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What is the difference between frame error rate (FER) and symbol error rate (SER)? I know the difference between frame error rate (FER) and bit error rate (BER) from the website below

https://www.quora.com/What-is-the-difference-between-frame-error-rate-FER-and-bit-error-rate-BER-in-LDPC-codes

In this website,it said

Bit is the unit that data transmit in the physical line. Frame is consist of many bits. A unit frame has more information than a unit bit.

However,the symbol consist of many bits too. e.g.symbol "0" = 000,symbol "7"=111,they all consist of three bits,so what is the difference between frame error rate (FER) and symbol error rate (SER)?

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Let's say that you transmit a signal using 4-PAM, with symbols $3,1,-1,-3$. The bits-to-symbols mapping is

$\begin{align*} 3 &\rightarrow 00 \\ 1 &\rightarrow 01 \\ -1 &\rightarrow 11 \\ -3 &\rightarrow 10.\end{align*}$

The symbol error rate (SER) is the probability that the receiver estimates the wrong symbol; for example, that it estimates $3$ when the transmitted symbol is $1$. Note that this is different than the bit error rate, because some symbol errors cause one bit error, while others cause two errors (for example, estimating $3$ when $-1$ was transmitted).

This is relevant for the physical layer in a communications system. However, in the data link layer (DLL), the "protocol data unit" is a frame; that is, the DLL can only deal with entire frames. A single bit error in an entire frame counts as a frame error; usually, the frame is discarded and a retransmission is requested. The error is typically detected by a CRC. The frame error rate is the ratio of errored frames to transmitted frames.

As an example, consider communication using 2-PAM, and a memoryless binary symmetric channel with probability of bit error $P_b = 1\times10^{-3}$. The DLL uses 64-bit frames. The probability that a frame has zero errors is $$(1-P_b)^{64} \approx 0.938,$$ which means that the probability of receiving an errored frame is $$\text{FER} \approx 1-0.938 = 0.062,$$ or one errored frame out of every 16 transmitted frames.

You can reduce the probability of handling an errored frame to the DLL using forward error correction in the physical layer. If you use a Hamming (7,4) FEC code, which can correct one single error in a 7-bit word, then the probability of bit error out of the decoder is reduced to $$P_d \approx \binom{7}{2} (1-P_b)^5 P_b^2 = 20.9 \times 10^{-6}.$$ We can say that the probability that a group of four bits ("nibble") is received with at least one error is approximately $P_d$.

Since a frame is made up of 16 nibbles, then the probability that a frame is received with at least one error is $$1-(1-P_d)^{16} \approx 334 \times 10^{-6},$$ or one frame in 3000.

(Here is some code that implement these ideas, and verify the results.)

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  • $\begingroup$ Shouldn't your $P_b$ have changed since the "bit error rate" on the physical channel in the coded system corresponds to only $\frac 47^{\text{th}}$ as much energy "per bit"? $\endgroup$ Sep 14 at 12:28
  • $\begingroup$ @DilipSarwate I'll take a look... I seem to remember I was assuming a BSC with fixed $P_b$, but I may be wrong. $\endgroup$
    – MBaz
    Sep 14 at 12:56
  • $\begingroup$ @DilipSarwate I am in fact assuming a BSC; in other words, I'm assuming energy increases by 7/4 when using Hamming. In my experience, most networking-oriented books take this approach. $\endgroup$
    – MBaz
    Sep 14 at 17:50
  • $\begingroup$ Well, then, I think that you are comparing apples and oranges. The output of the FEC system is nybbles of $4$ bits, and the probability that there is at least one error in the nybble is your $P_d$ (to a good approximation). There are $16$ nybbles in a $64$-bit DLL frame and so the probability of no bit errors in the DLL frame is $(1-P_d)^{16}\approx 0.99966$, not $(1-P_d)^{64}\approx 0.9987$ as you claim. This gives a much better FER than what your answer says. Finally, the comparison to the uncoded system is flawed because the $P_b$ there should be much smaller (larger SNR). $\endgroup$ Sep 14 at 19:33
  • $\begingroup$ I disagree with most of your assertions. Your calculations are way off any reasonable analysis of the problem. $\endgroup$ Sep 16 at 4:29
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The physical layer submit frames. Frames consists of symbols. Symbols consists of bits. So, FER, the probability any symbol would be in error, while SER is the probability any bit be in error. Obviously, BER, SER, and FER are related.

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