The values of w and v are w = 10 + (1/10) = 10.1 v = 50 + (2/10) = 50.2

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The values of w and v are
w = 10 + (1/10) = 10.1
v = 50 + (2/10) = 50.2

The Values Of W And V Are W 10 1 10 10 1 V 50 2 10 50 2 1 (149.25 KiB) Viewed 30 times

Q3. (30% of Assignment 1B mark) The response C(s) of the unity-feedback system shown below to a unit-step input does not meet desired transient specifications and requires a controller. The desired transient specifications for the system are overshoot percentage of approximately 7.5% and settling time of 0.4 seconds (using 2% criterion). plant R(S) 1 ) s(s+w)(s + v) c(8) The values of w and v are given by w = 10 + Lo and v = 50 + where the values of p and q are the two last digits of your student number. For example, if your student number is 12345678, p = 7 and q = 8, therefore w = 10.7 and v = 50.8. (a) Show that the desired dominant closed-loop pole locations should be around s = -10 j12.1285 (6) Use the root-locus angle criterion to prove that the desired dominant closed-loop values do not lie on the root-locus of the uncompensated system (c) Using root-locus methods, design a lead-compensator in the form shown below to achieve the desired transient specifications s + 21 Ge(s) = S + P1