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For communication systems, there is usually a target minimum rate $R$ bits per second. An outage occurs if the actual rate ever falls below $R$. Since $\frac{1}{2}\text{log}_2(1+\text{SNR})$, sometimes outage is also talked about in terms of $\text{SNR}$ and received signal power needing to be greater than some threshold $P$.

My question is: Say I have a very small packet of five symbols, and during the second symbol I experience an outage. The other four symbols received power was all greater than $P$. Does this mean that my entire packet was unable to be decoded?

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For example, if I use the encoding scheme where: 0 = 00000, 1 = 11111, then I have 1 bit per 5 channel uses. So I can use the formula: $\frac{1}{2}\text{log}_2(1+\text{SNR})=\frac{1}{5}$ bits per channel use, and can solve for the required $\text{SNR}=0.3195$. Does this mean that all five symbols should have $\text{SNR}\geq0.3195$ in order to successfully receive the single bit?

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  • $\begingroup$ Making sure I understand your formula - it looks like the Shannon channel capacity equation C = B log2(1 +SNR). Where does the 1/2 come in in your use? From the capacity formula this is suggesting your channel bandwidth is 1/2 and the bound on the achievable error free rate in that case is 1/5 bits per second? $\endgroup$
    – Dan Boschen
    Dec 12, 2019 at 15:47
  • $\begingroup$ But to your final question: No. The simple counter example is to consider one bit with significantly less SNR (and clearly be in error) while at least three of the bits are significantly higher SNR (and clearly error free). $\endgroup$
    – Dan Boschen
    Dec 12, 2019 at 15:53
  • $\begingroup$ I'm reading out of this book Equation 5.7 web.stanford.edu/~dntse/Chapters_PDF/… $\endgroup$
    – Engineer
    Dec 12, 2019 at 16:16

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No. It means 1 out of 5 symbols was not able to be decoded. Beyond that it depends on your encoding scheme and criteria for deciding if that packet is usable. For example what if you used an encoding scheme with BPSK symbols such that 1 1 1 1 1 = "1" and -1 -1 -1 -1 -1 = "0"? (You could replace those with other codes as well 1 1 -1 -1 1 = "1" and -1 -1 1 1 -1 = "0" for example). Typically the packet is longer and error correction encoding is used so that individual symbol errors would not cause the entire packet to be lost.

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  • $\begingroup$ I understand that bit errors can be corrected by codes but I'm trying to ask about the rate equation. If I'm trying to understand how the equation is used in the literature to justify packet drops. I see many papers using SINR or rate thresholds as the basis to determine if a packet is dropped or not but my intuition also was similar to yours that it should just mean that single bit is in error not the entire packet $\endgroup$
    – Engineer
    Dec 12, 2019 at 15:05
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    $\begingroup$ To your edit, your could receive the single bit in our example if 3 out of the 5 bits are correct--- therefore you can have entirely different SNR's for each bit. Your formula appears to be the Shannon-Hartley theorem which would be averaged received signal power NOT a requirement on each symbol. This is also a bound on what is possible in a single channel but not what is achieved. $\endgroup$
    – Dan Boschen
    Dec 12, 2019 at 15:38

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