# Modulated Light and Doppler effect

1. May 13, 2008

### Deane

I'm trying to detect a remote moving target. I know that if I send out a light beam against a distant target that is moving towards or away from me that the reflected light will be frequency shifted by a small amount. Now what if I amplitude modulated the light with an audio frequency. Would that frequency also be changed by the Doppler effect? If yes, what is the formula to determine the amount of shift of the audio tone? Thanks. (I have been trying to find this answer for a while.)

2. May 13, 2008

### pzlded

Reflection does not change light velocity, thus AM frequency should be unchanged. AM radio station signals that bounce off the ionosphere are intelligible.

When coherent light bounces off a distant object, the reflected light is slightly out of phase from the incident light. The greater the phase shift, the greater the distance to the object.

Object speed is determined by comparing the object's distance taken at one time with the object's distance taken at another time.

Once superconductive circuits are available within chips, phase shift determinations may be more accurate.

3. May 13, 2008

### Deane

Thanks for thinking about my question. However I really want to know if the reflected wave's modulation is subject to the Doppler effect. Let me give an example. Assume an object is stationary and I am travelling towards the object. I transmit a continuous light beam that is modulated in intensity by a 1000 Hz. signal at the object and I monitor the reflected light.
I am pretty sure that there will be a Doppler shift of the modulating frequency but I am not sure. If this is true, what formula do I use to determine the frequency shift?

4. May 13, 2008

### pzlded

Think of photons as raindrops. The faster a car travels through the drops, the greater the number of drops that hit the windshield.

Suppose that a detector that moves at the speed of light toward the light source; will detect double AM emission frequency?

5. May 13, 2008

### Redbelly98

Staff Emeritus
Can you explain the basis for the conclusion in that sentence? It seems to me that the AM modulation frequency would be Doppler shifted.

Succesive peaks of the AM modulated signal will travel different round-trip distances, so the period of the return signal will be different than that of the outgoing signal.

6. May 13, 2008

### pzlded

Sorry, I hit the send button too early; leaving a half baked entry. The last sentence should be:

"Suppose that a detector moves at the speed of light toward a light source. Will the detected AM frequency be doubled?"

In the case of a detector moving at light speed, there are no round trips from the detector to the light source, then back to the detector. A photon from the detector will arrive at the same time as the detector.

7. May 14, 2008

### swapnilster

In case of reflection and refraction of light,its the wavelength of light and not the frequency that changes along with the velocity.May be anyone could look at the problem from that angle.

8. May 14, 2008

### Redbelly98

Staff Emeritus
I think we are talking strictly in-air propagation in this discussion, weren't we? (Correct me if that's wrong.) The velocity of the light may safely be treated as a constant, and the object being measured is nonrelativistic.

For moving targets, wavelength and frequency will change upon reflection. I can explain why in more detail if anybody wishes.

9. May 15, 2008

### Deane

Yes, this is not a trick question and has real world application. Classical physics only. Light traveling through air, modulated with a 1 KHz. signal. What is the formula for the Doppler Effect and does it apply to the modulated echo?

10. May 15, 2008

### Redbelly98

Staff Emeritus
Yes, it applies to the modulation as well as the carrier frequency. Suppose for argument's sake that the peaks in the modulation occur every 1 million cycles of the carrier. This relation will not change upon reflection, so any shift in carrier means a shift in modulation as well.

To derive the value of the shift (skip to end if you just want the final result), first take the reference frame of the "moving" object. Here the incident and reflected signals both have the same frequency, since reflection from a stationary object will not shift the frequency.

However, in the measurement frame where the object is moving, the incident and reflected beams will have equal magnitude, but opposite sign, frequency shifts. Suppose the object is moving toward you. Then from the object frame's viewpoint, you are now moving in the same direction of the incident beam, and in the opposite direction as the reflected beam. Hence a red-shift for incident, and blue-shift for reflected beam frequencies.

From the observer's viewpoint, there is a doubling of the shift compared to the situation where the moving object simply emits a beam at a given frequency:

$$\frac{\Delta f}{f}=2\frac{v}{c}$$

where we are ignoring terms of order $$(v/c)^2$$

For an object travelling 100 mph, I calculate a 0.3 ppm (parts-per-million) frequency shift. For a 1 kHz modulation, that would be only a 0.0003 Hz shift. You probably want to use significantly higher than audio frequencies.

11. May 15, 2008

### pzlded

Light with 1 Hz amplitude modulation, has 1x10$$^{6}$$ meters between cycles. A stationary detector will detect the 1 Hz modulation.

A detector that moves .5x10x10$$^{6}$$ meters toward the emitter during a second will detect 1.5 Hz modulation. In other words, the Doppler effect results from velocity relative to carrier speed, not modulation speed.

Doppler shifts occur in refracted light. Light travels through a vacuum 1.5 x as fast as through glass, thus modulation frequency will be Doppler shifted within the glass.

F = $$\frac{1x10^{6}}{(1-.5)x10^{6}}$$

12. May 15, 2008

### pzlded

Opps, I put a - instead of a plus in the equation in the previous post.

F = $$\frac{1x10^{6}}{(1+.5)x10^{6}}$$