Normally for fair comparison we assume equal transmission power and equal rate. If rate can not be made equal how can be compare two different rate channel codes?
Dear Saqib Ejaz, In this situation apply joint optimization: source and channel decoding are to be compared for the decoding of variable length codes.
Dear Sir, thanks in first place. i try to elaborate my question in more details as follows,
"i have one channel code having rate 1/2 and other channel code has rate 1/4. is their any way or any scenario (or under which condition) i can present the bit error rate BER performance curves of corresponding channel codes on the same plot (BER vs Eb/N0)?"
As i use SNR per bit which is "Eb/N0=code rate*SNR"
so if code rate is different Eb/N0 will be different and the two curves can not be drawn on the same plot for a fair comparison. However, is it logically correct to assume that both channel codes are transmitted at same power and I disregard the code rate factor and plot BER curves for both codes vs. SNR instead of Eb/N0?
Please correct me if i am wrong. thank you for your time.
Dear Saqib Ejaz, With my research experience I suggested you for formulation of the problem. No problem, you are always welcomed.
The correct way is to compare for energy (Eb/N0) required for a transmission of the useful information bits, i.e. you need to normalize to compensate for the parity bits.
For example you send coded bits over the same channel and the receiver has the same SNR for both codes and each received bit has the same Eb/N0 but this for the coded bits so you have to double / quadruple this value if you want to look at the information bits as in the 1/4 code you "discard" 3 bits whereas in the other one you "discard" only 1.
This also corresponds to the fact that that the 1/4 code needs twice as much time to transmit the same information over the same channel.
I hope it makes any sense.
Dear Mr. Vit, i greatly appreciate your reply. In terms of mathematics or when it comes to simulation point of view " normalize to compensate for the parity bits" as you said, is equivalent to " SNR per bit which is "Eb/N0=code Rate*SNR" this is what actually i and i think most of people doing research in channel coding do.
what if i do not consider the CODE RATE factor and plot my BER curves against only SNR instead of Eb/N0. I know this will violate the fair comparison condition (equal rate scenario) but is there any other option so i can compare two different channel codes having different code rate?
actually I think it is SNR/coderate - i.e. the 1/4 code needs twice as much energy compared to the 1/2
I think you can compare two codes for the same SNR if you clearlyjustify why you do so. Sometimes it even makes sense.
For instance if you know this is your channel, this is your modulation scheme and you have a maximum radiated power so you know receiver's SNR and you want to see what code gives you the desired target coded BER required for a target PER/QoS. In other words you system is given, you do want to make it work, you do not really mind data-rate too much you only want to make work because for instance the 1/2 would not work.
SNR per bit= Code rate*SNR, is the actual formula. but you exactly got the situation i had the similar notion about this problem yet i needed to discuss if my intuitive thinking was in right direction. many thanks for your endorsement of underlying theory.
Dear Vit Sipal, according to your formula you suggested can we re write as
SNR required for channel code of code rate(1/2)=2*SNR required for channel code of code rate(1/4)
i.e. the 1/4 code needs twice as much energy compared to the 1/2. Kindly correct me i get this wrong. thanks for your time.
The comparison of two different rate channel codes has to be put into context:
a) power limited (or multi-user interference limited) system: the winner is the code that requires the lowest Eb/No (or SNR per info bit = SNR / code_rate) needed to achieve the required error rate.
b) bandwidth limited system: the higher code rate wins (e.g., rate 1/2 beats 1/4)
c) BOTH power and bandwidth constraints are significant: then you need to find a way to tradeoff power and bandwidth requirements. Maybe you can find a higher level abstraction of the system to identify the best.
The complexity here is one reason why many modern systems dynamically adapt the modulation and coding to the channel.
interesting answer...so how can we relate the energy consumption and code rate? In other words, a channel code having low code rate will be more energy efficient or higher code rate in a power limited channels?
SNR required for channel code of code rate(1/2)=2*SNR required for channel code of code rate(1/4)
or is it incorrect relation?
Details in this word file.
Sorry for one correction in my question, SNR per bit=Eb/N0=SNR/code rate. If i reduce this conversation to a one point can we say that ,
"the 1/4 code needs twice as much energy compared to the 1/2", as suggested by Mr. Vit Sipal. Mathematically,
SNR required for channel code of code rate(1/2)=2*SNR required for channel code of code rate(1/4)
In other words, the words written and equation presented agree to each other. Maybe this is will make everything clear.
Dear Mr. Saqib Ejaz, I am not doing research in this field but through my research experience I may suggest you to go for transformation techniques.
With regards to Vit Sipal response:
Yes, SNR_per_bit = SNR / coderate
Yes, *IF* the SNRs are the same then SNR_per_bit (or energy per bit) will need to be twice as much for rate 1/4 as rate 1/2. However, this doesn't happen for coherent communications with a well-designed rate=1/4 code. Typically, the SNR requirement drops enough so that the SNR_per_bit is equal to better than that of the rate=1/2 code. From a practical point of view, sometimes the SNR requirement doesn't drop due to overriding signal acquisition and/or tracking issues.
More generally: The coding work that I'm familiar with usually expresses these SNR metrics in terms of received energy per bit time and per symbol time, Eb and Es, respectively. Without a specific type of modulation in mind, coherent binary antipodal signalling is generally assumed, such that:
SNR = 2 * E_s / N_0 = 2 * R * E_b / N_0,
where R is the code rate of the channel code and N_0 is the one-sided power spectral density of the noise.
see "Channel Codes" by W. Ryan and Shu Lin or
"Error Control Coding" by Shu Lin and Daniel Costello
whether the curve between snr(db) and ber gives the same meaning with the curve between eb/n0 and ber?
Satheesh Monikandan Balakrishnan , For two different systems I think it is when log2(M_a)*R_a = log2(M_b)*R_b, where M is the modulation order and R is the code rate. Than you will have for both curves an equal horizontal shift between SNR and Eb/N0 of 10log10(log2(M)*R).
In case your question is about when the SNR and Eb/N0 curves will be equal then it is when log2(M)*R = 1.
I hope it help.
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