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Hi all, I am having problems using the combination of conv and deconv function. I have a large number of random processes which I convolve with its fliped version, then I deconvolve the result obtained from convolution with the fliped version of random process, the procedure works for all but some specific cases. The code below should help you understand the problem. %%%%%%% Code starts %%%%%%%%%%%%%% t_bit-6; r_bit=1/t_bit; fc1= 2*r_bit; n=4; input_stream1=[0 1 0 1]; i=1; j=1; % Value of m depends on the order of the modulation for m=1:n:(length(input_stream1)-n+1) k=1; % Extracting number of bits to be represented in a symbol bit_stream1=input_stream1(1,m:(m+n-1)); dec1=bin_to_dec(bit_stream1); % Calculating phase angle depending on the decimal equivalent theta1(j)1* ((2*pi)/M); for t=(m-1)*t_bit*(1/n):1/(20*fc1):((m+(n-1))*t_bit*(1/n))-(1/ (20*fc1)) % Formulae to find of the M-PSK signal s1(k)=amp1*cos(2*pi*fc1*t + theta1(j)); y(i)=t; i=i+1; k=k+1; end % Convolution fliped_s1=fliplr(s1); r_s1=conv(s1,fliped_s1); % Deconvolution s_1onv(r_s1,fliped_s1); diff = s1 - s_1 end %%%%%%%%% Code ends%%%%%%%%%%%%%%%%% The diff in this case should be zero but it is not, this problem occurs for the input_stream1= 0 1 0 1 and 1 1 0 1 for all other 4 bit combinations the diff is rightly zero. I am trying to find out as to why am I not getting the diff as 0 for all cases. Please can someone help me find out the problem. Thanking you all in anticipation Brijraj |
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conv and deconv functions
Started by ●June 16, 2004
Reply by ●June 18, 20042004-06-18
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Hi,
The deconv function is
slightly trick. I will give you a hint and you can actually try and see if it
helps.
Deconv is implemented by
long division in Matlab. For this a filter is implemented with the arguments
being fed as the coefficients of the numerator and denominator with the
data being 1. Thus, the deconv is actually an impulse output of a filter
with the arguments as coefficients of numerator and denominator. Now if the
first coefficient is zero, it may not be accepted and maybe the function
will alter the polynomial to overcome the situation resulting in an undesired
output. This might explain for the [0 1 0 1] part and maybe the second input of
[1 1 0 1] may have something to do with consistency of poles and zeros of the
filter. Moreover, deconv is sensitive to noise and gives undesirable outputs in
presence of noise.
An alternate solution could
be by taking fft and ifft. Since convolution will be multiplication in frequency
domain, things maybe a lot easier then.
In case you find a solution,
then do share it on the group.
Cheers,
-Snehamoy Banerjee
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