Why do i need to divided by log2(M) * Ts instead of Ts ? For the symbol error rate of 8-PSK

I saw this code of 8-PSK in the internet,and i don't know why do i need to divided by log2(M) * Ts instead of Ts

code:

M=8;
r=sqrt(3);
Ts=10^4;
SNR_dB=[-3,0,3,6,9];
Ne=zeros(1,length(SNR_dB));
y=zeros(1,5);
theta=zeros(1,M);
Eav=r^2;
Eavb=Eav/log2(M);
for ii=1:length(SNR_dB)
    SNR=10^(SNR_dB(ii)/10);
    No=Eavb/SNR;
    for jj=1:M
        theta(jj)=((jj-1)*2*pi/M);
    end
    for t=1:Ts
        s=randi(M)-1;
        s_bin=dec2bin(s,3)-'0';
        s_new=r*cos(theta(s+1))+1i*r*sin(theta(s+1));
        n=randn(1,2)*sqrt(No/2);
        r0=real(s_new)+n(1);%
        r1=imag(s_new)+n(2);
        
        theta_r=angle(r0+r1*1i);
        
        if -1*pi<theta_r && theta_r <0
            theta_r=theta_r+2*pi;
        end
        
        if (theta(1)-(pi/M)+(2*pi))<theta_r ||theta_r <(theta(1)+(pi/M))
            s_hat=0;
            s_hat_bin=dec2bin(s_hat,3)-'0';
        end
        
        detect
        for k=2:M
            if (theta(k)-(pi/M))<theta_r && theta_r<(theta(k)+(pi/M))
                s_hat=k-1;
                s_hat_bin=dec2bin(s_hat,3)-'0';
            end
        end
        
        for n=1:log2(M)
            if s_bin(n)~=s_hat_bin(n)
                Ne(1,ii)=Ne(1,ii)+1;
            end
        end
        y(1,ii)=Ne(1,ii)./(log2(M)*Ts)%why do i need to divided by log2(M) * Ts instead of  Ts 
    end
end
semilogy(SNR_dB,y,'-r');
grid on

Because the error probability is : total error numuber/total number,in this code,i create Ts number,and turn them into the the 3bits binary code,i mean 0=>000,7=>111.So in fact,i create Ts numbers instead of log2(M)*Ts.

And i modify these code

for n=1:log2(M)
    if s_bin(n)~=s_hat_bin(n)
       Ne(1,ii)=Ne(1,ii)+1;
    end
end
 y(1,ii)=Ne(1,ii)./(log2(M)*Ts)

to

 if s_bin~=s_hat_bin
     Ne(1,ii)=Ne(1,ii)+1
 end
y(1,ii)=Ne(1,ii)./Ts

The simulation become wrong ,why?

The second code i can explain that i compare the nth bit of s_bin with nth bit of s_hat_bin,if their bit are not equal,then the error number will plus one.

The third code i can explain that i compare the s_bin with nth bit of s_hat_bin,i mean like s_bin=000,s_hat_bin = 001,then s_bin is not equal to s_hat_bin so plus 1.

Or is the second code for bit error rate and the third code for the symbol error rate?

Can anyone tell me?

Answer 1

[KALYAN ACHARJYA]

Why do i need to divided by log2(M) * Ts instead of Ts ?

Please check the below expression and discuss with your communication teacher.

Detail here

[yang-En Hsiao]

As you said ,Ts=Tb *log2(M),it means that three bits is equal to 1 symbol,however,in the begining of the code ,i said

for t=1:Ts
        s=randi(M)-1;
        s_bin=dec2bin(s,3)-'0';

I means i create Ts symbol ,and just let them turn into three bits for each symbol,and in this code ,it said Ts *log2(M),it is symbol number times log2(M),but the formula you mention is bit number times log2(M),so that is different .

And the second code,it said

for n=1:log2(M)
    if s_bin(n)~=s_hat_bin(n)
       Ne(1,ii)=Ne(1,ii)+1;
    end
end
 y(1,ii)=Ne(1,ii)./(log2(M)*Ts)

so when the s_bin=001.s_hat_bin=010,the Ne will become 2,because we have two different bits,but there just one different for the symbol,because s_bin=001=1,s_hat_bin=010=2,symbol 1 and symbol 2 are different