Closed Loop Transfer Function Query

Question

I have a block diagram as given in the image below.

The essence of my problem is that my solution differs from "the solution". I am really confused and doubting myself. I presume mine is correct but am happy to be corrected.

This is what I get:

$$ G_{c}(z) = \dfrac{Kz(z+1)}{z(z-0.25)(z+0.5)+K(z+1)} $$

This is "the solution":

$$ G_{c}(z) = \dfrac{Kz(z+1)}{(z-0.25)(z+0.5)+K(z+1)} $$

Which one is right? Any help would be greatly appreciated. Note: the solution, in my opinion, has a missing $z$ in the denominator. P.S. the question is to find the corresponding closed-loop transfer function.

Answer 1

The solution that you are given is correct. To see why this you first have to identify what the closed loop will look like. If the open loop is just given by $G(z)$ and the negative feedback loop has a gain $H(z)$, as shown in the figure below, then the closed loop can be expressed as,

$$ \frac{Y(z)}{U(z)} = \frac{G(z)}{1 + G(z)\, H(z)}. \tag{1} $$

In your case $G(z)$ and $H(z)$ can be expressed as,

$$ G(z) = \frac{K\, \left(1 + z^{-1}\right)}{\left(1 - \frac14\,z^{-1}\right) \left(1 + \frac12\,z^{-1}\right)}, \tag{2} $$

$$ H(z) = z^{-1}. \tag{3} $$

Substituting $(2)$ and $(3)$ into $(1)$ yields,

$$ \frac{Y(z)}{U(z)} = \frac{\frac{K\, \left(1 + z^{-1}\right)}{\left(1 - \frac14\,z^{-1}\right) \left(1 + \frac12\,z^{-1}\right)}}{1 + \frac{K\, \left(1 + z^{-1}\right)}{\left(1 - \frac14\,z^{-1}\right) \left(1 + \frac12\,z^{-1}\right)}\, z^{-1}} = \frac{K\, \left(1 + z^{-1}\right)}{\left(1 - \frac14\,z^{-1}\right) \left(1 + \frac12\,z^{-1}\right) + K\, \left(1 + z^{-1}\right)\, z^{-1}}. \tag{4} $$

However the final expression is given in $z$ and not in $z^{-1}$. Equation $(4)$ can be written into this form by multiplying both the numerator and denominator by $z$ to the highest power of $z^{-1}$ of the current numerator or denominator of $(4)$, which in this case is two. Doing this yields,

$$ \frac{Y(z)}{U(z)} = \frac{K \left(1 + z^{-1}\right)}{\left(1 - \frac14 z^{-1}\right) \left(1 + \frac12 z^{-1}\right) + K \left(1 + z^{-1}\right) z^{-1}} \frac{z^2}{z^2} = \frac{K \left(z + 1\right) z}{\left(z - \frac14\right) \left(z + \frac12\right) + K \left(z + 1\right)}. \tag{5} $$

Without knowing the steps that you have taken it is hard to say where you went wrong, but I suspect that you might have made some mistake when considering $H(z)$.