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大小: 47.9MB文件類型: .zip金幣: 1下載: 0 次發(fā)布日期: 2023-07-29
- 語言: Matlab
- 標(biāo)簽: 調(diào)頻廣播??matlab??
資源簡(jiǎn)介
調(diào)頻發(fā)射機(jī)通過調(diào)頻來調(diào)制音頻輸入。它的范圍是在美國(guó)廣播調(diào)頻廣播波段88.1-107.9兆赫。您可以使用調(diào)頻發(fā)射機(jī)頻率,間隔為100KHz,但我建議使用奇數(shù)頻率,以減少干擾廣播調(diào)頻電臺(tái)的機(jī)會(huì)。廣播調(diào)頻頻段分為200KHz頻段。這是一個(gè)相對(duì)較大的帶寬,因此它也被稱為寬帶調(diào)頻,而不是窄帶調(diào)頻,可以低到5千赫。每個(gè)通道的帶寬約為150KHz,盡管在這個(gè)范圍之外會(huì)有側(cè)帶泄漏。在調(diào)頻無線電中,信息是通過調(diào)制載波頻率進(jìn)行編碼的,
代碼片段和文件信息
%?clc;
%?clear;
%?close?all;
function?Demodulation(s_toutput_file)
%%%%%%%%%%%%%%%%%%%%%%%%%??sample?rate??%%%%%%%%%%%%%%%%%%%%%%%%%
ExtensionFactor=3;??????????????????????????%?raise?sample?rate?3?times
FS=44100;???????????????????????????????????%?sample?rate?of?wave?file
fs=FS*ExtensionFactor;??????????????????????%?sampling?rate
ts=1/fs;
%%%%%%%%%%%%%%%%%%??receive?the?signal:?s_t??%%%%%%%%%%%%%%%%%%%%
%?load?s_t_matlab;????????????????????????????%?s_t:?input?from?data-file
[tmpN]=size(s_t);??????????????????????????%?points?of?samples
MusicLen=N/ExtensionFactor;?????????????????
t=-N/2*ts:ts:(N/2-1)*ts;
%%%%%%%%%%%%%%%%%%??Obtain?s1_t?and?s2_t?from?s_t?%%%%%%%%%%%%%%%
%通過頻譜圖確定濾波器的大致帶寬值B
Nfft=2^ceil(log2(N)+2);
df=fs/Nfft;fk=(-Nfft/2:Nfft/2-1)*df;
s_f=ts*fftshift(fft(s_tNfft));
figure;plot(fkabs(s_f));
%?LPF得到L+R信號(hào)s1_t
B0=15000;???????????????????????????????????%由s_t頻譜圖得到
h=fir1(40[B0]/(fs/2));?????????????????????%?考慮41階?
s1_t=filter(h1s_t);???????????????????????%?demodulated?wave
%窄帶濾波得到導(dǎo)頻信號(hào)
B1=15000;???????????????????????????????????%?B1-B2(Hz)?BPF
B2=23000;???????????????????????????????????%?
hbpf=fir1(40[B1B2]/(fs/2));???????????????%?考慮41階
wbpf=filter(hbpf1s_t);????????????????????%?信號(hào)通過系統(tǒng)
f_dao=19000*2;
%帶通濾波得到L-R的DSB信號(hào)
B3=23000;???????????????????????????????????%?B3-B4(Hz)?BPF
B4=53000;??????????????????????????????????%?
hbpf1=fir1(40[B3B4]/(fs/2));??????????????%?考慮41階
wbpf1=filter(hbpf1s_t);???????????????????%?信號(hào)通過系統(tǒng)
%DSB想干解調(diào)得到L-R信號(hào)
s2_t0=wbpf1*2*cos(2*pi*f_dao);
B5=20000;???????????????????????????????????%由s_t頻譜圖得到
h=fir1(40[B5]/(fs/2));?????????????????????%?考慮41階?
s2_t=filter(h1s2_t0);???????????????????????%?demodulated?wave
????
%%%%%%%%%%%%%%%%%%%%%%??de-mix??%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
rL_t=(s1_t+s2_t)/2;
rR_t=(s1_t-s2_t)/2;
%%%%%%%%%%%%%%%%%%%%%%??Wave?file??%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
X=zeros(MusicLen2);
for?i=1:MusicLen
????X(i1)=rL_t(ExtensionFactor*i);?????????%?restore?44.1kHz?sample?rate?????
????X(i2)=rR_t(ExtensionFactor*i);????????
end
%?WAVWRITE(XFS‘recved_music.wav‘);
audiowrite(output_fileXFS);
disp(‘receivedmusic?signal?generated?!‘);
return;
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2018-10-10?15:39??源碼\
?????文件????????2301??2017-03-02?15:35??源碼\Demodulation.m
?????文件?????????578??2018-10-13?13:47??源碼\Modulation.m
?????文件?????1769516??2017-03-02?15:28??源碼\recved_music.wav
?????文件??????589584??2011-10-16?21:20??源碼\sending_music.wav
?????文件?????1769516??2017-03-01?16:59??源碼\speech.wav
?????文件?????9836561??2017-03-07?10:16??源碼\s_t_matlab.mat
?????文件?????????157??2018-10-13?13:46??源碼\TotalRun.m
?????目錄???????????0??2018-10-10?15:39??演示\
?????文件????36156306??2017-03-07?14:45??演示\10441調(diào)頻立體聲廣播MATLAB仿真.mp4
?????文件?????1769516??2017-03-08?10:20??演示\recved_music.wav
?????文件??????????64??2018-10-10?16:34??說明.txt
?????目錄???????????0??2018-10-10?15:38??需求\
?????文件???????24064??2017-02-28?22:00??需求\主要內(nèi)容及基本要求.doc
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