If I have a 20Mhz bandwidth provided by my 2.4ghz router then will my internet speed will be 20M symbols per sec which is equal to 20×4= 80Mbps (considering 16 qam modulator with 4 bits per symbol) Is it true?
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1\$\begingroup\$ Payload data rate will be smaller - overall data rate (including formatting bytes) will be 80 Mbps. \$\endgroup\$– Andy akaMay 12, 2020 at 7:28
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The successful symbol rate depends on the channel settling, the phase flatness, etc
RF channels tend to have very steep filtering (at the post-down-conversion frequencies where multipole Intermediate Frequency active filters ) to allow packing lots of channels and lots of users into a few hundred MHZ of spectrum (such as 2.4GHz WiFi).
Thus the precise successful symbol rate depends on
the interference from adjacent channels,
reduced phase warping if adjacent channels are UNUSED on purpose to allow moderate IF filtering
the silicon area devoted to IF active filters, where phase-compensation requires MORE opamps and other silicon-area-hogs such as ratiod-capacitors to accurate place the Zeros
If you have a one-pole filter with Tau timeconstant, being a 1-pole allows 6.28 settling Tau in the bandwidth. At 8dB per Tau (a Neper), you have settling to 48dB error flow which allows dense constellations. For a cheap filter.
But one-pole filters prevent dense channel allocation, so are not used.
answered May 12, 2020 at 7:57
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\$\begingroup\$ the silicon area devoted to IF active filters In the last century indeed we did things like that. These days nearly all receivers are zero-IF or low-IF and directly feed the downconverted signal into an ADC. Also not sure how this answer will help OP. \$\endgroup\$ May 12, 2020 at 8:04