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 $$P?$$

  • \$\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


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


though the second notation is more popular. So we have


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



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

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.


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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