Hi ACICY,
I think you just wanted to confirm if the transfer function placed is apt or not.
You made a good try, but seems that the direct path in the numerator is missed.
After looking at it, i feel the transfer function is 1.
Here is how it is,
Consider the adder in the top path as a node from left. It takes in two paths, one directly from X(z) and other from the path of delay, and multiplier. Implies the equation at that node is
X(z) + (z^-1)*0.25*X(z) % Let this be S(z)
Now lets see at the adder at the second half of the circuit, one path is the resultant sum of first path, the other path is the Y(z) , delay, multipler. Impies the equation at that node is
% Inward flow to the adder
S(z) + (z^-1)*Y(z)*(-0.25) => S(z) - 0.25*(z^-1)*Y(z)
% The outward flow of the adder is Y(z)
% Equating the both
S(z) - 0.25*(z^-1)*Y(z) = Y(z)
=> S(z) = Y(z)*(1+(0.25*(z^-1)))
=> X(z) + (z^-1)*0.25*X(z) = Y(z)*(1+(0.25*(z^-1)))
=> X(z)*(1+(0.25*(z^-1))) = Y(z)*(1+(0.25*(z^-1)))
=> X(z) = Y(z)
=> Y(z)/X(z) = 1
=> H(z) = 1
Hope this helps.
Regards,
Sriram
