Polyphase Decomposition in Matlab
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