In the most simple terms, the police radar sends out a certain frequency. The higher (if the target is moving towards the LEO) the frequency returned, the faster they are moving

So the radar sends out a freq. A. It hits the target and gets returned as another freq. then the second return returns yet another frequency. So the higher the frequency between the first shot and the second, the higher the speed.

Am I close?

2. Yeah...

You have the radar wave. Think of it as on a freq graph, ie sine wave. As it hits the target, the target "smashes" it and compresses the wave in regards to time -> increased frequency.

3. Nice

4. ## Cool

Cool post asianfire. Haven't heard this discussed in some time.

5. You have the right idea Asianfire, in that a higher frequency (doppler tone), the higher the speed of the target. But to clarify how that doppler tone happens, when the radar signal bounces off a moving object, its frequency increases (if the target is approaching), or decreases (if the target is moving away). This is called the Doppler shift, and is very slight (only a few thousand hertz out of the tens of gigahertz of the original signal). But radar guns use a nifty, and downright simple technique to extract that doppler shift. The same antenna is used to simultaneously transmit and receive. The received signal (bounced back from the target) gets mixed with the transmitted signal, and the difference in frequency between these two signals causes a tone to be generated (similar to the whistle you get on an AM radio when it isn't perfectly tuned). The frequency of the tone (the Doppler tone) is a direct function of the original radar frequency and the speed of the target. Therefore, the gun simply measures the frequency of the doppler tone and converts it to a target speed.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•