Exam Solutions 91-99

Exam Solutions 91-99 - 0.1 0.2 0.3 0.4 0.5 0.6 0.5 1 1.5 0.1 0.2 0.3 0.4 0.5 0.6 0.02 0.04 0.06 0.08 step input amplitude 1 Time(sec Amplitude Ramp

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Unformatted text preview: 0.1 0.2 0.3 0.4 0.5 0.6 0.5 1 1.5 0.1 0.2 0.3 0.4 0.5 0.6 0.02 0.04 0.06 0.08 step input amplitude 1 Time (sec) Amplitude Ramp input amplitude 0.1 Time (sec) Amplitude Control Systems (SC3020et): Instruction stencil no. 2 due October 16 th , 2003 Name: Student Number: 0.1 Partial-fraction expansion Give the partial-fraction expansion of the following transfer function G ( s ) = ( s + 3) ( s + 1) 2 · ( s + 2) (1) 0.2 Sketching of the Nyquist plot The following system is given: G ( s ) = s + 1 ( s- 1) 2 (2) Sketch the Nyquist plot of G ( s ) and use the plot to determine for which values of K the closed- loop system is stable. 0.3 Root Locus Sketch the root locus for the characteristic equation of the system represented by L ( s ) = s + 1 s ( s + 1)( s + 2) (3) Compute all the characteristics that are involved, i.e. centroids, crossing imaginary axis, angle of departure and angle of arrival, location of breakaway points. Determine also the value of the root locus gain for which the complex conjugate poles have a damping ratio of 0.5. 0.4 Design Lead Compensator A numerically controlled machine has a transfer function given by G ( s ) = 1 s ( s + 1) (4) Performance specifications of the system in the unity feedback configuration are satisfied if the closed loop poles are located at s =- 1 ± j √ 3 . 1. Show that this specification cannot be achieved by choosing proportional control alone. 2. Design a lead compensator D ( s ) = K s + z s + p that will meet the specification Control Systems (SC3020et) ( http://www.dcsc.tudelft.nl/sc3020et/index.html ) Instruction stencil no. 2: answers 1 Problem 1: Partial-fraction expansion (see textbook p. 98) The transfer function G ( s ) = ( s +3) ( s +1) 2 · ( s +2) can be rewritten to G ( s ) = C 1 s +2 + C 2 s +1 + C 3 ( s +1) 2 The constants C 1 , C 2 , and C 3 are computed as follows C 1 = ( s + 2) · G ( s ) | s = − 2 = s +3 ( s +1) 2 | s = − 2 = 1 ( − 1) 2 = 1 C 2 = d ds ( s + 1) 2 · G ( s ) | s = − 1 = d ds s +3 s +2 | s = − 1 = − 1 ( s +2) 2 | s = − 1 = − 1 1 2 = 1 C 3 = ( s + 1) 2 · G ( s ) | s = − 1 = s +3 s +2 | s = − 1 = 2 1 = 2 Substitution of C 1 , C 2 , and C 3 results in G ( s ) = 1 ( s +2)- 1 ( s +1) + 2 ( s +1) 2 Problem 2: Sketching of the Nyquist plot The problem is to sketch the Nyquist plot of the transfer function G ( s ) = s + 1 ( s- 1) 2 The location of the poles and zeros are illustrated in the figure below. The Nyquist contour encircles the entire right half plane (see fig. 6.16 of the textbook). Every point on the Nyquist contour has a gain and argument when substituted in the transfer function G ( s ) and therefore represents a point in the complex plane....
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This note was uploaded on 08/13/2010 for the course EE EE 302 taught by Professor Taufik during the Spring '10 term at Cal Poly.

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Exam Solutions 91-99 - 0.1 0.2 0.3 0.4 0.5 0.6 0.5 1 1.5 0.1 0.2 0.3 0.4 0.5 0.6 0.02 0.04 0.06 0.08 step input amplitude 1 Time(sec Amplitude Ramp

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