Question

Question 2: Poles and zeros, and how they affect the transient response modes This question tests your understanding of the effects of nearby poles and ze- ros to the response mode of a specified pole (or conjugate pair). You should also understand what a pole-zero-gain plot is and how it is specified by, and specifies a transfer function A unit impulse function is applied to the system described by the transfer function G(8) = K (8+) (s + 1)(8 +2) (a) Draw the pole-zero-gain plot for G(S) (assume for now that 0 << < 1). (b) Use the graphical method studied in lectures to determine the form of the system response as the zero moves from the origin into the left half-plane. (c) For illustrative purposes, consider the system response for some or all of the following values (i) z = 0.25, K = 8, (ii) z= 0.5, K = 4, (iii) z =1, K=2, (iv) z = 1.5, K = 1, (v) z= 2, K = 1, and (vi) z = 3, K = 2/3. (d) If you were free to choose the zero location, what would be the signifi- cance of these results in terms of a control design ? Consider especially the time taken for the system impulse response to decay to an insignif- icant value.

Answer 1

Given that - Gy(s) = K (S+Z) IS+1)(5+2) ②. for above given function Ges) pole = -1, -2 Zero = -Z Cosz (1) » Zero roszę !) 6. Adding a LHP zero to the transfer function make the step response faster (decrease the rise time of the peak time).

O. ① z=0.25 k=8 O- Zero * -pole Gy(s) = 81 5+0.25) (5+1)(5+2) -1 -0.251 = @(S-(-0.25)) (S-(-))(5-(-2)) Z-0.5 Kit. GCS) = 40 (S+O'S) (5+1 (5+2) -2 -1 -0.5 Z =1 K=2 G(S) = 29 (5+1) (5+1) (5+2) * u. Z=1.5 KSI (15): 11S+1.5) (5+1)(5+2) -1.5

0 Z=2 K=1 Ges) = 1.(5+2) (5+1) (5+2) * Z =3 k= 2/3 GCS) = 2/3(5+3) (5+1) (5+2) @ If , I am free to choose the location of Zero's then I will set it at origin. Because any function is very stable when zero located at e z=0