A general block diagram of a feedback controlled process with a disturbance in the output is:

enter image description here

If we assume that the following blocks are equal to 1, and therefore we remove them:

$$ K_m = G_v = G_m = 1$$

We have the following expression in frequency domain:

$$Y = G_dD + G_pG_c(Y_{sp} - Y)$$

If we rearrange this we reach:

$$ Y = \frac{G_pG_c}{1+G_pG_c}Y_{sp} + \frac{G_d}{1+G_pG_c} D$$

All books insist that to check stability, we have to check the roots of the characteristic equations alone, i.e., the roots of $1 + G_pG_c$, since this appears in both denominators. However, shouldn't we also check the actual numerators? For example, if the disturbance is unstable, and:

$$G_d = \frac{1}{s-1}$$

Then the system is not stable at all. It does not matter what the roots of $1 + G_pG_c$ are, since we are adding a positive root to the denominator. It would be: $$Y = \frac{G_pG_c}{1+G_pG_c}Y_{sp} + \frac{1}{(s-1)(1+G_pG_c)} D$$ The literature i've seen always refers only to the characteristic equation and not the numerator... is there a reason for this? Shouldn't we always substitute the transfer functions first?

  • $\begingroup$ As far as $G_D$, it's a matter of where you draw the boundary for analyzing stability. If you ask is the system composed of {$G_D$ and the loop} stable? Then it's as you say. If you're just asking about the loop, $G_D$ isn't part of the loop. Another way to see loop stability, is that instability happens when going around the circle produces a 360deg phase at unity gain. We start with already 180 deg phase from the minus sign in the operator on the left, and so the product of everything else around the loop must be another 180 deg, which is where {everything else} = -1 comes from $\endgroup$
    – Pete W
    Commented Mar 9 at 20:33
  • $\begingroup$ or rather - instability begins at that condition ... $\endgroup$
    – Pete W
    Commented Mar 9 at 20:39
  • $\begingroup$ Now lets say ${G_C}{G_P}$ had an unstable pole, while $1/(1+{G_C}{G_P})$ had stable poles? If so, the expression for closed loop response vs set-point, ${G_C}{G_P}/(1+{G_C}{G_P})$ would, as a whole, also be stable. Algebraically, you could show it by multiplying both the numerator and denominator by the poles of ${G_C}{G_P}$ to show that. $\endgroup$
    – Pete W
    Commented Mar 10 at 2:54
  • $\begingroup$ It can be noted that in this control architecture one doesn't have any way of influencing the poles $G_d$. It can also be noted that an unstable $G_d$ can also be modeled using a stable $G_d$ but having the signal $D$ grow exponentially, just as long $Gd\,D$ yields the same signal. $\endgroup$
    – fibonatic
    Commented Mar 10 at 11:50

1 Answer 1


For example, if the disturbance is unstable... Gd = 1/(s-1)

In that case, the (s-1)^(-1) would be part of the denominator after simplification; and hence part of the characteristic equation.

The literature i've seen always refers only to the characteristic equation and not the numerator.

No, $G_d$ is a ratio of polynomials and itself contains a numerator and a denominator. The denominator needs to be shifted to the main denominator.


$$ \frac{(s+1)/\color{red}{(s+4)}}{(s+2)/(s+7)} = \frac{(s+1)(s+7)}{\color{red}{(s+4)}(s+2)} $$

All books insist that to check stability, ...

If they are well written books, they will most likely contain some fine print at the beginning of the chapter or some kind of foot note that says something like: "The individual blocks in the block diagram are assumed to be stable systems". Since you have not given a specific (reputable) reference text book, it not possible to know what the author's stated assumptions were.


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