Find the allowable regions in the s-plane for the poles of a transfer function of a
standard second-order system if the system response requirements are tr less than or equal to 0.6
seconds, M p less than or equal to 10%, and ts less than or equal to 3 seconds. Use the definitions given
tr is about 1.8/蠅n
ts=4.6/蟽
Mp=e^(-pi*味/sqrt(1-味))
I have no idea how to answer... please tell me how to start this??
thanks to whoever answers this|||The characteristic eqn of a 2nd order system is:
s虏 + 2味蠅ns + 蠅n虏 = 0
蠅n = 1.8/tr = 1.8/0.6 = 3 rad/s
MPO = e^[-蟺味 /鈭?1 - 味)] = 0.1
蟺味 /鈭?1 - 味) = ln10 = 2.3
蟺虏味虏 = 2.3虏 (1 - 味 ), =%26gt; 味虏 = 0.537(1 - 味), =%26gt; 味虏 + 0.537味 - 0.537 = 0,
味 = [ -0.537 +- 鈭歿0.28858 + 4(0.537)}]/2 = -0.2686 + 0.78 = 0.5118
s虏 + 2(0.5118)3s + 3虏 = 0, s = [-3.07126 +- 鈭歿9.4326 - 4*9}]/2
s = -1.535 +- j2.577, =%26gt; complex pole pair in the left-half plane.
but since, settling time, ts = 4.6/蟽
蟽 = 4.6/ts = 4.6/3 = 1.53
s = -蟽 + j蠅d
蠅d = 蠅n鈭?1 - 味) = 3鈭?1 - 0.5118) = 3鈭?.488 = 2.095
Therefore, the allowable regions of the left-half s-plane
is the rectangle defined by:
s = -1.53 +- j2.095, and s = -1.535 +- j2.577
Note:
rise time: tr
maximum % overshoot: MPO
settling time: ts
damping factor: 味
natural angular frequency: 蠅n
damped frequency: 蠅d
2nd order transfer function poles: s
dissipation parameter: 蟽
I hope this helped.
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