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?

  • $\begingroup$ 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
    Commented Feb 19, 2022 at 9:16

1 Answer 1


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 $f>f_D$)!


$f_D$ 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 $J_0(0) = 1$.

  • $\begingroup$ Yes, I was wrong. I confused the notations in my notes. Let's delete the wrong comment. I will check the book. $\endgroup$
    – AlexTP
    Commented 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
    Commented Feb 18, 2022 at 16:21
  • $\begingroup$ @AlexTP Thanks for looking into that! $\endgroup$ Commented Feb 19, 2022 at 19:59

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