bpsk调制matlab代码（bp算法matlab代码）[20240421更新]

今天给各位分享bpsk调制matlab代码的知识，其中也会对bp算法matlab代码进行解释，如果能碰巧解决你现在面临的问题，别忘了关注本站，现在开始吧！

本文目录一览：

• 1、用matlab进行BPSK调制和解调的仿真
• 2、如何用Matlab编写码率为1/2的卷积码+BPSK调制
• 3、matlab编程实现最大似然解调 采用BPSK调制 并画出误码率曲线
• 4、求matlab高手解答
• 5、用MATLAB求BPSK、BFSK、BASK的解调和调制程序
• 6、随机产生100个数字信号．对其进行bpsk调制 其载波频率为100hz 试用matlab画出 急求

用matlab进行BPSK调制和解调的仿真

function output_frame = demodulation(input_modu, index)

% demodulation for IEEE802.11a

% Input: input_modu, complex values representing constellation points

% index

% Output: output_frame, output bit stream (data unit is one bit)

% In this version, increase the quatilization levels into 8.

% note: Matlab index starts from 1

Q_length=length(input_modu);

QAM_input_I = real(input_modu);

QAM_input_Q = imag(input_modu);

output_frame = zeros(1,length(input_modu)*index);

switch index

case 1,

BPSK_Demodu_I = [0 1]; %f(m)=(m+1)/2 + 1, so I=-1 --- 1, I=1 --- 2

idx = find(QAM_input_I1);

QAM_input_I(idx) = 1;

idx = find(QAM_input_I-1);

QAM_input_I(idx) = -1;

output_frame = BPSK_Demodu_I(round((QAM_input_I+1)/2) + 1);

case 2,

QPSK_Demodu_IQ = [0 1]; %f(m)=(m+1)/2 + 1, so I=-1 --- 1, I=1 --- 2

idx = find(QAM_input_I1);

QAM_input_I(idx) = 1;

idx = find(QAM_input_I-1);

QAM_input_I(idx) = -1;

idx = find(QAM_input_Q1);

QAM_input_Q(idx) = 1;

idx = find(QAM_input_Q-1);

QAM_input_Q(idx) = -1;

output_frame(1:2:end) = QPSK_Demodu_IQ(round((QAM_input_I+1)/2) + 1);

output_frame(2:2:end) = QPSK_Demodu_IQ(round((QAM_input_Q+1)/2) + 1);

case 3,

remapping=[0 0 0;0 0 1;0 1 0;0 1 1;1 0 0;1 0 1;1 1 0;1 1 1].';

for i=1:Q_length

phase_det=[2QAM_input_I(i)0QAM_input_Q(i) 0QAM_input_I(i)QAM_input_I(i)20QAM_input_Q(i) QAM_input_I(i)-20QAM_input_Q(i) -2QAM_input_I(i)QAM_input_I(i)00QAM_input_Q(i) QAM_input_I(i)-2QAM_input_Q(i)0 QAM_input_I(i)0-2QAM_input_I(i)QAM_input_Q(i)0 2QAM_input_I(i)QAM_input_Q(i)0 0QAM_input_I(i)QAM_input_I(i)2QAM_input_Q(i)0]

a=find(phase_det);

output_frame((1+(i-1)*3):(3+(i-1)*3))=remapping((1+(a-1)*3):(3+(a-1)*3));

end

case 4,

QAM_16_Demodu_IQ = [0 1 3 2]; %f(m)=(m+3)/2 + 1, so I=-3 --- 1, I=1 --- 3

idx = find(QAM_input_I3);

QAM_input_I(idx) = 3;

idx = find(QAM_input_I-3);

QAM_input_I(idx) = -3;

idx = find(QAM_input_Q3);

QAM_input_Q(idx) = 3;

idx = find(QAM_input_Q-3);

QAM_input_Q(idx) = -3;

tmp = round((QAM_input_I+3)/2) + 1;

output_frame(1:4:end) = bitget(QAM_16_Demodu_IQ(tmp),2);

output_frame(2:4:end) = bitget(QAM_16_Demodu_IQ(tmp),1);

tmp = round((QAM_input_Q+3)/2) + 1;

output_frame(3:4:end) = bitget(QAM_16_Demodu_IQ(tmp),2);

output_frame(4:4:end) = bitget(QAM_16_Demodu_IQ(tmp),1);

case 5,

remapping=[0 0 0 0 0;0 0 0 0 1;0 0 0 1 0;0 0 0 1 1;0 0 1 0 0;0 0 1 0 1;0 0 1 1 0;0 0 1 1 1;

0 1 0 0 0;0 1 0 0 1;0 1 0 1 0;0 1 0 1 1;0 1 1 0 0;0 1 1 0 1;0 1 1 1 0;0 1 1 1 1;

1 0 0 0 0;1 0 0 0 1;1 0 0 1 0;1 0 0 1 1;1 0 1 0 0;1 0 1 0 1;1 0 1 1 0;1 0 1 1 1;

1 1 0 0 0;1 1 0 0 1;1 1 0 1 0;1 1 0 1 1;1 1 1 0 0;1 1 1 0 1;1 1 1 1 0;1 1 1 1 1].';

for i=1:Q_length

phase_det=[4QAM_input_I(i)0QAM_input_Q(i)QAM_input_Q(i)2;2QAM_input_I(i)QAM_input_I(i)40QAM_input_Q(i)QAM_input_Q(i)2;0QAM_input_I(i)QAM_input_I(i)20QAM_input_Q(i)QAM_input_Q(i)2;4QAM_input_I(i)2QAM_input_Q(i)QAM_input_Q(i)4;

2QAM_input_I(i)QAM_input_I(i)42QAM_input_Q(i)QAM_input_Q(i)4;0QAM_input_I(i)QAM_input_I(i)22QAM_input_Q(i)QAM_input_Q(i)4;2QAM_input_I(i)QAM_input_I(i)44QAM_input_Q(i);0QAM_input_I(i)QAM_input_I(i)24QAM_input_Q(i);

QAM_input_I(i)-40QAM_input_Q(i)QAM_input_Q(i)2;-4QAM_input_I(i)QAM_input_I(i)-20QAM_input_Q(i)QAM_input_Q(i)2;-2QAM_input_I(i)QAM_input_I(i)00QAM_input_Q(i)QAM_input_Q(i)2;QAM_input_I(i)-42QAM_input_Q(i)QAM_input_Q(i)4;

-4QAM_input_I(i)QAM_input_I(i)-22QAM_input_Q(i)QAM_input_Q(i)4;-2QAM_input_I(i)QAM_input_I(i)02QAM_input_Q(i)QAM_input_Q(i)4;-4QAM_input_I(i)QAM_input_I(i)-24QAM_input_Q(i);-2QAM_input_I(i)QAM_input_I(i)04QAM_input_Q(i);

QAM_input_I(i)-4-2QAM_input_Q(i)QAM_input_Q(i)0;-4QAM_input_I(i)QAM_input_I(i)-2-2QAM_input_Q(i)QAM_input_Q(i)0;-2QAM_input_I(i)QAM_input_I(i)00QAM_input_Q(i)QAM_input_Q(i)0;QAM_input_I(i)-4-4QAM_input_Q(i)QAM_input_Q(i)-2;

-4QAM_input_I(i)QAM_input_I(i)-2-4QAM_input_Q(i)QAM_input_Q(i)-2;-2QAM_input_I(i)QAM_input_I(i)0-4QAM_input_Q(i)QAM_input_Q(i)-2;-4QAM_input_I(i)QAM_input_I(i)-2QAM_input_Q(i)-4;-2QAM_input_I(i)QAM_input_I(i)0QAM_input_Q(i)-4;

4QAM_input_I(i)-2QAM_input_Q(i)QAM_input_Q(i)0;2QAM_input_I(i)QAM_input_I(i)4-2QAM_input_Q(i)QAM_input_Q(i)0;0QAM_input_I(i)QAM_input_I(i)20QAM_input_Q(i)QAM_input_Q(i)0;4QAM_input_I(i)-4QAM_input_Q(i)QAM_input_Q(i)-2;

2QAM_input_I(i)QAM_input_I(i)4-4QAM_input_Q(i)QAM_input_Q(i)-2;0QAM_input_I(i)QAM_input_I(i)2-4QAM_input_Q(i)QAM_input_Q(i)-2;2QAM_input_I(i)QAM_input_I(i)4QAM_input_Q(i)-4;0QAM_input_I(i)QAM_input_I(i)2QAM_input_Q(i)-4];

a=find(phase_det);

output_frame((1+(i-1)*5):(5+(i-1)*5))=remapping((1+(a-1)*5):(5+(a-1)*5));

end %5+i 3+i 1+i 5+3*i 3+3*i 1+3*i 3+5*i 1+5*i -5+i -3+i -1+i -5+3*i -3+3*i -1+3*i -3+5*i -1+5*i -5-i -3-i -1-i -5-3*i -3-3*i -1-3*i -3-5*i -1-5*i 5-i 3-i 1-i 5-3*i 3-3*i 1-3*i 3-5*i 1-5*i

case 6,

QAM_64_Demodu_IQ = [0 1 3 2 6 7 5 4]; %f(m)=(m+7)/2 + 1, so I=-7 --- 1, I=1 --- 5

idx = find(QAM_input_I7);

QAM_input_I(idx) = 7;

idx = find(QAM_input_I-7);

QAM_input_I(idx) = -7;

idx = find(QAM_input_Q7);

QAM_input_Q(idx) = 7;

idx = find(QAM_input_Q-7);

QAM_input_Q(idx) = -7;

tmp = round((QAM_input_I+7)/2) + 1;

output_frame(1:6:end) = bitget(QAM_64_Demodu_IQ(tmp),3);

output_frame(2:6:end) = bitget(QAM_64_Demodu_IQ(tmp),2);

output_frame(3:6:end) = bitget(QAM_64_Demodu_IQ(tmp),1);

tmp = round((QAM_input_Q+7)/2) + 1;

output_frame(4:6:end) = bitget(QAM_64_Demodu_IQ(tmp),3);

output_frame(5:6:end) = bitget(QAM_64_Demodu_IQ(tmp),2);

output_frame(6:6:end) = bitget(QAM_64_Demodu_IQ(tmp),1);

end

如何用Matlab编写码率为1/2的卷积码+BPSK调制

如何用Matlab编写码率为1/2的卷积码+BPSK调制

可以根据K0和Fm算出总的时间,然后除以采样点数,就是载频的时间间隔吧,我觉得是这样的,你试试。

matlab编程实现最大似然解调 采用BPSK调制 并画出误码率曲线

您好，您这样：

clc;

clear;

fc=4800;fs=12000;fb=2400;

%要调制的数字信号

a=randint(1,12,2);%随机产生12个“0”，“1”

s=zeros(1,60);

for i=1:12

for j=1:5

if(a(i)==0)

s(j+(i-1)*5)=0;

else

s(j+(i-1)*5)=1;

end

end

end

plot(s);xlabel('基带信号');

figure

pwelch(s);%功率谱

figure

%波形成形滤波器(平方根升余弦滚降)

h=firrcos(14,1200,1200,4800,'sqrt');

figure

stem(h);xlabel('成形滤波器的单位冲击响应');

[H,W]=freqz(h,1);

H=abs(H);

figure

plot(H);xlabel('成形滤波器的频率响应');

s=fftfilt(h,s);

figure

plot(s);xlabel('通过成形滤波器后的基带信号');

figure

pwelch(s);%经波形成形滤波器后的功率谱

%已调信号

e=dmod(a,4800,2400,12000,'psk',2);%调制

figure

plot(e);xlabel('已调信号');

enoise=e+randn(1,60);%enoise=e+.1*randn(1,60)不同功率的高斯白噪声

aa=ddemod(enoise,4800,2400,12000,'psk',2);%解调

figure

stem(aa);xlabel('解调后的数字信号');%解调后的数字信号

p=symerr(a,aa)/12 %误码率

%误码率曲线

figure

r=-6:3:12;

rr=10.^(r/10);

pe1=1/2*exp(-rr);%相干解调的误码率曲线

hold on

plot(r,pe1,'r');grid on;

pe2=(1-1/2*erfc(sqrt(rr))).*erfc(sqrt(rr));%差分相干解调的误码率曲线

plot(r,pe2,'b');xlabel('bpsk,dpsk误码率曲线');

set(gca,'XTick',-6:3:18);

求matlab高手解答

很明显是你这句r = u + sgma .* randn(size(u));的u与sgma .* randn(size(u))，或者sgma .* randn(size(u))没法计算，

由于不知道你的GSMconv_EN(s);函数式什么，不知道u的行数与列数，这里关键是你要给的输入让sgma与u的行列恰当。

要求r = u + sgma .* randn(size(u));这句要求sgma的行列与u的行列一样。你好好检查下，应该可以查出来的。

希望对你有帮组

用MATLAB求BPSK、BFSK、BASK的解调和调制程序

【PDF】通信原理matlab

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-基于Matlab仿真计算 曾峰 zf0579@sina.com You can ...3.1.1 BASK 本节主要讨论幅度调制技术,从最简单的二...0.999; %Frequency Offset df = 10; %Source BFSK...比特数.当M等于2,这时PSK通常被称为二进制相移键控(BPSK...

百度搜下。

随机产生100个数字信号．对其进行bpsk调制 其载波频率为100hz 试用matlab画出 急求

FSK是在数字信号调制中使用较典型的一种调制方式,其利用载波的频率变化来传递中二进制数据用伪随机序列PNSeq仿真1对应100 Hz 正弦波. 幅度A 均为

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