which configuration has greater range, sf8 bw 125 khz or sf8 500 khz? . It is clear that the data rate is higher with Bw 500 khz, but I am still not clear which of the 2 cases has greater scope
Greater scope, not sure what you mean by that ?
However, the lower the bandwidth the greater the range\distance potential for LoRa.
Is this related to The Things Network (TTN), that uses a bandwidth of 125 Khz as standard ?
i meant the distance they can cover. I understood that sf8 and bw 125 Khz cover more distance between end-device and gateway.
Thank you very much for the quick answer.
Best regards
For US915 uplinks, I see for TTN:
- 903.9 - SF7BW125 to SF10BW125
- 904.1 - SF7BW125 to SF10BW125
- 904.3 - SF7BW125 to SF10BW125
- 904.5 - SF7BW125 to SF10BW125
- 904.7 - SF7BW125 to SF10BW125
- 904.9 - SF7BW125 to SF10BW125
- 905.1 - SF7BW125 to SF10BW125
- 905.3 - SF7BW125 to SF10BW125
- 904.6 - SF8BW500
So, it seems the last one is a valid choice too? (The 1.0.2 Regional Parameters lists that as DR4, while SF8BW125 is DR2. Some other regions have some non-BW125 data rates as well.)
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With as smaller bandwidth you have a bigger range (distance) you can reach. So with the same SF8, 125 kHz will reach a bigger distance than with SF8 using 500 kHz bandwidth.
More important is that 125KHz BW isn’t employed at all for downlinks in US.
Hello
But according to the Shannon theorem, used in the spread spectrum theory:
C= B* log2(1+S/N)
where:
where
C is the channel capacity in bits per second, a theoretical upper bound on the net bit rate (information rate, sometimes denoted excluding error-correction codes;
B is the bandwidth of the channel in hertz (passband bandwidth in case of a bandpass signal);
S is the average received signal power over the bandwidth (in case of a carrier-modulated passband transmission, often denoted C), measured in watts (or volts squared);
N is the average power of the noise and interference over the bandwidth, measured in watts (or volts squared); and
S/N is the signal-to-noise ratio (SNR) or the carrier-to-noise ratio (CNR) of the communication signal to the noise and interference at the receiver (expressed as a linear power ratio, not as logarithmic decibels).
the definition above is given in wikipedia
If the distance between end-device and gateway is increased, then S / N ----> 0 and can make the following approximation::
C=B* S/N =====> B/C=N/S
longer distance S / N -----> 0
f I increase the distance between both teams and want to keep constant C, which is the capacity of the channel in bps; must increase B, which is the bandwidth of the channel.
This makes me think; that with SF8 and BW500 Khz it should have a higher data rate and even greater range distance than in the case of SF8 and BW 125 Khz.
I am confused with this subject
Best regards
The sensitivity level of SF8 BW 500kHz is basically 6dB higher than SF8 125kHz, so the maximum theoretical range is divided by 4.
Your reasoning about Shannon theorem is incorrect: this is a formula for the maximum channel capacity, it does take into account the distance. Shannon simply gives an upper bound on the data rate you can get: so higher bandwidth means higher data rate. If you want to look at the range you need to look at the link budget, and here when the bandwidth is multiplied by 2, you lose 3dB which translate in a division by 2 of the max distance.
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It is like dragging a rake over ground. the bigger the rake you drag (where the size of the rake equals bandwidth) the more difficult it is to drag forwards and the earlier you are exhausted from dragging. The measure of exhausting affects the distance (range) you can drag the take.
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Then you have to work out why your logic is incorrect.
The higher bandwidths do have a higher data rate, but the range is shorter, and its very easy to demonstrate this in practice.
The sensitivity in the Rx is:
S = -174 + 10xlog10(BW) + NF + SNRlimit [3]
There it is clear that increasing the BW increases the level of sensitivity of Rx and therefore it is reasonable for the range to be shorter.
Thank you very much Clams
Thank you very much for helping me with this issue.