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# lecture9 - Real The poles of 2 2 1 2 = s s s G are j s ±-=...

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1 AE 383 System Dynamics Lecture Notes 9 (July 22 nd 2006) Poles and the Stability of Linear Systems Remember that the natural response of a dynamic system is given by the complementary solution (the homogeneous solution) of the differential equation. ) ( t f cx x b x a = + + & & & the homogeneous equation is 0 = + + cx x b x a & & & To find the complementary solution we find the routs of the “characteristic equation”. roots the are , 0 2 1 2 s s c bs as = + + then t s t s c e c e c x 2 1 2 1 - - + = Note that the characteristic equation is also given as the denominator of the transfer function. c bs as s f s x s G function Transfer + + = = = 2 1 ) ( ) ( ) ( The roots of denominator polynomial are called the ‘poles’ of the transfer function. The s-plane (complex plane) plot of the poles of a transfer function is very useful for visualizing the character of the natural response of the system. For example the pole of 5 1 ) ( + = s s G is at 5 - = s Imaginary

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Unformatted text preview: Real The poles of 2 2 1 ) ( 2 + + = s s s G are j s ±-= 1 2 , 1 x x 2 Note that each point on the s-plane represents an exponential function e st If any pole of the transfer function lies to the right of the imaginary axis, i.e. in the right half of the s-plane, the real part of that pole will be positive. This means that one exponential factor in the natural response will tend toward infinity with increasing time. In this case the system is said to be dynamically unstable. On the other hand, if all the poles are in the left half-plane , all the exponentials will die out with increasing time. As a conclusion, a necessary and sufficient condition for a linear dynamic system to be dynamically stable is that all its poles lie in the left half of the s-plane....
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