Is the number of poles of an open loop transfer function the same after closing the loop? Why? My textbook says that it is, but if OLTF is
$$G(s)$$
and CLTF is
$$\frac{G(s)} {1+G(s)H(s)}$$
I don't see why the number of poles would have to be the same.
 A: This statement is only true for the case that the sensor transfer function, $H(s) = 1$, which can in many cases be assumed.
Then, given the open loop transfer function $G(s)$, the closed loop transfer function $G_c(s)$ is
$$
G_c(s) = \frac{G(s)}{1 + G(s) H(s)} = \frac{G(s)}{1 + G(s)}
$$
Now write $G(s) = N(s)/D(s)$, where $N(s)$ is the numerator polynomial and $D(s)$ the denominator polynomial. Then, write $G_c(s)$ as
$$
G_c(s) = \frac{\frac{N(s)}{D(s)}}{1 + \frac{N(s)}{D(s)}} = \frac{N(s)}{D(s) + N(s)}
$$
So in the denominator of $G_c(s)$, you are effectively adding two polynomials, namely $D(s)$ and $N(s)$. Let $n_D$ denote the degree of $D(s)$ and $n_N$ the degree of $N(s)$.
For $G(s)$ to be causal (and therefore to exist in "real-life"), $n_N \leq n_D$ is necessary. So if you add $N(s)$ to $D(s)$, its degree cannot increase.
Example: Lets take an example with $n_N = 1$ and $n_D = 2$:
$$
\begin{align}
N(s) &= s + 1 \\
D(s) &= s^2 + 5 s + 1
\end{align}
$$
So you get
$$
\begin{align}
D(s) + N(s) &= (s^2 + 5 s + 1) + (s + 1) \\
&= s^2 + (5 + 1)s + (1 + 1) \\
&= s^2 + 6 s + 2
\end{align}
$$
So $D(s) + N(s)$ is still a second order polynomial, just as $D(s)$, so they have the same number of roots and therefore, $G_c(s)$ has the same number of poles as $G(s)$.
