Modulated Light and Doppler effect

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Discussion Overview

The discussion revolves around the effects of the Doppler effect on amplitude-modulated light when reflecting off a moving target. Participants explore whether the modulation frequency of the light is subject to Doppler shifting and seek to understand the relevant formulas for calculating any frequency shifts. The scope includes theoretical considerations and practical implications in the context of light propagation in air.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the amplitude modulation frequency of light would change due to the Doppler effect when reflecting off a moving target.
  • Another participant asserts that reflection does not change light velocity, suggesting that the AM frequency should remain unchanged.
  • A different participant proposes that the modulation frequency would indeed experience a Doppler shift, seeking clarification on the formula to determine this shift.
  • One analogy compares photons to raindrops, suggesting that a moving detector would experience a different frequency due to the varying distances traveled by successive peaks of the AM signal.
  • Another participant emphasizes that in the case of a detector moving at light speed, the detected AM frequency could be perceived as doubled.
  • Some participants argue that while the wavelength of light changes upon reflection, the frequency may also be affected, particularly in the context of non-relativistic speeds.
  • One participant provides a formula for calculating the frequency shift, indicating that the shift applies to both the modulation and carrier frequency, and discusses the implications of the observer's frame of reference.
  • Another participant discusses the relationship between the Doppler effect and modulation frequency, suggesting that the effect arises from the relative velocity to the carrier speed rather than the modulation speed.

Areas of Agreement / Disagreement

Participants express differing views on whether the modulation frequency is affected by the Doppler effect, with some asserting it remains unchanged while others argue it is subject to shifting. The discussion remains unresolved, with multiple competing perspectives presented.

Contextual Notes

Participants reference classical physics and assume non-relativistic speeds in their discussions. There are mentions of potential limitations in understanding the effects of reflection and refraction on frequency shifts, as well as the need for clarity on the conditions under which these effects apply.

Deane
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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.)
 
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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.
 
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 traveling 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?
 
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?
 
pzlded said:
Reflection does not change light velocity, thus AM frequency should be unchanged.

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.
 
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.
 
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.
 
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.
 
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
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:

<br /> \frac{\Delta f}{f}=2\frac{v}{c}<br />

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

For an object traveling 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
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
Opps, I put a - instead of a plus in the equation in the previous post.

F = \frac{1x10^{6}}{(1+.5)x10^{6}}
 

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