I modulated two channels using fmmod() method and demodulate that modulated signal using the fmdemod() method. How can we get back those two channels that are modulated? How can we do this?
My code is:
Fs = 8000; % Sampling rate of signal
Fc = 3000; % Carrier frequency
t = [0:Fs-1]'/Fs; % Sampling times
s1 = sin(2*pi*300*t)+2*sin(2*pi*600*t); % Channel 1
s2 = sin(2*pi*150*t)+2*sin(2*pi*900*t); % Channel 2
x = [s1,s2]; % Two-channel signal
dev = 50; % Frequency deviation in modulated signal
y = fmmod(x,Fc,Fs,dev); % Modulate both channels.
z = fmdemod(y,Fc,Fs,dev); % Demodulate both channels.
figure(1);
plot(z);
What type of value does variable z contain? How do I get back channels from the fmdemod method?
Not sure to correctly understand your question, but you can get the channel 1 by doing:
x(1:end,1) % or x(1:end,2) for channel 2
For the value z, you can get the first demodulated channel by doing:
z(1:end,1) % or z(1:end,2) for the second demodulated channel
Is it what you need?
Related
I have a matlab code to Frequency modulation and demodulation a signal. My code is work well for modulation part. My message signal is m and modulated signal is u, code plot the message signal and its integral in one graph for plotting 1.
Then signal modulated with carrier and program plots the modulated signal in time domain for plotting 2.
After that, by the help of some code blocks program find the frequency spectrum of modulated signal and message signal to plot graph of them for plotting 3.
In demodulation part program make some fundamental calculation for FM detection, then to obtain message signal it uses filter.
Last part program plots the graph of recovered signal with message signal to compare them.
I summarized all code because ı do not know whre is the problem.
My problem about plotting 3 when I make zoom graph 3 I see some phase foldings or like it.Graph is not symmetric according to y-axis.
I did not solve this problem, I research about them and I decided to use unwrap(). Although I tried a lot, I could not be successful. How can I get rid of this phase folding with unwrap() function. Thank you.
My matlab code is ;
ts = 0.0001;% Sampling interval
t0 = 0.15; % Duration
t = 0:ts:t0;% define time vector
%% OTHER PARAMETERS
fc = 200; % Carrier signal frequency
kf =50; % Frequency deviation constant
fs = 1/ts; % Sampling frequency
%% MESSAGE SIGNAL SIMPLY
m = 1*(t<t0/3)-2*(t<2*t0/3).*(t>=t0/3);
%% Integration of m
int_m(1) = 0;
for k =1:length(m)-1
int_m(k+1) = int_m(k) + m(k)*ts;
end
%% PLOTTING 1
figure; subplot(211); % Message signal
plot(t,m);grid on;xlabel('time');ylabel('Amplitude');
title('m(t)');ylim([-3 2]);
subplot(212);plot(t,int_m);% Integral of message signal
grid on; xlabel('time');ylabel('Amplitude');title('integral of m(t)');
ylim([-0.07 0.07]);
%% FM MODULATED SIGNAL
u = cos(2*pi*fc*t + 2*pi*kf*int_m);
%% PLOTTING 2
figure; plot(t,u); % Modulated signal in time domain
grid on;xlabel('time');
ylabel('Amplitude');title('FM :u(t)');
ylim([-1.2 1.2]);
%% FINDING FREQUENCY SPECTRUM AND PLOTTING 3
% Frequency spectrum of m(t)
f=linspace(-1/(2*ts),1/(2*ts),length(t));
M=fftshift(fft(m))./length(t); % Taking fourier transform for m(t)
U=fftshift(fft(u))./length(t); % Taking fourier transform for u(t)
figure;subplot(211); % Frequence spectrum of m(t)
plot(f,abs(M)); grid;
xlabel('Frequency in Hz');xlim([-500 500]);
ylabel('Amplitude');title('Double sided Magnitude spectrum of m(t)');
subplot(212);plot(f,abs(U)); % Frequency spectrum of u(t)
grid;xlabel('Frequency in Hz');xlim([-500 500]);
ylabel('Amplitude');title('Double sided Magnitude spectrum of u(t)');
%% DEMODULATION (Using Differentiator)
dem = diff(u);
dem = [0,dem];
rect_dem = abs(dem);
%% Filtering out High Frequencies
N = 80; % Order of Filter
Wn = 1.e-2; % Pass Band Edge Frequency.
a = fir1(N,Wn);% Return Numerator of Low Pass FIR filter
b = 1; % Denominator of Low Pass FIR Filter
rec = filter(a,b,rect_dem);
%% Finding frequency Response of Signals
fl = length(t);
fl = 2^ceil(log2(fl));
f = (-fl/2:fl/2-1)/(fl*1.e-4);
mF = fftshift(fft(m,fl)); % Frequency Response of Message Signal
fmF = fftshift(fft(u,fl)); % Frequency Response of FM Signal
rect_demF = fftshift(fft(rect_dem,fl));% Frequency Response of Rectified FM Signal
recF = fftshift(fft(rec,fl)); % Frequency Response of Recovered Message Signal
%% PLOTTING 4
figure;subplot(211);plot(t,m);grid on;
xlabel('time');ylabel('Amplitude');
title('m(t)');ylim([-3 2]);
subplot(212);plot(t,rec);
title('Recovered Signal');xlabel('{\it t} (sec)');
ylabel('m(t)');grid;
My problem is in this third graph to show well I put big picture
k = -(length(X)-1)/2:1:length(X)/2;
Your k is not symmetric.
If you work with symmetric k is it working fine?
I want to upsample by 5 a signal in frequency domain, and then plot (stem) it. I figured how to upsample,
Fk=(1/5)*upsample(ak_new,5);
now this creates a vector that is 5 times bigger than the original one, and I need to take the inverse Fourier series of this signal
Fn=(Fk*(exp((1i*2*pi/N*n'*n))));
where n is a sample vector (-1000:1000), as you can see, I can't make the transformation since n is not the same size as Fk anymore. How can I solve this?
You need to plot your upsampled signal in frequency domain, on a similarly "upsampled" frequency vector.
If your initial frequency vector was -1000, -999, -998 ..., now it should be -1000, -999.8, -999.6.
Here is a simple example:
Fs = 2000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 2000; % Length of signal
t = (0:L-1)*T; % Time vector
S = sin(2*pi*400*t); % Signal
Y = fft(S);
ak_new = fftshift(abs(Y/L)); % Initial signal in frequency domain
Fk = upsample(ak_new,5); % Upsampled signal
f = (Fs/L)*(-L/2: 1 :L/2-1); % Initial frequency vector
fup = (Fs/L)*(-L/2:(1/5):L/2-1/5); % Upsampled frequency vector
subplot(2, 1, 1);
stem(f, ak_new);
title('Before upsampling');
subplot(2, 1, 2);
stem(fup, Fk);
title('After upsampling');
I am trying to implement a basic DFT algorithm on Matlab.
I simply use in phase and quadrature components of a sine wave with phase modulation(increasing frequency a.k.a chirp). I do compare my results with fft command of Matlab. My code gives the same results whenever there is no phase modulation(pure sine). Whenever I add chirp modulation, results differ. For example, when I use a chirp with some bandwidth around a carrier, the expected results should be a frequency distribution of chirp bandwidth starting from carrier frequency. However, I get a copy of that result backwards starting from carrier frequency as well. You can use my code below without modifying anything. Figure 5 is my result and figure 6 is the expected result. Carrier is 256 Hz with a 10Hz bandwidth of chirp. You can see the code below. The important part is for loop where I take dft of my signal. Also uou can see my dft result below.
close all;
clear all;
%% signal generation
t = (0:0.0001:1); % 1 second window
f = 256; %freq of input signal in hertz
bw = 10; % bandwidth sweep of signal
phaseInput = 2*pi*t*bw.*t;
signalInput = sin(2*pi*f*t + phaseInput); %input signal
inphase = sin(2*pi*f*t).*cos(phaseInput); %inphase component
quadrature = cos(2*pi*f*t).*sin(phaseInput); %quadrature component
figure
plot(t,signalInput,'b',t,inphase,'g',t,quadrature,'r');
title('Input Signal');
xlabel('Time in seconds');
ylabel('Amplitude');
%% sampling signal previously generated
Fs = 1024; %sampling freq
Ts = (0:1/Fs:1);%sample times for 1 second window
sPhase = 2*pi*Ts*bw.*Ts;
sI = sin(2*pi*f*Ts).*cos(sPhase);
sQ = cos(2*pi*f*Ts).*sin(sPhase);
hold on;
plot(Ts,sI+sQ,'b*',Ts,sI,'g*',Ts,sQ,'r*');
fftSize = Fs; %Using all samples in dft
sampleIdx = (0:1:fftSize-1)';
sampledI = sI(1:fftSize)';
sampledQ = sQ(1:fftSize)';
figure;
plot(sampleIdx,sampledI,sampleIdx,sampledQ);
title('Sampled IQ Components');
%% DFT Calculation
dftI = zeros(fftSize,1);
dftQ = zeros(fftSize,1);
for w = 0:fftSize-1
%exp(-2*pi*w*t) = cos(2*pi*w*t) - i*sin(2*pi*w*t)
cI = cos(2*pi*w*sampleIdx/fftSize); %correlation cos
cQ = -sin(2*pi*w*sampleIdx/fftSize); %correlation sin
dftI(w+1) = sum(sampledI.*cI - sampledQ.*cQ); %
dftQ(w+1) = sum(sampledI.*cQ + sampledQ.*cI);
end;
figure;
plot(Fs*sampleIdx/fftSize,dftI);
title('DFT Inphase');
xlabel('Hertz');
figure
plot(Fs*sampleIdx/fftSize,dftQ);
title('DFT Quadrature');
xlabel('Hertz');
figure;
plot(Fs*sampleIdx/fftSize,sqrt(dftQ.^2+dftI.^2));
%% For comparison
sampledInput = sin(2*pi*f*Ts + sPhase);
Y = fft(sampledInput(1:1024),1024);
Pyy = Y.*conj(Y)/1024;
f = (0:1023);
figure;
plot(f,Pyy)
title('Power spectral density')
xlabel('Frequency (Hz)')
the reason lies in the fact that two different signals will definitely give your two different frequency spectrums. check out the code below, you will find that the input of the dft algorithm you actually gave is sampledI+jsampledQ. as a result, what you are doing here is NOT merely decomposing your original signal into In-phase and quadrature components, instead, you are doing Hilbert transform here -- to change a real signal into a complex one.
cI = cos(2*pi*w*sampleIdx/fftSize); %correlation cos
cQ = -sin(2*pi*w*sampleIdx/fftSize); %correlation sin
dftI(w+1) = sum(sampledI.*cI - sampledQ.*cQ); %
dftQ(w+1) = sum(sampledI.*cQ + sampledQ.*cI);
so the sampledInput for comparison should be sampledInput = sI+1i*sQ;.
i want to make a continious sinwave which will change his frequency 3 times, like in the photo. I tryied with the fmmod function but i can't make it work, so is there any other way , thanks in advance.
here is my code
Fs = 44100; % Sampling rate of signal
Fc = 3000; % Carrier frequency
t = [0:Fs-1]'/Fs; % Sampling times
x = sin(2*pi*4000*t)+sin(2*pi*5000*t)+sin(2*pi*8000*t); % my signal
dev = 50; % Frequency deviation in modulated signal
y = fmmod(x,Fc,Fs,dev);
plot(y)
Frequency modulation without fmmod:
Fs = 44100; % Sampling rate of signal
T=1/Fs; % period
t = 0:T:0.01; % sample times
% amplitude of message signal
vm = 5
% amplitude of carrier signal
vc =5
% message frequency
fm =500
% carrier freq
fc = 5e3
% modulation index
m =5
figure
y = vc*sin(2*pi*fc*t+m.*cos(2*pi*fm*t));
plot(t,y);
I am trying to extract a sinusoid which itself has a speed which changes sinusiodially. The form of this is approximately sin (a(sin(b*t))), a+b are constant.
This is what I'm currently trying, however it doesnt give me a nice sin graph as I hope for.
Fs = 100; % Sampling rate of signal
Fc = 2*pi; % Carrier frequency
t = [0:(20*(Fs-1))]'/Fs; % Sampling times
s1 = sin(11*sin(t)); % Channel 1, this generates the signal
x = [s1];
dev = 50; % Frequency deviation in modulated signal
z = fmdemod(x,Fc,Fs,fm); % Demodulate both channels.
plot(z);
Thank you for your help.
There is a bug in your code, instead of:
z = fmdemod(x,Fc,Fs,fm);
You should have:
z = fmdemod(x,Fc,Fs,dev);
Also to see a nice sine graph you need to plot s1.
It looks like you are not creating a FM signal that is modulated correctly, so you can not demodulate it correctly as well using fmdemod. Here is an example that does it correctly:
Fs = 8000; % Sampling rate of signal
Fc = 3000; % Carrier frequency
t = [0:Fs]'/Fs; % Sampling times
s1 = sin(2*pi*300*t)+2*sin(2*pi*600*t); % Channel 1
s2 = sin(2*pi*150*t)+2*sin(2*pi*900*t); % Channel 2
x = [s1,s2]; % Two-channel signal
dev = 50; % Frequency deviation in modulated signal
y = fmmod(x,Fc,Fs,dev); % Modulate both channels.
z = fmdemod(y,Fc,Fs,dev); % Demodulate both channels.
If you find thr answers useful you can both up-vote them and accept them, thanks.