide a more accurate speed reading despite the Doppler effect.

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