Polyphase Decomposition in Matlab I created the lowpass filter 

h = firls(100, [0 .2 .25 1], [ 1 1 0 0],[.001 60]);


My goal is to get four filters  $H_{k}(z) = H(W_{4}^{k}z)  k =0,1,2,3$

From my knowledge to get the polyphase components you would do the following

$H_{0}(z) = \sum h(4n)z^{-n}$
$H_{1}(z) = \sum h(4n+1)z^{-n}$
$H_{2}(z) = \sum h(4n+2)z^{-n}$
$H_{3}(z) = \sum h(4n+3)z^{-n}$  

To do this in matlab i just did\

 h0=h(1:4:end);
 h1=h(2:4:end);
 h2=h(3:4:end);
 h3=h(4:4:end);


This doesn't seem to produce the right results.... If anyone can clarify how to do polyphase decomposition in mat lab or tell me what I am doing wrong I would appreciate it

EDIT:

I have a signal x[n] that is a FDM (frequency Division Multiplex) signal made up of four real valued baseband signals interpolated by a value of of 4 and modulated using the complex carrier $e^{j \frac{2*pi*k}{4}n}$  thefour base band signals are added to make the FDM signal x[n].

My goal is to estimate each of the 4 base band signals using the polyphase components of the lowpass filter.  The matlab code I am using is:

h = firls(100, [0 ,2 ,.25 1], [ 1 1 0 0],[.001 60]);
h0=h(1:4:end);
h1=h(2:4:end);
h2=h(3:4:end);
h3=h(4:4:end);

%first baseband signal
y0 = conv(x,h0);  
y0 = downsample(y0, 4);  
y0 = abs(fft(y0));  


Here are the 4 estimated baseband signals I get compared to the actual


The polyphase components don't seem to be changing anything

I also tried setting up the filters like this

% Lowpass Filter
h0 = firls(100, [0 .2 .25 1], [1  1 0  0],[.001 60]);
h1 = firls(100, [0 .2 .25 1], [1  1 0  0],[.001 60]);
h2 = firls(100, [0 .2 .25 1], [1  1 0  0],[.001 60]);
h3 = firls(100, [0 .2 .25 1], [1  1 0  0],[.001 60]);
figure(1)
impz(h);
figure(2);
freqz(h);

% PolyPhase Componets
h0(2:4:end)=0;
h0(3:4:end)=0;
h0(4:4:end)=0;

h1(1:4:end)=0;
h1(3:4:end)=0;
h1(4:4:end)=0;

h2(1:4:end)=0;
h2(2:4:end)=0;
h2(4:4:end)=0;

h3(1:4:end)=0;
h3(2:4:end)=0;
h3(3:4:end)=0;


and ended up with these plots



still seems to be something wrong

(NOTE: Orange is actual signals.  Blue is my estimate.)
It seems I was missing the concept all together.

The phrase $ H_{k}(z)=H(W_{4}^kz) k= 0,1,2,3$ given the lowpass filter H(z) represents the following:

$$H(z) = h_{0} + h_{1}+h_{2}+h_{3}$$
$$H_{0}(z) = h_{0} + h_{1}e^{\frac{-j2\pi*0*1}{4}}+h_{2}e^{\frac{-j2\pi*0*2}{4}}+h_{3}e^{\frac{-j2\pi*0*3}{4}}$$
$$H_{1}(z) = h_{0} + h_{1}e^{\frac{-j2\pi*1*1}{4}}+h_{2}e^{\frac{-j2\pi*1*2}{4}}+h_{3}e^{\frac{-j2\pi*1*3}{4}}$$
$$H_{2}(z) = h_{0} + h_{1}e^{\frac{-j2\pi*2*1}{4}}+h_{2}e^{\frac{-j2\pi*2*2}{4}}+h_{3}e^{\frac{-j2\pi*2*3}{4}}$$
$$H_{3}(z) = h_{0} + h_{1}e^{\frac{-j2\pi*3*1}{4}}+h_{2}e^{\frac{-j2\pi*3*2}{4}}+h_{3}e^{\frac{-j2\pi*3*3}{4}}$$

$H_{0},H_{1},H_{2},H_{3}$ represent the polyphase components of filter $H(z)$

To get these componets in mat lab i used 

% Lowpass Filter
N = 100;
h = firls(N, [0 .2 .25 1], [1  1 0  0],[.001 .0001]);

% PolyPhase Componets
for i=1:length(h),
h0(i)=h(i);
end

for i=1:length(h),
h1(i)=h(i)*exp((-j*2*pi*((i-1)))/4);
end

for i=1:length(h),
h2(i)=h(i)*exp((-j*2*pi*(2*(i-1)))/4);
end

for i=1:length(h),
h3(i)=h(i)*exp((-j*2*pi*(3*(i-1)))/4);
end


using these components I was able to get the right estimations