%M=2,h=1/2,GMSK，N=512,g=[17,15],Ns=16,2倍采樣,L=3,H with 4 6 and 8 taps, 5 iterations
clc
clear
EbN0=[0:2:10];
ber1 = [6.8369e-002  2.7778e-002  7.2165e-003  2.0825e-003  6.4205e-004  1.3168e-004]; % H with 4 taps
% ber2 = [7.1955e-002  2.4486e-002  6.9854e-003  2.0764e-003  5.6302e-004
% 1.1854e-004]; % H with 6 taps
ber2 = [8.3563e-002  3.5679e-002  7.5714e-003  1.9477e-003  5.9810e-004  1.2020e-004]; % H with 8 taps
% semilogy(EbN0,ber1,'k-o',EbN0,ber2,'k-s',EbN0,ber3,'k->','LineWidth',2);
% ber3 = [6.0140e-002  1.6994e-002  4.5214e-003  1.4522e-003  4.2761e-004  5.5992e-005]; % ML
semilogy(EbN0,ber1,'k-o',EbN0,ber2,'k-s','LineWidth',2);
grid
xlabel('E_b/N_0(dB)');
ylabel('BER');
% legend('H has 4 taps','H has 6 taps','H has 8 taps');
legend('H_2_n has 4 taps','H_2_n has 8 taps');