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通信原理課程習題: 1.以2^10個二進制PRBS作為消息碼,載波頻率為2.4GHz,利用MATLAB畫出QPSK調制波形的頻譜圖和星座圖,其中碼元速率為50MBaud/s 2.在上述生成的QPSK信號中引入高斯白噪聲,調節SNR至10dB(或者其他認為合適的值)然后重新生成星座圖,并計算誤碼率。(選做)

資源截圖

代碼片段和文件信息

%1.以2^10個二進制PRBS作為消息碼,載波頻率為2.4GHz,利用MATLAB畫出QPSK調制波形的頻譜圖和星座圖,其中碼元速率為50MBaud/s
%2.在上述生成的QPSK信號中引入高斯白噪聲,調節SNR至10dB(或者其他認為合適的值)然后重新生成星座圖,并計算誤碼率。(選做)
%3.要求代碼分成以下幾個部分:
%Designed?By?:
function?[]?=?homework3()
%%?生成偽隨機碼
pngen?=?commsrc.pn(‘NumBitsOut‘2^10);?%?創建偽隨機噪聲發生包
data?=?generate(pngen);?%?生成偽隨機序列
fm?=?50;?%?波特率20MHz
Tm?=?1/fm;?%?碼元周期
N_sample?=?399;?%?采樣次數為3的倍數
fs=fm*N_sample;??%采樣頻率
N?=?N_sample*length(data);?%?總采樣點個數
dt=1/fs;?%?時域采樣間隔
df=fs/N;?%?頻域采樣間隔
T?=?N*dt;?%?截斷時間
t?=?linspace(0TN);?%?時域橫坐標
Bs?=fs/2;?%?系統帶寬
f?=?linspace(-BsBsN);?%?頻域橫坐標
fc=2.4e3;?%?載波2.4GHz
SNR_DB=(0:1:12);?%?逐步改變信噪比
%%?生成兩路調制信號
NRZ?=?[];???????????%并行信號
NRZ_0?=?[];??????????%采樣后的第一路串行信號
nrz_0?=?[];?????????%采樣前的第一路串行信號
NRZ_1?=?[];??????????%采樣后的第二路串行信號
nrz_1?=?[];?????????%采樣前的第二路串行信號
for?k=1:length(data)??????%?采樣的同時也是對信號進行NRZ與RZ變換?
????for?j=1:(N_sample)???????%?NRZ變換
????????NRZ((k-1)*N_sample+j)=data(k);
????end???
end
for?k=1:(length(data)-1)??????%?采樣的同時也是對信號進行串并行變化
????if?(mod(k?2)==0)
????????for?j=1:(N_sample)???????%??串行第一路信號采樣
????????????if?data(k)==1
?????????????????NRZ_0((k-1)*N_sample+j)=1;
?????????????????nrz_0(k)=1;
????????????else
?????????????????NRZ_0((k-1)*N_sample+j)=-1;
??????????????????nrz_0(k)=-1;
????????????end
????????end
????????for?j=1:(N_sample)???????%??串行第二路信號采樣
????????????if?data(k-1)==1
?????????????????NRZ_1((k-1)*N_sample+j)=1;
?????????????????nrz_1(k)=1;
????????????else
?????????????????NRZ_1((k-1)*N_sample+j)=-1;
?????????????????nrz_1(k)=-1;
????????????end
????????end
????else
????????for?j=1:(N_sample)???????%??串行第一路信號采樣
????????????if?data(k)==1
?????????????????NRZ_1((k-1)*N_sample+j)=1;
?????????????????nrz_1(k)=1;
????????????else
?????????????????NRZ_1((k-1)*N_sample+j)=-1;
?????????????????nrz_1(k)=-1;
????????????end
????????end
????????for?j=1:(N_sample)???????%??串行第二路信號采樣
????????????if?data(k+1)==1
?????????????????NRZ_0((k-1)*N_sample+j)=1;
?????????????????nrz_0(k)=1;
????????????else
?????????????????NRZ_0((k-1)*N_sample+j)=-1;
?????????????????nrz_0(k)=-1;
????????????end
????????end
????end????
end
for?j=1:(N_sample)???????%?串行信號修正(補充對最后一個碼元的采樣)
????if?data(length(data))==1
????????NRZ_0((length(data)-1)*N_sample+j)=1;
????????nrz_0(length(data))=1;
????else
????????NRZ_0((length(data)-1)*N_sample+j)=-1;
????????nrz_0(length(data))=-1;
????end
????if?data(length(data)-1)==1
????????NRZ_1((length(data)-1)*N_sample+j)=1;
????????nrz_1(length(data))=1;
????else
????????NRZ_1((length(data)-1)*N_sample+j)=-1;
????????nrz_1(length(data))=-1;
end
end
%%?生成QPSK信號
mNRZ_0=NRZ_0.*sin(2*pi*fc*t);????????%?第一路調制后的載波信號
mNRZ_1=NRZ_1.*cos(2*pi*fc*t);???????%?第二路調制后的載波信號
QPSK?=?mNRZ_1-mNRZ_0;????????%?生成QPSK調制信號

figure(1)???????%?繪制波形圖
subplot(3?3?1);plot((0:1:N_sample*length(data)-1)*dt/N_sample?NRZ);grid?on;axis([0?length(data)*dt?-2?2]);title(‘并行信號波形‘);xlabel(‘S‘)
subplot(3?3?2);plot((0:1:N_sample*length(data)-1)*d

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