4 ms·

Surely it's a different narrow frequency range? Otherwise sunlight would act as a signal jammer, a source of noise that drowned out our wireless signals?

by deogeo 7y ago

itcrowd 7y ago
Indeed, the radiation coming from the sun in the visible spectrum is much larger than the microwave radiation coming off it [1]. Note: the microwave spectrum ends at ~300 GHz or 1 mm wavelength, that is 1 million nanometers, so the scale on the x-axis is wayyy too short in the image. The actual value is very, very low. [1] https://upload.wikimedia.org/wikipedia/commons/0/0d/EffectiveTemperature_300dpi_e.png https://upload.wikimedia.org/wikipedia/commons/0/0d/Effectiv...
Dylan16807 7y ago
You can filter out a stable amount pretty easily but oops I just checked the blackbody math again and an exponent flipped when I changed one of the ranges while everything else stayed almost the same, I screwed up and ignore everything I said.
- deogeo 7y ago
  Hehe, happens :) But, suppose the sun did interfere - how would you filter this out? Due to the black-body nature of the source, the amplitude would be stable, but the phase wouldn't be - you'd get entirely uncorrelated white noise, right? How can you filter that?
  - Dylan16807 7y ago
    I've never really looked into how the encoding is typically done on those bands. You don't need a lot of bits per Hz when you have 14GHz of space, so nothing particularly elaborate with phase is strictly necessary. But yes it would definitely hurt speeds, and you have a good point in that it's something to consider as a sanity check!
  - itcrowd 7y ago
    Good question. A good but short answer is given on Wikipedia (https://en.wikipedia.org/wiki/Matched_filter#Matched_filter_in_digital_communications https://en.wikipedia.org/wiki/Matched_filter#Matched_filter_...). Any undergraduate textbook on digital signal processing should have details on this topic. edit: to add to this, the capacity C ("throughput") of the communications channel is linear with bandwidth B (in Hertz) but only logarithmic (log_2) with noise power N. This was derived by Shannon (https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem#Statement_of_the_theorem https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theore...) and in equation form: C = B*log_2(1+S/N) Example: signal to noise ratio is 1 (S/N = 1). Result: C = B Now we increase the bandwidth of our channel by a factor of 10, keeping the signal power equal. This will lead to a factor 10 more noise power (S/N = 1/10). Result: C = 10*B*log_2(1 + 1/10) = 1.38*B So we can improve the channel capacity by increasing the bandwidth even in a noisy channel.

Surely it's a different narrow frequency range? Otherwise sunlight would act as a signal jammer, a source of noise that drowned out our wireless signals?

by deogeo 7y ago

Surely it's a different narrow frequency range? Otherwise sunlight would act as a signal jammer, a source of noise that drowned out our wireless signals?

itcrowd 7y ago

Indeed, the radiation coming from the sun in the visible spectrum is much larger than the microwave radiation coming off it [1]. Note: the microwave spectrum ends at ~300 GHz or 1 mm wavelength, that is 1 million nanometers, so the scale on the x-axis is wayyy too short in the image. The actual value is very, very low. [1] https://upload.wikimedia.org/wikipedia/commons/0/0d/EffectiveTemperature_300dpi_e.png https://upload.wikimedia.org/wikipedia/commons/0/0d/Effectiv...

Dylan16807 7y ago

You can filter out a stable amount pretty easily but oops I just checked the blackbody math again and an exponent flipped when I changed one of the ranges while everything else stayed almost the same, I screwed up and ignore everything I said.

deogeo 7y ago
Hehe, happens :) But, suppose the sun did interfere - how would you filter this out? Due to the black-body nature of the source, the amplitude would be stable, but the phase wouldn't be - you'd get entirely uncorrelated white noise, right? How can you filter that?
- Dylan16807 7y ago
  I've never really looked into how the encoding is typically done on those bands. You don't need a lot of bits per Hz when you have 14GHz of space, so nothing particularly elaborate with phase is strictly necessary. But yes it would definitely hurt speeds, and you have a good point in that it's something to consider as a sanity check!
- itcrowd 7y ago
  Good question. A good but short answer is given on Wikipedia (https://en.wikipedia.org/wiki/Matched_filter#Matched_filter_in_digital_communications https://en.wikipedia.org/wiki/Matched_filter#Matched_filter_...). Any undergraduate textbook on digital signal processing should have details on this topic. edit: to add to this, the capacity C ("throughput") of the communications channel is linear with bandwidth B (in Hertz) but only logarithmic (log_2) with noise power N. This was derived by Shannon (https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem#Statement_of_the_theorem https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theore...) and in equation form: C = B*log_2(1+S/N) Example: signal to noise ratio is 1 (S/N = 1). Result: C = B Now we increase the bandwidth of our channel by a factor of 10, keeping the signal power equal. This will lead to a factor 10 more noise power (S/N = 1/10). Result: C = 10*B*log_2(1 + 1/10) = 1.38*B So we can improve the channel capacity by increasing the bandwidth even in a noisy channel.