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We know that the signal attenuates out with distance and according to the channel transfer function or frequency response, the signal frequency components attenuate to different values based on frequency.

The effect of distance is widely acknowledged & used in practice as a path-loss model which is exponentially decreasing in distance but frequency component is relatively less discussed.

It is very embarrassing to find that the everyday used wireless channel's transfer function is not known to most of us!

I searched all over the web with all naive keywords but it seems the spectrum of the channel transfer function is not there anywhere not even in many text books!

On the similar lines, what is the typical conclusion about the spectrum? are high frequencies better for some reason, so that we always modulate to higher frequencies or it is just that we have available bands there?

--P.S.--
After a fundamental clarification by user5108_Dan, It seems to be widely acknowledged that very-high-frequency communication is highly-challenging (I am not sure if is it just the dsp/hardware challenge!). See for eg, http://en.wikipedia.org/wiki/Extremely_high_frequency

I am thinking that the channel transfer function if exists, becomes crucial to know before we decide our band of transmission and transmit power.

OR, Is this freq. response completely application specific that it includes antenna characteristics and other hardware in the chain? Even in that case, it appears to me that it helps the designer in choosing the right hardware. is n't it?

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  • $\begingroup$ This part of your question does not make sense to me: are high frequencies better for something that we always modulate to higher frequencies ?? Care to rephrase it? Or re-state it? Are you saying Are high frequencies better for some reason, so that we always modulate to higher frequencies? $\endgroup$ – Peter K.♦ Nov 21 '13 at 12:36
  • $\begingroup$ @Peter: Thanks for the clarification. Yes I meant that exactly. editing now to rephrase some... $\endgroup$ – Loves Probability Nov 22 '13 at 4:40
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Let me try to clarify something that I suspect you misunderstand.

In terms free space propagation, there is nothing in the laws of physics that say a high frequency signal will attenuate more than a low frequency signal. In free space, a 1 MHz signal attenuates at the same rate as visible light.

Here on earth, the atmosphere, moisture in particular, will cause high frequencies to attenuate more, but in general, atmospheric affects are minimal.

It is certainly true however than when calculating the path loss for a wireless link, the Friis Transmission Equation predicts that the "loss" is proportional the square of the frequency.

http://en.wikipedia.org/wiki/Friis_transmission_equation

To understand this apparent contradiction, you need to understand how the Friis equation was developed. The basic idea is that if you have a perfect isotropic radiator at the center of a sphere, then each square meter of that sphere will receive the same amount of energy, regardless of frequency.

On the other hand, the amount of energy an antenna can receive, is a strong function of its physical size, which is what the Friis equation is all about.

The Friis equation predicts the amount of energy that will pass through a square on the surface of that sphere, but the size of that square is one wavelength, not 1 meter. Consequently, less energy passes through the square at high frequencies because the square is smaller. This is a way to normalize, if you will, the effects of antenna size. Everything is based on a wavelength.

Said a bit differently, if you could build an ideal antenna with infinite bandwidth, then the gain of that antenna would be proportional to frequency, which would cancel out the path "losses" due to frequency. Antennas however are inherently narrow band devices.

Regarding your question as to whether high frequencies are better. There is nothing better about higher frequencies. The attenuation due to moisture is worse, the loss through building walls is worse, the hardware is more expensive and more difficult to design, and the interference due to multipath propagation is worse, but that is where the available spectrum is.

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  • $\begingroup$ Though I wish I would appreciate you by name, @user5108 , Thanks for the responsible response clarifying one of my fundamental misunderstanding. However, having understood what you said so far, en.wikipedia.org/wiki/Extremely_high_frequency#Propagation kind of explanations confuse me a bit. I hope you can comment on it. Editing my Q to include this point. $\endgroup$ – Loves Probability Nov 22 '13 at 4:54
  • $\begingroup$ When you said high frequency, I thought you meant 5 GHz, or something in that range. I don't have any experience in the 30 - 300 GHz range. Only governments and large corporations have enough money for that band. $\endgroup$ – user5108_Dan Nov 22 '13 at 11:50
  • $\begingroup$ Well, the antennas are usually much smaller at high frequencies. Would be difficult to make a pocket-sized phone if it used a kHz frequency range. $\endgroup$ – jan Nov 22 '13 at 13:11
  • $\begingroup$ @user5108_Dan Actually, my intention in showing >3GHz is only to highlight that there is a difference that comes with frequency. The effective atmospheric channel, including practical surroundings, I doubt is going to result attenuation as a function frequency. $\endgroup$ – Loves Probability Nov 22 '13 at 14:03

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