Interpretation of Clarke's Doppler power spectral density

Question

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 $\displaystyle S(f) = \frac{1}{\pi f_d \sqrt{1 - \left(\frac{f}{f_d}\right)^2}}$

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.

Answer 1

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 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.

Answer 2

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.