Differences between two closed-loop systems What could I say about stability of those systems by looking at the scheme?

System 1


System 2


$y_r(s)$ is input, $a(s)$, $b(s)$, $p(s)$, $q(s)$ and $r(s)$ are some polynomials and $y$ is output.

I calculated tranfer $\frac{y}{y_r}$ and it's the same for both systems.

$$\frac{y}{y_r} = \frac{r(s)b(s)}{a(s)p(s)+b(s)q(s)}$$

What makes the difference between stability and behavior of those systems? 



Some progress:

I've edited the schemes. There is something, that obviosly differs in those systems. It's $e_{ss}$ - stady state error.

For the first system is:
$$e_{ss1}=y_r(s)\frac{r(s)}{p(s)}-y(s)\frac{q(s)}{p(s)}$$

For the second:
$$e_{ss2}=y_r(s)r(s)-y(s)q(s)$$

Does it tell me something about stability?



I guess this form is more usefull:

$$e_{ss1}=y_r(s)\frac{r(s)a(s)}{a(s)p(s)+b(s)q(s)}$$
$$e_{ss2}=y_r(s)\frac{r(s)a(s)p(s)}{a(s)p(s)+b(s)q(s)}$$

The second system's steady state error has higher order numinator than denominator. Does it mean, the signal's higher frequencies would be infinitely amplified?
There are more than just one way to approach the solution:

1) You can find the steady state error by finding the equation of the denominator of the transfer function.

In system 1: 

$$D(s) = a(s)p(s) + b(s)q(s)$$

The order of the system is determined by this order of $D(s)$. Solving for $D(s)$ will give you the poles of the system. If the poles 'solutions of $D(s)$' all are negative. That means if all the values of the solution to $D(s)$ are in the left half plane $&lt;0$. This means the system is stable.

2) The error in time domain is defined as "input - output".
That is:

$$e(t) = y_r(t) - y(t)$$

Now, to compute the steady state error, that is time at infinity 'after a long time'.
The equation will be: $$ e_{ss} = \lim_{t \to \infty}  y_r(t) - y(t) $$
That is simply taking the limit as t goes to infinity.

Now, using the final value theorem to compute the steady state error in the $s$ domain the equation will be:

$$E_{ss} = \lim_{ s \to 0} E$$

That is in your terms

$$E_{ss} = \lim_{s \to 0} s\cdot e_{ss1} \quad(1)$$

$$E_{ss} = \lim_{s \to 0} s\cdot e_{ss2} \quad(2)$$

Therefore; you can note that the result depends on the 'Type of the system and the Input'.
Type of the system: is how many poles at zero do you have?
That is writing:

$$a(s)p(s) + b(s)q(s) = s^n (s+a_0)(s\cdot a_1) \cdot ... \cdot(s\cdot a_{n-1})$$

the value n represents the order of the system.


Type zero means no poles to zero.
Type I means one pole at zero, etc.


Now if the input is a unit step that is $$y_r(s) = \dfrac{1}{s}$$ 
this will cancel out with the $s$ as specified in equations $(1)$ and $(2)$. The equation will be:

$$ e_{ss1} = \lim_{s \to 0} \dfrac{r(s)a(s)}{a(s)p(s)+b(s)q(s)} $$

Now, looking at the type of system: 


If it is of type one or higher, it is clear now that the $e_{ss1}$ goes to
infinity yielding unstable system.
If type zero, it will have a certain value, yet still stable.


These are the techniques, and the answer highly depends on the type of the system or even the characteristic equation that is the equation of the denominator of the transfer function.