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In a typical closed loop system, with G(s) as the forward function, and H(s) as the feedback function, why is the transfer function used to calculate the bode plot G(s)H(s) instead of G(s) / 1+G(s)H(s)?

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  • \$\begingroup\$ Short answer is you can use the CLTF. But, if the objective is to design a controller, it's often convenient to base the design on the phase margin and/or gain margin which are readily derived from the OLTF. Furthermore, if the closed loop happens to be unstable then the frequency response (Bode plot) can't be measured practically, whereas the open loop may well be stable and hence measurable. \$\endgroup\$ – Chu Aug 8 '18 at 7:16
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Let me try to give you a short insight of why that is done. \$G(s)H(s)\$ is called the loop gain. You don't need to find the whole system transfer function (i.e closed loop) to have a 'feel' for what the stability looks like.

Now, with the loop gain you can analyze the relative stability of the system. Think about it, when the magintude of \$G(s)H(s)\$ is -1, the denominator in the closed loop transfer function becomes zero and the transfer function gives you and unbounded result.

So by plotting the bode plot of \$G(s)H(s)\$, you can see how close to instability the system is (how close you are to 1 or 0dB when the phase is -180 degrees or how close to -180 when the gain is 0 dB; this means -1 in the denominator). Those two tests done with the loop gain are the phase and gain margin tests.

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