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6

A simple, "non-technical" way of thinking of it is the fact that the Doppler frequency is proportional to $\cos\theta$. The amplitudes of cosine, however, are not uniformly distributed, but are heavily weighted towards $\pm 1$. Example plot to demonstrate, using Python/Pylab code: theta = linspace(0, 2*pi, 1001) x = cos(theta) hist(x) More rigor can be ...


5

In addition to Carlos answer, I want to correct your general understanding: What I understand of Doppler spread is that the relative motion between Transmitter (TX) and Receiver (RX) change the exposing time of signal. In rapport to a constant-distance TX-RX, a moving toward each other TX-RX "compresses" signal in time (signal takes less time to ...


4

Short answer; I'll try to come back and improve this later. Recall that a matched filter is equivalent to a sliding correlator. At each time instant, the sliding correlator multiplies the last $T$ (where $T$ is the pulse duration) of the input signal by the conjugate of the pulse shape and coherently integrates the result over the pulse duration. If there ...


4

Bottom Line First What you suggest will only work if your signal is at baseband and not at a carrier, which is usually the best way to do any simulation work regardless. If your signal is not at baseband first multiply by $e^{-j2\pi f_c t}$ where $f_c$ is the carrier frequency and low pass filter out the higher image (which will now be centered at $-2f_c$) ...


4

For lower-cardinality PSKs like the QPSK, the "classical" way is to take the signal to the $M$th power, $M$ being the number of constellation points. From the shape of an $M$-PSK modulation, it's clear that the constellation points are $e^{j2\pi \frac mM},\, m =0,1,\ldots,M-1$, and putting that to $M$th power yields $${\left(e^{j2\pi \frac mM}\...


3

Does radio communication have to account for the doppler effect? Yes. Would be pretty terrible if that wasn't the case: RADAR wouldn't work! Do moving objects such as planes and rockets have to account for this, Yes. Phones in cars and trains, too. Your 5G NR phone of the future operating above 60 GHz, will have to do that at walking speeds, too. or is ...


3

Part of your misunderstanding comes from the fact that there are many ways in which the radar signal processing chain is implemented. Depending on the type of radar, targets of interest, hardware, etc., some methods are more appropriate than others. We will consider pulsed-Doppler radar here. In the chain you describe: In modern pulse-Doppler systems using ...


3

Radar designer here: It sounds like you’re talking about pulse-Doppler (PD) radar systems. For PD radars, the process is essentially as you described: Generate a waveform (typically at IF) and then convert to RF Transmit the waveform at RF Receive the waveform at RF, and eventually mix it down to IF. Apply IQ demodulation (for digital receivers a Hilbert ...


3

The ambiguity function is used in radar systems to get the distance and the relative speed of a moving object with respect to the transmitter. Is called ambiguity because it also tells about the ability to distinguish objects that are close between them and with a similar vector velocity. The ideal ambiguity function is a Delta function in both domains. ...


2

Two suggestions that may independently, or combined, help to solve your problem. Consider using a different window/weighting function for your FFT processing. There are window functions with low uniform sidelobes such as the Chebyshev window. This window function allows for selection of a constant sideline level response. It sounds like you are using a ...


2

frequency shift of about 890 kHz (supposing my math is correct) $$\begin{align} \Delta f_\text{Doppler} &= f_0 \frac vc\\ &=800\cdot 10^6\,\text{Hz}\frac{3.4\cdot10^2\frac{\text m}{\text s}}{3\cdot 10^8\frac{\text m}{\text s}}\\ &\approx 800\cdot 10^6\,\text{Hz}\cdot 1.13\cdot 10^{-6}\\ &\approx 906 \,\text{Hz} \end{align}$$ So, your shift ...


2

So this is something we run into with the radar community on a pretty regular basis. For LFMs, the solution is a bit trivial, but for coded waveforms it becomes a real issue. For us radar folk, we're dealing with targets that are at escape velocity (right around 11 km/s typically). I actually have some pending intellectual property being released via my ...


2

I refer to the graph you presented at your original post. The interpretation of the graph is as follows: the wider the yellow part of the graph, the more immune to Doppler your system is. Normally it is desirable to have the system immune to the relative speed between source and target, for example when you do not know the targets speed. Therefore, the more ...


2

This is what I would try: Ignore the $n(t)$ and $i(t)$ since you know nothing about them. I am assuming that $x$ is complex and the rest of the variables are all real. Take the complex natural log of both sides: $$ \ln \left( | x(t) | \right) + j \arg( x(t) ) = \ln(A) + j ( a t^2 + b t + c ) $$ Separate the real and imaginary parts. Using the imaginary ...


2

Without understanding more detailed results on what i(t) is, one solution is to estimate either the phase or frequency first, and then pass that result through a filter. An approach to do the frequency estimation digitally directly on the I (real) and Q (imaginary) samples of x(t) is the cross product frequency discriminator. After hard limiting x(t) to ...


2

Actually the PRI also changes. There is a complete explanation about this topic in "Cobbold - Foundations of Biomedical Ultrasound" chapter 10 (Pulsed Methods for Flow Velocity Estimation and Imaging). Consider a case where PRI of emitted pulse is 1 s and the wave speed is 9 m/s and the object is 10 m away and moving with speed of 1 m/s toward the antenna. ...


2

When there is relative motion between the transmitter and receiver the frequency seen by the receiver is not $F_c$, so let's say you incur a Doppler shift of $\Delta_f$, then the shifted signal frqeuency is $F_c - \Delta _f$ now when you downconvert with $F_c$ the frequency offset of $\Delta_f$ still remains in the signal. That means the baseband spectrum "...


2

The second way is how it is done. In the fast-time (or range-bin dimension) you are right to perform the matched filter. However, Doppler information is gathered from sampling pulse-to-pulse. The MTI filter essentially subtracts pulses so that if they have similar phase, or the phase is not changing at all, then there is little to no Doppler being generated ...


1

Answer : Down-converting from $f_c$ to DC involves multiplying with locally generated $f_c$. Which would create 2 copies of the original rx signal at $f_c$. One at DC and one at $2f_c$. Then you pass this though a Lowpass filter to suppress the copy at $2f_c$. What you are left with it the copy at DC ($f=0$). When there is a Doppler shift in the rx signal ...


1

The simple answer is down-conversion is just a frequency translation. Consider the downcoversion explained by the cosine product rule: $$\cos(\alpha)\cos(\beta) = \frac{1}{2}\cos(\alpha+\beta) + \frac{1}{2}cos(\alpha-\beta)$$ If we are down-converting a we are interested in the difference, which we get by low pass filtering. So consider the down-conversion ...


1

When you downconvert your radar signal, it is based on the center frequency of your transmitted signal, $f_c$ (assuming that is the frequency you choose to downconvert with). However, if your received signal ($f_r$) has a doppler shift on it, the signal moves away from your center frequency. So the f on the right side of your equation becomes the difference ...


1

Yes, the Doppler effect applies to any electromagnetic wave. No matter whether that is an actual Radar pulse, or a microwave oven falling from an office tower, or a passenger aircraft's transponder, or an LPWAN device. But, this possibility is especially limited for LPWAN devices, as I'll explain below. Low Doppler shift due to low carrier frequency The ...


1

This is my second answer. Technically it is introducing a high pass filter, but a very simple one with interesting properties for the OP's signal type. Starting with your signal: $$ x(t)=Ae^{j(at^2+bt+c)}+n(t)+i(t) $$ For a chosen value of $d$ (a lag time): $$ x(t-d)=Ae^{j(a(t-d)^2+b(t-d)+c)}+n(t-d)+i(t-d) $$ Define the filtered signal as the ...


1

You need to resample the signal to simulate the Doppler induced dilation. The resampling factor is I=Td/Ts where Td = Duration of the signal after dilation Ts = Actual duration of the transmitted signal In Matlab you can use the function resample(), but you need to find the resampling parameters P and Q from I. Also, I=1+v/c for signal expansion and I=1-v/...


1

If you examine a contour plot of the Ambiguity function - the lines connect equal values of the Ambiguity function. This means that if you travel along one of these lines (i.e. keeping the same value of the function) then your system will not be able to distinguish between the signal that have this combination of Doppler and time delay. Note that the values ...


1

frequency is the derivative of phase over time; phase hence frequency's integral over time. And that's all the knowledge you need, since baudrate gives you the time between consecutive symbols.


1

If we want to know the direction of a target, or a velocity toward or away from the radar, we must use the IQ demodulation technic to find a negative velocity, right? No, whether you do that on the original passband signal, or on an intermediate frequency or in baseband after IQ demodulation doesn't matter, mathematically all these three are equivalent. It'...


1

If the only reason for the buffer is so your algorithm has something to process, then the simplest way to implement this is to keep your current algorithm but use a shorter buffer. If the larger buffer was present for other reasons (i.e. data arrives in large chunks), then keep that larger buffer, but feed it into a shorter buffer that is given to your ...


1

These are interesting results. Your comment "The X axis represents the distance travelled in metres (from 3.5m away to 0.5m away from the sound source). " seems to contradict "With this explanation, '0' represents 0.5m from sound source, while '3.0' represents 3.5m from sound source." The latter seems to be the one that fits your data better. What is more ...


1

Your code is (probably) ok. But your way of plotting is slightly wrong. The complete spectrum of the baseband PSK signal would include two peaks one at positive frequency $f_0=1000$ and the other at negative frequency $f=-1000$. When you IQ modulate this signal with $f_m=3000$ you will create two twins one at $f_{11}=2000$ and $f_{12} = 4000$ and the other ...


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