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Peter K. ♦
Peter K. ♦
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We have the closed loop transfer function:

T(s)=L(s)/(1+L(s))$$T(s)=\frac{L(s)}{1+L(s)}$$

So as far as I understand we check along the jω$j\omega$-axis on the Bode plot whether L(s)=-1$L(s)=-1$, cause that's when T(s)$T(s)$ has pole on the axis and that leads to instability. 

My question is, why don't we need to check the entire right half plane whether a gain/phase modification leads to L(s)=-1$L(s)=-1$ there? It doesn't make sense from my standpoint, cause we don't want any poles on the right half plane.

We have the closed loop transfer function:

T(s)=L(s)/(1+L(s))

So as far as I understand we check along the jω-axis on the Bode plot whether L(s)=-1, cause that's when T(s) has pole on the axis and that leads to instability. My question is, why don't we need to check the entire right half plane whether a gain/phase modification leads to L(s)=-1 there? It doesn't make sense from my standpoint, cause we don't want any poles on the right half plane.

We have the closed loop transfer function:

$$T(s)=\frac{L(s)}{1+L(s)}$$

So as far as I understand we check along the $j\omega$-axis on the Bode plot whether $L(s)=-1$, cause that's when $T(s)$ has pole on the axis and that leads to instability. 

My question is, why don't we need to check the entire right half plane whether a gain/phase modification leads to $L(s)=-1$ there? It doesn't make sense from my standpoint, cause we don't want any poles on the right half plane.

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Kakukk777
Kakukk777
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How does the phase/gain margin method work?

We have the closed loop transfer function:

T(s)=L(s)/(1+L(s))

So as far as I understand we check along the jω-axis on the Bode plot whether L(s)=-1, cause that's when T(s) has pole on the axis and that leads to instability. My question is, why don't we need to check the entire right half plane whether a gain/phase modification leads to L(s)=-1 there? It doesn't make sense from my standpoint, cause we don't want any poles on the right half plane.