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I would like to determine the following transfer functions: $$T_1=\frac{y^\wedge}{d^\wedge}$$ $$T_2=\frac{y^\wedge}{r^\wedge}$$ $$T_3=\frac{e^\wedge}{d^\wedge}$$ $$T_4=\frac{e^\wedge}{r^\wedge}$$ In the following system: $$\begin{bmatrix}d^\wedge\\r^\wedge\end{bmatrix} =\begin{bmatrix}1&-C\\ P&1\\\end{bmatrix} \begin{bmatrix}u^\wedge\\e^\wedge\end{bmatrix} $$whence $$\begin{bmatrix}u^\wedge\\e^\wedge\end{bmatrix} =\begin{bmatrix}1&-C\\ P&1\\\end{bmatrix}^{-1} \begin{bmatrix}d^\wedge\\r^\wedge\end{bmatrix}=\begin{bmatrix}(1+CP)^{-1}&C(1+PC)^{-1}\\ -P(1+CP)^{-1}&(1+PC)^{-1}\\\end{bmatrix} \begin{bmatrix}d^\wedge\\r^\wedge\end{bmatrix} $$

I therefore wrote: $$\frac{u^\wedge}{d^\wedge}=(1+CP)^{-1}$$ $$\frac{u^\wedge}{r^\wedge}=C(1+CP)^{-1}$$ $$\frac{e^\wedge}{d^\wedge}=-P(1+CP)^{-1}$$ $$\frac{e^\wedge}{r^\wedge}=(1+PC)^{-1}$$

Now, may I use the relation $$y^\wedge=u^\wedge P$$ hence multiply both $$\frac{u^\wedge}{d^\wedge}=(1+CP)^{-1} and \frac{u^\wedge}{r^\wedge}=C(1+CP)^{-1}$$ by $$P?$$

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  • \$\begingroup\$ It is a bit hard to answer your question "may I use \$\hat{y} = \hat{u}P\$?" because you have not stated what \$\hat{y}\$ is. \$\endgroup\$
    – SomeEE
    Commented Feb 25, 2014 at 15:35
  • \$\begingroup\$ @MathEE It is very difficult and cumbersome to ask questions here, as I do not have sufficient credit for adding attachments, which would have rendered this far simpler. y is simply the output, u is the input and P is the plant let's say. \$\endgroup\$
    – peripatein
    Commented Feb 25, 2014 at 17:03
  • \$\begingroup\$ \hat{y} ⇒ \$\hat{y}\$ \$\endgroup\$
    – jippie
    Commented Feb 25, 2014 at 20:14

2 Answers 2

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Your initial 4 transfer functions look correct (assuming the stated inverses exist). Then, if you are using scalar transfer functions,

$$\hat{y}(t)=\hat{u}(t)P=P\hat{u}(t),$$

though the second notation is more popular. So we have

$$\frac{\hat{y}}{\hat{u}}=P.$$

Then you can derive T1 as

$$T_1 = \frac{\hat{y}}{\hat{d}}=\frac{\hat{y}}{\hat{u}}\frac{\hat{u}}{\hat{d}}=P(1+CP)^{-1}$$

and do the same for T2:

$$T_2 = \frac{\hat{y}}{\hat{r}}=\frac{\hat{y}}{\hat{u}}\frac{\hat{u}}{\hat{r}}=PC(1+PC)^{-1}.$$

So, the answer to your question is yes, you can. The answer in the non-scalar case is actually also yes, but at the condition that you pay attention to inverses (they must exist) and to the notation you use. As for instance

$$\hat{y}(t)=P\hat{u}(t)\neq\hat{u}(t)P,$$

and

$$C(1+PC)^{-1}\neq C(1+CP)^{-1}.$$

Note that you wrote both in your question. If you were treating a non-scalar problem, you should also use I to define the identity matrix.

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There isn't anything that stops you from making assumptions. But however, it should be a meaningful. Say it shouldn't be like multiplying or dividing it by 0. Also you should notice that $$ u = (1+CP)^{-1}d + C(1+CP)^{-1}r $$ so when you say that \$\frac{u}{d} = (1+CP)^{-1}\$, it just means that you are making \$r = 0\$. Similarly with \$\frac{u}{r}\$ as well. I hope this helps.

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