Root locus and the step response

TopicRoot locus and the step response

Title

A gain in a unity-feedback loop

ExampleW08.1

Step1 / 5

Mode

Given

The gain $K$ drives the plant $G(s) = \frac{1}{s(s + 2)}$ under unity negative feedback. For $K = 4$ , find the closed loop's $\omega_n$ , $\zeta$ and poles, its step response's overshoot, peak, rise and settling times, and the open loop's phase margin.

$G(s) = \frac{1}{s(s + 2)}$
unity negative feedback
the gain $K = 4$
settling within $2\ \%$ of the final value; rise from $10\ \%$ to $90\ \%$

The loop at

K = 4

: closed-loop poles, step response and phase margin

Quantity	Value
$\omega_n$ (rad/s)	2
$\zeta$	0.5
$\sigma = -\zeta\omega_n$
$\omega_d$ (rad/s)
overshoot (%)
$T_p$ (s)
$T_r$ (s)
$T_s$ (s)
$\omega_c$ (rad/s)
PM (°)

Close the loop

Under unity feedback

\frac{Y(s)}{R(s)} = \frac{KG(s)}{1 + KG(s)} = \frac{K}{s^2 + 2s + K}

. At

K = 4

this is

s^2 + 2s + 4

, so

\omega_n^2 = 4

and

2\zeta\omega_n = 2

\omega_n = 2

and

\zeta = 0.5

Sim

The loop on its root locus

Drag a closed-loop pole along the locus, or slide $K$ , and watch the step response, the Bode plot and the Nyquist plot follow. The red pen marks the overshoot and the settling time.

Assumed: supplied here, not in the Materials: $K$ tuned from $0.5$ to $20$ in steps of $0.5$

Fig.The loop as the figure draws it: the gain and the plant under unity negative feedback.

j\omega

\sigma

Fig.The root locus: drag a closed-loop pole along it to set the gain. Crosses are the open-loop poles.

K

4.0

$\omega_n$ (rad/s)	2.000
$\zeta$	0.500
$\sigma$	−1.000
$\omega_d$ (rad/s)	1.732
overshoot (%)	16.303
$T_p$ (s)	1.814
$T_r$ (s)	0.819
$T_s$ (s)	4.038
$\omega_c$ (rad/s)	1.572
PM (°)	51.827

At $K$ = 4.0 the response overshoots by 16.303 % and settles in 4.038 s.

Root locus and the step response

Worked examples

A gain in a unity-feedback loop

Close the loop

The loop on its root locus