Matlab simulation problem about ifft

My problem is that i cannot reconstruct to the input signal (random binary signal) x(t) by using ifft from frequency domain to time domain.Even a square wave, i cannot reconstruct to x(t) after Y = fft(A,nfft).please someone help

clear; 

%This is an input signal, x(t) 

for i = 0:9 
A(i*10+1:i*10+10) = (rand(1)>0.5); 
end 
X = (1:10*(9+1)); 

%Fourier transform of a square wave pulse with time interval 1 ns each by 
%taking 512 point. 

T = 1e-09; 
Fs = 1/T; 
nfft = 512; 
Y = fft(A,nfft); 
Y = Y(1:nfft/2); 
mx = Y; 
f = (0:nfft/2-1)*Fs/nfft; 

%Plotting random binary signal,x(t) 
figure(1); 
subplot(2,1,1); stem(X/10,A); 
xlim([0 15]); ylim([-1.5 1.5]); 
title('Random binary Signal'); 

%Plotting fourier transform of random binary signal 
subplot(2,1,2); plot(f,mx); 
title('Fourier Transform of a random binary');

Answer 1

[Wan]

anyone help please i only need to solve the domain problems from frequency to time. 10 points in time domain but 512 points in frequency domain

Answer 2

[Rick Rosson]

In your code, I see where you have defined A as a function of X, but I do not see where you have defined x as a function of t. In fact, there is no vector of time values at all. Furthermore, the code computes Y as the FT of A whereas the block diagram shows that Y is the output of a system with transfer function of H. Also, the code makes no use of the ifft function, which is the main focus of your question.

Please re-write your code so that it represents what the block diagram shows. You may want to start by defining x as a function of t where t is a vector of time values with increment of 1 ns.

HTH.

Answer 3

[Rick Rosson]

Here is a start:

   dt = 1e-9;
   Fs = 1/dt;
   N = 10;
   t = dt*(0:N-1)';

   x = (rand(10,1) > 0.5);

   X = fft(x);
   dF = Fs/N;

   f = ...

   H = ...

   Y = H.*X;

   y = ifft(Y);

   ...

   ...

HTH.

Answer 4

[Wan]

for the input signal x(t), i have to take 10 point to represent 1 ns, and each ns only represent one or zero. if i want to have 10ns signal,that means there are ten binary 1 or 0

Answer 5

[Rick Rosson]

Let me introduce some commonly used terminology that I think will help you out. Based on what you are saying, you want to have a binary signal x( t ), consisting of randomly-generated 0's and 1's. The symbol time of your signal will be 1 nanoseconds per symbol, corresponding to a symbol rate of 1 x 10^9 symbols per second. The signal will have an oversampling rate of 10 samples per symbol, so that the sampling rate will be 10 x 10^9 samples per second, corresponding to a sampling time of 0.1 nanoseconds per sample.

Does that make sense?

Here is some code that should help you get started:

   % Oversampling rate (in samples per symbol):
   oversamplingRate = 10;

   % Symbol time (in seconds per symbol):
   symbolTime = 1e-9;

   % Number of symbols:
   N = 200;

   % Create binary signal:
   b = double(rand(1,N) > 0.5);
   x = repmat(b,oversamplingRate,1);
   x = x(:);

   % Simulation start and stop times (in seconds):
   startTime = 0;
   stopTime = startTime + N*symbolTime;

   % Sampling time (in seconds per sample):
   dt = symbolTime/oversamplingRate;

   % Sampling rate (in samples per second):
   Fs = 1/dt;

   % Vector of time values (in seconds):
   t = (startTime:dt:stopTime-dt)';

   figure;
   stem((0:N-1)',b);
   ylim([-0.2 1.2]);
   xlabel('Index number');

   figure;
   stem(t/1e-9,x);
   ylim([-0.2 1.2]);
   xlabel('Time (in nanoseconds)');

I hope that helps.

Rick

[Wan]

thanks a lot, Rick. But i have the fourier transform of x(t) output like this, and i don't think this is a correct transform of x(t).

i used your code to design the time domain, here is my code:

clear;

%%%%%%%%%%%%%%% Random binary signal, input x(t)%%%%%%%%%%%%%%

startTime = 0; stopTime = 3e-9; oversamplingRate = 10; symbolTime = 1e-9; N = (stopTime - startTime)/symbolTime; x = double(rand(1,N) > 0.5); x = repmat(x,oversamplingRate,1); x = x(:)'; dt = symbolTime/oversamplingRate; Fs = 1/dt; t = startTime:dt:stopTime-dt;

%%%%%%%%%%%%%%% Fourier transform of binary signal, X(w) %%%%%%%%%%%%%%%%

nfft = 512; X = fft(x,nfft); X = X(1:nfft/2); Xw = (X); % representing the foruier transform,Xw f = (0:nfft/2-1)*Fs/nfft;

%%%%%%%%%%%%%% Plotting binary signal x(t) %%%%%%%%%%%%%% figure(1); subplot(211);stem(t,x); xlim([0 4e-9]);ylim([-1 1.5]); %% axis limit title('Random Binary signal,x(t)'); xlabel('Time'); ylabel('Amplitude');

%%%%%%%%%%%%%% Plotting fourier transform X(w) %%%%%%%%%%%%%%%%%%%% subplot(212); plot(f,Xw); title('Fourier Transform of input signal, X(w)'); xlabel('Frequency'); ylabel('Amplitude');

[Rick Rosson]

1. Please format your code so that it is more readable.
2. I modified my code a little bit since you tried it, please check the new version and try it again. The modification is fairly superficial - it is mainly to make the code a bit more robust.

[Rick Rosson]

Remember, that Xw is complex-valued. When you call the plot function:

   plot(f,Xw);

MATLAB will plot only the real-part of Xw while completely ignoring the imaginary-part. Please take a look at the MATLAB Command Window and you should see a Warning message to this effect.

What you should do instead is to plot the magnitude of Xw and/or the phase. Please try:

   plot(f,abs(Xw));    % magnitude response

and/or

   plot(f,angle(Xw);   % phase response

I think that may resolve the issue.

Here are the two plots: