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schematic

simulate this circuit – Schematic created using CircuitLab

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

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

Proof: enter image description here] But

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

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  • \$\begingroup\$ What work have you done so far? \$\endgroup\$ – TimWescott Jul 25 at 4:55
  • \$\begingroup\$ As previous posts - you've done no research. This is page 1 of the chapter on root locus, and easily found on the web. \$\endgroup\$ – Chu Jul 25 at 7:24
  • \$\begingroup\$ @Chu plz see edits \$\endgroup\$ – SUNITA GUPTA Jul 25 at 10:44
  • \$\begingroup\$ Is T(s)=KN/(N+KD) supposed to be the CLTF? If so, it's wrong. \$\endgroup\$ – Chu Jul 25 at 11:54
  • \$\begingroup\$ 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\$. \$\endgroup\$ – sarthak Jul 25 at 14:57
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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.

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  • \$\begingroup\$ 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} \$\endgroup\$ – SUNITA GUPTA Jul 25 at 13:55
  • \$\begingroup\$ 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. \$\endgroup\$ – Chu Jul 25 at 15:42
  • \$\begingroup\$ How can a TF peak to infinity .... try 1/(s+1) with s=-1 \$\endgroup\$ – Chu Jul 25 at 15:44
  • \$\begingroup\$ 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. \$\endgroup\$ – Chu Jul 25 at 15:48
  • \$\begingroup\$ Ok consider loop gain = K(s+1)/(s+2)s \$\endgroup\$ – SUNITA GUPTA Jul 25 at 15:58

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