# Why does Nyquist plot only need loop gain but not the entire closed loop transfer function?

I was trying to compare a Bode plot and a Nyquist plot of the same system and realized that I was probably comparing apples to oranges

Bode plot is plotting the entire closed loop transfer function, whereas the Nyquist plot is only plotting the loop gain Given the above system, the loop gain is defined as $C(s) \times P(s)$, whereas the closed loop transfer function is defined as $$\frac{1}{1+C(s) \times P(s)}$$

Why does Nyquist only care about the loop gain (but not the entire closed loop tf) and does it make sense to compare Bode with Nyquist?

• Actually, the closed-loop transfer function is $$\frac{C(s)P(s)}{1+C(s)P(s)}$$ – Zulu Apr 13 '15 at 18:48
• the loop gain is $$-C(s)\cdot P(s)$$ because of the subtraction – endolith Mar 1 '16 at 16:25

The Nyquist plot is useful for employing the Nyquist stability criterion. In summary, loop gain encirclements of the point $(-1, 0)$ on the Nyquist plot indicate instability. Unless the system was already unstable (i.e., has RHP poles), in which case a counter-clockwise encirclement must be made for each RHP pole.
• A system is unstable if it has any right-half plane poles. The poles of the fraction $\frac{N(s)}{D(s)}$ occur at the zeros of the denominator $D(s)$, so for the feedback system considered above whose closed-loop TF is $T=\frac{C(s)P(s)}{1+C(s)P(s)}$, the poles occur when $1+C(s)P(s)=0$. If all of these zeros occur on the left-half plane of $s$, the closed-loop TF will be stable; if any occur on the right-half plane, it will be unstable. Thus, the open-loop TF gives you all the information needed to know whether the closed-loop TF will be stable. This is very useful indeed. – Zulu Feb 21 '16 at 23:41