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Channel Capacity in AWGN

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gama

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awgn channel capacity

Hi,

in AWGN, we have channel capacity equation:
C = (½)log(1+SNR),
but if we use QAM16,
then the SNR means Eb/No or Es/No or Eav/No?

and even more if we are in Linear-filtering channel,
e.g, h(t)=0.8δ(t)-0.48δ(t-T)+0.36δ(t-2T),
then how we change the C equation or not?

thx.
gama
 

channel capacity qpsk

i guess the channel capacity = C=B log(1+S/N) not 1/2 log(1+s/n)
where S=Eb*Rb ;Rb is the bit rate
N=No*B ;where No is the noise power spectral density & B is the band width of channel
so i guess if you used the Es u shall use the Rs which = Rb/k ;where k=log (base2) M where K is the number of bits per symbol
& Es then is equal to Eb *Log(base 2) M which will get you back to Eb*Rb
note that this is valid for grey coding only ie: 1 bit error only allowed in the symbol
 
channel capacity awgn

The capacity C=B log(1+S/N) is measured in [bits/sec] while C=1/2 log(1+s/n) is in [bits/sec/Hz]. No difference.

Concerning the multipath situation, the capacity does change for sure. But, for your 16QAM, I do not have any idea. In general, however, you can transform the original channel to parallel channel (for example, OFDM technique) and get the capacity of this parallel channel as sum of individual channels.

Following reference might be helpful:
Digital Communication
by Edward A. Lee, David G. Messerschmitt
(I am sure it answers your first quesiton)
 

awgn equation

No Comment!

In an M-ary system :
Es/N0 (dB)=Eb/N0(dB) +log2(k) ,where k=code rate×log2(M)

(when coding is not used ,code rate is equal to 1)

Es/N0(dB)=10log10(Tsymbol/Tsampling) + SNR(dB)

and finally : C=B log(1+SNR)

It's that simple!
:D:D:D
 
channel capacity vs snr

amihomo said:
No Comment!

In an M-ary system :
Es/N0 (dB)=Eb/N0(dB) +log2(k) ,where k=code rate×log2(M)

(when coding is not used ,code rate is equal to 1)

Es/N0(dB)=10log10(Tsymbol/Tsampling) + SNR(dB)

and finally : C=B log(1+SNR)

It's that simple!
:D:D:D
Click to expand...

Sorry, but you ignore the informational meaning of C=B log(s+SNR).

Take Tsymbol=Tsampling and no channel coding for example, your equation goes unbounded when EbNo goes to infinity. This is not true since we know for example QPSK the capacity is limited to 2bits/sec/Hz no matter how large EbN0 (SNR) is.
 

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capacity awgn channel

Hi,

I think this thesis is one of the best in this field.

Regards
 
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capacity of awgn

changfa said:
amihomo said:
No Comment!

In an M-ary system :
1) Es/N0 (dB)=Eb/N0(dB) +log2(k) ,where k=code rate×log2(M)

(when coding is not used ,code rate is equal to 1)

2) Es/N0(dB)=10log10(Tsymbol/Tsampling) + SNR(dB) , for complex input signals

3) and finally : C=B log(1+SNR)

It's that simple!
:D:D:D
Click to expand...

Sorry, but you ignore the informational meaning of C=B log(s+SNR).

Take Tsymbol=Tsampling and no channel coding for example, your equation goes unbounded when EbNo goes to infinity. This is not true since we know for example QPSK the capacity is limited to 2bits/sec/Hz no matter how large EbN0 (SNR) is.
Click to expand...

Hi,
Obviously there is no doubt that equation (1) and (3) are correct , to prove equation (2) :

Es/N0(dB)=10log10(S×Tsymbol/(N/Bn))
=10log10( (Tsymbol×Fs).(S/N) )
=10log10(Tsymbol/Tsampling) + 10log10(S/N )
=10log10(Tsymbol/Tsampling) + SNR(dB)

where :
S=input signal power ,(Watts)
N=noise power ,(Watts)
Bn=noise Bandwidth ,(Hz)
Fs=sampling frequency ,(Hz)

(note that: Bn=Fs=1/Tsampling)

what you said about Eb/N0->∞ also holds if we let SNR->∞ in eq. (3) . and this not related to eq. (1) and (2)
 
channel capacity snr

amihomo said:
changfa said:
amihomo said:
No Comment!

In an M-ary system :
1) Es/N0 (dB)=Eb/N0(dB) +log2(k) ,where k=code rate×log2(M)

(when coding is not used ,code rate is equal to 1)

2) Es/N0(dB)=10log10(Tsymbol/Tsampling) + SNR(dB) , for complex input signals

3) and finally : C=B log(1+SNR)

It's that simple!
:D:D:D
Click to expand...

Sorry, but you ignore the informational meaning of C=B log(s+SNR).

Take Tsymbol=Tsampling and no channel coding for example, your equation goes unbounded when EbNo goes to infinity. This is not true since we know for example QPSK the capacity is limited to 2bits/sec/Hz no matter how large EbN0 (SNR) is.
Click to expand...

Hi,
Obviously there is no doubt that equation (1) and (3) are correct , to prove equation (2) :

Es/N0(dB)=10log10(S×Tsymbol/(N/Bn))
=10log10( (Tsymbol×Fs).(S/N) )
=10log10(Tsymbol/Tsampling) + 10log10(S/N )
=10log10(Tsymbol/Tsampling) + SNR(dB)

where :
S=input signal power ,(Watts)
N=noise power ,(Watts)
Bn=noise Bandwidth ,(Hz)
Fs=sampling frequency ,(Hz)

(note that: Bn=Fs=1/Tsampling)

what you said about Eb/N0->∞ also holds if we let SNR->∞ in eq. (3) . and this not related to eq. (1) and (2)
Click to expand...

The problem is your logic. In your first post, it seems that you want prove (3) by (1) and (2), and this procedure does not hold for sure. For your second post, you want use (1) and (3) to prove (2), which itself has nothing to do with (3) at all.

The bottomline is: the famous C=B log(1+P/NB) bits/sec is defined for Gaussian channel and this value C is an intrinsic property of the channel. When we talk about a specific modulation scheme, we have to calculate its capacity curve (the above reference by some friend might help), which is definitely below the famous C~SNR curve.
 

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capacity awgn

Hi Changfa,
I didn't mean to prove eq (3) by (1) and (2), i meant to show how Eb/N0 and Es/N0 are related to the SNR in eq (3).

it is evident that C is an intrinsic property of the channel.As you know, by definition, channel capacity (C) is the maximum data rate at which reliable transmission of information over the channel is possible.And according to shannon's limit it is an upper bound for the reliable data rate we can achieve in any modulation scheme, i.e. , At rates R<C reliable transmission of information over the channel is possible .

regards
 

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capacity snr

The channel capacity is independent to the modulation format. If 16 QAM is used, it is more efficient than QPSK and more close to the channel capacity.
 

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