Hacker News Reader

Comment by tzs | Hacker News Reader

For the constant velocity case, the graphic shows the transmit pulse as having two cycles, and the received pulse as having four cycles of higher frequency, with the received pulse being the same length as the transmit pulse.

Shouldn't the received pulse have the same number of cycles as the transmit pules, and the received pulse be shorter than the transmit pulse (or longer in the case of the target moving away instead of toward the source)?

Or does something non-intuitive happen (probably because of special relativity) resulting in the received pulse gaining energy from the reflection, which is reflected (pun intended) as the pulse being longer than you'd expect classically?

View on HN

You are right that the graphic is a highly simplified cartoon of what the actual received wave looks like - I should have stated that in the post.

Regarding the question, it is true that the received pulse will be shorter or longer than the transmitted one (depending on whether the object is moving towards or away from the sensor) but since we are assuming that the speed of the object is significantly less than the speed of light, we're making the simplifying assumption that this difference is negligible. People sometimes refer to this as the "stop and hop" assumption since in practice we are assuming that the object "stops" at point x when the radio wave is emitted and then "hops" to x + dt*v once the wave has been received dt seconds later.

Regarding the number of cycles, you are again correct that the image represents a gross oversimplification. One reason for this is that since the carrier frequency is so much higher than the doppler frequency, the true graph of the received wave would not be noticeably different than the transmitted one. It is easier to plot a realistic graph of the demodulated wave which you can see in the plot titles "samples of the demodulated pulse".

I know that this doesn't directly address the question you're asking, but to give an idea of the order of magnitude of the effect: the Doppler shift in frequency rounds to f_carrier * 2v/c. In the case of anything with "reasonable" speed, 2v/c is going to be very small (~10e-6 for Mach 1), and thus you would be talking about very minute differences between the transmitted and received pulse in terms of either overall pulse length or number of wavefronts received vs. sent (for what it's worth my intuition is that the pulse length actually shortens, but either way it's not measurable by the receiver).

Reply on news.ycombinator.com