[SOLVED] Problem in Demodulaton

clc; clear all; close all;

% ----------------------------------------------------------------------
% ----------------------------- Modulator ----------------------------
databit = randint(20,1);

fc = 25000;
fs = fc * 4;
t = 0:1/fs:0.0015;
carrier = sin(2*pi*fc*t + pi);
% ----------------------------------------------------------------------
squ_msg  = [];
carrier_m = [];

for i = 1 : length(databit)
  if databit(i) == 0
    squ = square(2*pi*(1000/1.5)*t, 66.7);
  else
    squ = square(2*pi*(1000/1.5)*t, 33.3);
  end
  squ_msg   = [squ_msg squ];            
  carrier_m = [carrier_m carrier];      
end

% To make the square wave amplitudes range from 0 to 1
for i = 1:length(squ_msg)
  if squ_msg(i)~=1
    squ_msg(i)=0;
  end
end
% ----------------------------------------------------------------------

% up conversion
output = carrier_m .* squ_msg;

% ----------------------------------------------------------------------
% ----------------------------- Demodulator -------------------------
% down conversion
x = output .* carrier_m;

% Pass through a LPF
b = low_pass();
y = filter(b, x);

figure;
plot(y);
grid;

% Recover the bits 0 and 1 from y

function Hd = low_pass
%LOW_PASS Returns a discrete-time filter object.

%
% MATLAB Code
% Generated by MATLAB(R) 7.13 and the Signal Processing Toolbox 6.16.
%
% Generated on: 08-Mar-2013 18:41:03
%

% Equiripple Lowpass filter designed using the FIRPM function.

% All frequency values are in Hz.
Fs = 100000;  % Sampling Frequency

Fpass = 2000;            % Passband Frequency
Fstop = 6000;            % Stopband Frequency
Dpass = 0.057501127785;  % Passband Ripple
Dstop = 0.0001;          % Stopband Attenuation
dens  = 20;              % Density Factor

% Calculate the order from the parameters using FIRPMORD.
[N, Fo, Ao, W] = firpmord([Fpass, Fstop]/(Fs/2), [1 0], [Dpass, Dstop]);

% Calculate the coefficients using the FIRPM function.
b  = firpm(N, Fo, Ao, W, {dens});
Hd = dfilt.dffir(b);

% [EOF]