How can I prove that for any control system root locus plot starts from poles of OLTF and ends at zeroes of OLTF? simulate this circuit – Schematic created using CircuitLab

G(s)H(s)=N/D

T(s)=KN/(D+KN)

For K=0 T=0

For K=infinity T=infinity

then how can K=0 {condition where forward path gain becomes zero} be considered as condition where the poles of T(s) will be equal to poles of G(s)H(s) when the transfer function has become equal to zero how can any value of s make the transfer function peak to infinity { condition of poles } and how can characteristic equation be defined for case when T(s)=0

• What work have you done so far? – TimWescott Jul 25 at 4:55
• As previous posts - you've done no research. This is page 1 of the chapter on root locus, and easily found on the web. – Chu Jul 25 at 7:24
• @Chu plz see edits – SUNITA GUPTA Jul 25 at 10:44
• Is T(s)=KN/(N+KD) supposed to be the CLTF? If so, it's wrong. – Chu Jul 25 at 11:54
• You are going into too many unnecessary details. Root locus is to check the stability of a closed loop system. If $k=0$ there is no closed loop (no forward path) and no question of stability. So, yes we would care about the root locus only in the limit as $k \rightarrow 0$ and not when it is $0$. – sarthak Jul 25 at 14:57

The CLTF is: $$\ T(s)=\frac{KN(s)}{D(s)+KN(s)}\$$, and the characteristic equation (CE) is:

$$\\small D(s)+KN(s)=0\$$

As $$\\small K\$$ goes from zero to infinity, the roots of the CE trace out the root locus, which is the path of the closed loop poles in the complex s-plane.

For the start and end points of the locus, we let $$\\small K=0\$$ and $$\\small K\rightarrow\infty\$$, respectively, as follows:

When $$\\small K=0\$$, the CE is: $$\\small D(s)=0\$$, and the roots are thus the poles of the open loop TF.

When $$\\small K\rightarrow \infty\$$, $$\\small KD(s)>>N(s)\$$, and the CE is: $$\\small KN(s)=0\$$; hence $$\\small N(s)=0\$$, and the roots are the zeroes of the open loop TF.

• You mean K never becomes equal to zero but tend to zero because other wise "how can K=0 {condition where forward path gain becomes zero} be considered as condition where the poles of T(s) will be equal to poles of G(s)H(s) when the transfer function has become equal to zero how can any value of s make the transfer function peak to infinity { condition of poles } and how can characteristic equation be defined for case when T(s)=0" {mentioned in question} – SUNITA GUPTA Jul 25 at 13:55
• K=0 is a mathematical convenience. The root locus is defined on the characteristic equation, not on the CLTF. We let K tend to infinity because infinity is not a number. – Chu Jul 25 at 15:42
• How can a TF peak to infinity .... try 1/(s+1) with s=-1 – Chu Jul 25 at 15:44
• Why can't characteristic equation be defined when K=0? It's defined as N(s)+KD(s)=0, so we have N(s)=0 when K=0 --- what's the problem with that? e.g. N(s)=(s+a)=0 gives s=-a. Think before you type. – Chu Jul 25 at 15:48
• Ok consider loop gain = K(s+1)/(s+2)s – SUNITA GUPTA Jul 25 at 15:58