Suppose i have the system equation

$Y(s) = G(s)X(s)+ 3T(s)$

Then what is the transfer function of the system?

I know that the transfer function is $Y(s)/X(s)$, but i can't get that expression.

  • $\begingroup$ What is $T(s)$? In general such a system doesn't have a transfer function because it's not linear (i.e., its output does not only depend on the input $X(s)$). $\endgroup$
    – Matt L.
    Aug 27, 2018 at 7:21
  • $\begingroup$ T(s ) is another input signal. It could be a disturbance signal $\endgroup$
    – hola
    Aug 27, 2018 at 7:45
  • $\begingroup$ OK, but then you have to define what you mean by transfer function. Do you consider this system a multi-input single-output (MISO) system? $\endgroup$
    – Matt L.
    Aug 27, 2018 at 7:52
  • $\begingroup$ In my control theory book they just take G(s) as THE "transfer function" of this system, i thought it was odd too $\endgroup$
    – hola
    Aug 27, 2018 at 7:53
  • $\begingroup$ The chapter i am on now introduces open&closed loop systems and their transfer function. But they also include disturbance and sensor noise signals in system. $\endgroup$
    – hola
    Aug 27, 2018 at 7:58

1 Answer 1


If you consider $X(s)$ as the only input of the system then the transfer function only exists for $T(s)=0$, and it is given by $G(s)$. Strictly speaking, with $T(s)\neq 0$, the system cannot be characterized by a transfer function because it is no longer linear. Note that a linear system must produce an output $Y(s)=0$ for $X(s)=0$. However, this is not the case for $T(s)\neq 0$.

You could also treat that system as a (linear) multiple input single output (MISO) system with two inputs $X(s)$ and $T(s)$. In that case there are two transfer functions: $$H_X(s)=\frac{Y(s)}{X(s)}\Big|_{T(s)=0}=G(s)$$

and $$H_T(s)=\frac{Y(s)}{T(s)}\Big|_{X(s)=0}=3$$


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