Question

G (s) = K / (s (s + 2)), H (s) = 2 is given. Perform the stability analysis of the negative feedback closed loop system on the frequency plane. Calculate the dividend and phase share.

Answer 1

Closed-Loop transfer function:

E(S) = US) - HSY(S)

$Y(s)=G(s)E(s)$

$Y(s)=G(s)(U(s)-H(s)Y(s))$

$(1+G(s)H(s))Y(s)=G(s)U(s)$

$H_{cl}(s)=\frac{Y(s)}{U(s)}=\frac{G(s)}{1+G(s)H(s)}$

$H_{cl}(s)=\frac{G(s)}{1+G(s)H(s)}$

The Open loop gain is:

$G(s)H(s)=\frac{2K}{s(s+2)}$

The system shows a pole in s=-2

Im GH Plane w=0 00-= Re w=0+

By the Nyquist stability criterion:

-1 pole in the left side of the imaginary plane

-Characteristic equation has no poles on the right side of the imaginary plane.

The system is stable

The magnitude of the closed-loop is:

$H_{cl}(s)=\frac{\frac{K}{s(s+2)}}{1+2\frac{K}{s(s+2)}}=\frac{K}{s(s+2)+2K}$

$H_{cl}(s) =\frac{K}{s^2+2s+2K}$

$H_{cl}(j\omega ) =\frac{K}{2j\omega +(2K-\omega^2)}$

$H_{cl}(j\omega ) =\frac{K}{(2\omega)^2 +(2K-\omega^2)^2}((2K-\omega^2)-j2\omega )$

$\left |H_{cl}(j\omega ) \right | =\frac{K}{(2\omega)^2 +(2K-\omega^2)^2}\sqrt{(2K-\omega^2)^2+(2\omega)^2 }$

$\left |H_{cl}(j\omega ) \right | =\frac{K}{\sqrt{(2K-\omega^2)^2+(2\omega)^2 }}$

The phase is:

$\left \langle H_{cl}(j\omega ) \right \rangle =tan^{-1}\left (-\frac{2\omega }{2K-\omega^2} \right )$