3
$\begingroup$

I am reading Wireless Communications by Andrea Goldsmith. I have some issues in the derivation of Clarke's Doppler power spectral density given there.

1)For beginning the autocorrelation of the in phase component is given by: $$A_{r_I}(\tau)=P_rJ_0(2\pi f_D\tau)$$ where $f_D$ is the doppler frequency, $J_0$ is Bessel function of zeroth order. Now the author takes its Fourier transform to obtain the Power Spectral Density. The Fourier transform of $J_0(x)$ is given here as $2\DeclareMathOperator{\rect}{rect}\rect(\pi\zeta)/\sqrt{1-4\pi^2\zeta^2}$. Using this I got $$F[A_{r_I}]=\frac{P_r}{\pi f_D}\frac{\DeclareMathOperator{\rect}{rect}\rect(f/2f_D)}{\sqrt{1-f^2/{f_D}^2}}=\frac{P_r}{\pi f_D}\frac{1}{\sqrt{1-f^2/{f_D}^2}}~~~\text{for} |f|\geq f_D$$ However the author gets $$\frac{1}{\pi f_D}\frac{2P_r}{\sqrt{1-f^2/{f_D}^2}}~~~\text{for} |f|\geq f_D$$ I am using simple transformation properties, i.e. $F[g(at)]=\frac{1}{a}G(f/a)$

How this extra factor of 2 is coming into play in this equation?

  1. Then she takes the autocorrelation of received signal given by $$A_r(\tau)=A_{r_I}\cos(2\pi f_c \tau)+A_{r_I,r_Q}\sin(2\pi f_c \tau)$$ For Clarke's model $A_{r_I,r_Q}=0$ so the author just takes Fourier transform of first term which should turn out to be: $$F[A_{r_I}]*[0.5\delta(f-f_c)+0.5\delta(f+f_c)]$$ where * is convolution operator. Assuming $F[A_{r_I}]=S_{r_I}(f)$, I am writing $$F[A_r(\tau)]=0.5[S_{r_I}(f-f_c)+S_{r_I}(f+f_c)]$$ where as the author gets $$F[A_r(\tau)]=0.25[S_{r_I}(f-f_c)+S_{r_I}(f+f_c)]$$ Why we have an extra $1/2$ here?
  1. Furthermore, she writes the final expression as:

$$\frac{P_r}{2\pi f_D}\frac{1}{\sqrt{1-(|f-f_c|/f_D)^2}} \text{for}|f-f_c|\geq f_D$$ Why she has not considered the term $S_{r_I}(f+f_c)$ here in this expression?

$\endgroup$
1
  • $\begingroup$ For questions (1) and (2), the author is wrong. Please refer to Dan Boschen's answer for details. For question (3), I presume that the reason is that the other term does not provide new information. Indeed, in the paragraph that follows this expression, the author does mention $\pm f \pm f_c$. Next time, please define all terms and, if possible, include the extraction of the book. $\endgroup$
    – AlexTP
    Feb 19, 2022 at 9:16

1 Answer 1

1
$\begingroup$

First to clarify:

This PSD of the Doppler Spread as derived assumes isotropic scattering and omnidirectional antenna such that the phase as received is uniformly distributed over $(-\pi, \pi]$ and an absence of any strong direct path (Rayleigh fading).

For this case, the power spectral density of the Doppler spread would be given as:

$$S(f) = \frac{2P_R}{\pi f_D\sqrt{1 - (f/f_D)^2}}, |f|\le f_D$$

(And zero for $f>f_D$)!

Where

$f_D$ is the maximum Doppler spread

It is intuitive that $S(f)=0$ for $f>f_D$ when we consider that the transmitted signal would convolve with the Doppler spreading resulting in a spreading of the signal as received, and therefore the resulting signal would occupy $2F_d$ larger spectrum.

The factor of two must lie in the definition of what $P_R$ is. The subsequent derivation that the OP proceeds with seems correct in using the Fourier Transform of the Bessel function to get the power spectral density. The OP mentions "the in-phase component" while the fading signal as received would have in phase and quadrature components of interest. If $P_R$ represents only the in-phase component, then the total power would be given as $2P_R$ and the autocorrelation function as well should be:

$$A_{rr}(\tau)=2P_RJ_0(2\pi f_D\tau)$$

Given that the autocorrelation at $\tau=0$ is the variance and $J_0(0) = 1$.

$\endgroup$
3
  • $\begingroup$ Yes, I was wrong. I confused the notations in my notes. Let's delete the wrong comment. I will check the book. $\endgroup$
    – AlexTP
    Feb 18, 2022 at 11:43
  • 2
    $\begingroup$ I read the book. The $P_R$ is defined as you have presumed: $2 P_R = N \alpha^2$, where $N$ is the number of arrival angles, and $\alpha^2$ is the received power within an arrival angle. The book is not well written and contains many annoying mistakes. $\endgroup$
    – AlexTP
    Feb 18, 2022 at 16:21
  • $\begingroup$ @AlexTP Thanks for looking into that! $\endgroup$ Feb 19, 2022 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.