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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 propagate), then signal is "expanded" in frequency domain. Similarly, a moving away RX-TX "expands" signal in time and "compresses" its spectrum. In short, that is scaling Fourier Transform. These two extreme cases set the left and right bounds of spreading an original frequency between $-f_d$ and $+f_d$ where $f_d$ is max Doppler spread.

In looking at the Clarke model, it is just multiple propagation model with rich scattering environment and equal angle of arrival. (link for more details Clarke model)

If I understand well, there are two assumptions which are reasonale in urban environment:

  • Rayleigh fading
  • equal angle of arrival, or equal receiver sensitivity

I have followed the math from the original article, it seems ok. The final Doppler power spectrum is then $\displaystyle S(f) = \frac{1}{\pi f_d \sqrt{1 - \left(\frac{f}{f_d}\right)^2}}$

Clarke Doppler power spectrum

What I don't understand is that why energy is concentrated to the two extreme spread frequency $-f_d$ and $f_d$ while angles of arrival are uniform. Is there any physical interpretation ? What am I missing from the famous Clarke model ? Personally, this model seems well-model the typical urban environment.

R. H. Clarke, A Statistical Theory of Mobile-Radio Reception, The Bell System Technical Journal, July/Aug 1968, p. 957ff

Answers Although the answer of Carlos captures the most fundamental mathematical part, the real answer is in his comment about "mapping between angle and frequency". Moreover, the answer of Maximilian is interesting too.

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  • $\begingroup$ Given a constant velocity and no multipath you would expect a constant frequency offset for Doppler.. $\endgroup$ – Dan Boschen Apr 3 '17 at 10:36
  • $\begingroup$ Thank Dan, it is correct. But it is not why I ask for help. $\endgroup$ – AlexTP Apr 3 '17 at 10:51
  • $\begingroup$ Sorry I misunderstood you question; I think Carlos answered it below but what were you expecting to see for the energy other than the plot you show? $\endgroup$ – Dan Boschen Apr 3 '17 at 11:01
  • $\begingroup$ Yes, Carlos did answer my question, but in his comment. It is the mapping between angle of arrival $\theta$ and frequency $f$. $\endgroup$ – AlexTP Apr 3 '17 at 20:21
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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)

histogram of cosine amplitudes

More rigor can be seen by noting that \begin{align} f &= f_d \cos\theta\\ \theta &= \cos^{-1}\left(\frac{f}{f_d}\right) \end{align} and the power received at any angle is proportional to a small angle increment $d\theta$:

$$ P(\theta) \propto d\theta = \frac{-1}{f_d\sqrt{1-\left(\frac{f}{f_d}\right)^2}} df $$

And the total power can be determined by integrating the above quantity, which is identically what defines a power spectral density.

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  • $\begingroup$ Thank you. The math is clear, but does not answer my question. My question is what is the physical interpretation of this cos distribution, or how the LoS and 180° reflexion captures most energy physically. $\endgroup$ – AlexTP Apr 3 '17 at 10:48
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    $\begingroup$ The LoS and 180 degree reflections do not capture the most energy-- this is not claimed by the model. It only appears to be so because the plot with the singularities is versus frequency, not angle. The nonlinear mapping between frequency and angle is why the singularities appear. $\endgroup$ – Robert L. Apr 3 '17 at 10:53
  • $\begingroup$ Thanks again, this is what I want to ask, the "nonlinear mapping". Do you agree that I can say, in another language, if we take the same amount of bandwidth $\Delta f$ to integrate, the closer to extremities of Doppler spectrum this band is, the more angle $d\Theta$ we accumulate, and that is the reason why we have more power at extremities ? $\endgroup$ – AlexTP Apr 3 '17 at 11:54
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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 propagate), then signal is "expanded" in frequency domain. Similarly, a moving away RX-TX "expands" signal in time and "compresses" its spectrum. In short, that is scaling Fourier Transform.

Your understanding is correct in the wideband-sense. However, Clarke's model refers to the narrow-band situation, where the Doppler spread is given by $f_d=f_c \frac{v}{c}$. In a wideband situation, you dont have a carrier frequency. In Clarkes model, you assume that the bandwidth $\Delta f$ of the signal is much smaller than $f_c$ and the signal is concentrated in $f_c\pm\frac{\Delta f}{2}$. In Clarke's model, each frequency experiences the same shift, i.e. X_{out}(f)=X_{in}(f-df), where $df$ is the instaneous shift, $X_{in},X_{out}$ are the Fourier transforms of transmit and received signal. This is approximately correct, as long as $\Delta f<<f_c$. In your wide-band model, each frequency experiences a shift that is proportional to the frequency, i.e. $X_{out}(f)=X_{in}(\alpha f)$ with $\alpha=\frac{v}{c}$.

EDIT: Let me explain a bit more in mathematical terms:

In general, given a sine wave with frequency $f$ which is sent to a receiver, where TX and RX have a relative speed of $v$, the sine wave is received with a frequency $f(1\pm-\frac{v}{c})$ (sign depending on direction of movement).

The narrowband assumption now says that a transmit signal is located around a carrier frequency $f_c\pm\Delta f$ where $2\Delta f<<f_c$ is the bandwidth of the signal (I use $2\Delta f$ as the bandwidth for notation simplicity). Now, assume a sine wave with frequency $f_c-\Delta f$ is transmitted. So, the received sine wave has a frequency $$f_{out}=f_{in}\left(1-\frac{v}{c}\right)=f_c\left(1-\frac{v}{c}\right)-\Delta f\left(1-\frac{v}{c}\right)=f_c-\Delta f - f_c\frac{v}{c} - \Delta_f\frac{v}{c}\approx f_{in}-f_c\frac{v}{c}$$ where the approximation comes from $\Delta f << f_c$. As you see, the shift of the frequency is not depending on the actual frequency relative to the carrier frequency. This is the narowband assumption.

I dont want to say that the Doppler spread effect does not change the bandwidth of a signal. In fact, it spreads a signal by $f_D=f_c\frac{v}{c}$. However, the important distinction I want to point out is that in narrowband you can assume all frequencies experiences the same shift, whereas in wideband, the shift depends on the actual frequency. Clarke's model holds for the narrow-band case, as it describes the distribution of the frequency shift, when a sine wave with any frequency (within the bandwidth) is sent into the system.

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  • $\begingroup$ Thanks. However I don't get it. How are these cases wideband and narrowband different ? Can I say in narrowband case $X_{out}(f_1) = X_{in}(\alpha f_1), X_{out}(f_2) = X_{in}(\alpha f_2)$ and the $f_1 - f_2 \approx \alpha (f_1 - f_2)$ because $abs(f_1 - f_2) << 1$ ?. Let me state my opinion in another way, are you telling me that Doppler spread does not change the bandwidth $\Delta f$ of signal ? My opinion is that the band do be expanded, but the expansion is not significant because of the narrowband nature (or condition) $\Delta f << f_c$. $\endgroup$ – AlexTP Apr 3 '17 at 12:33
  • $\begingroup$ @AlexTP I've added some math. derivation, maybe this makes it more clear? $\endgroup$ – Maximilian Matthé Apr 3 '17 at 12:46
  • $\begingroup$ Thanks. I see what you mean now. Indeed, we have told the same story because $f_{out} = f_{in}(1-\frac{v}{c}) $ is still scaling operation, but the point approximation to constant frequency shift is very interesting. Could you please elaborate on "Clarke's model holds for the narrow-band case, as it describes the distribution of the frequency shift" ? I understand that without narrowband assumption, the Doppler spectrum formula is not as what I cited. $\endgroup$ – AlexTP Apr 3 '17 at 14:22
  • $\begingroup$ What I want to know is that if the support of the Doppler spectrum depends on the narrowband assumption. Because I understood that for a given frequency, each angle of arrivial $\theta$ creates a $f_{out}(\theta)$. The LoS and 180 degree reflexion path create two extremities of Doppler spectrum, and the support should independent to nature of transmitted signal. What will change is only the power spectrum itself. $\endgroup$ – AlexTP Apr 3 '17 at 14:22
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    $\begingroup$ Each angle of arrival creates a different frequency shift $f(\theta)$, that is $f_{out}=f_{in}+f(\theta)$. In other ways, I think you can understand the Doppler spectrum as a probability density function of the experienced doppler shift, when the angle of arrival is equally distributed. I.e., it is most probable, that the shift is $\pm f_D$, but not very probable that the shift is $0$. $\endgroup$ – Maximilian Matthé Apr 3 '17 at 18:52

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