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2) Suppose this unity feedback system for the mirror segments of a telescope has the loop transfer function, (8 marks) $L(s) = G_c(s)G(s) = \frac{K}{s(s^2 + 2s + 5)}$ a) Find the asymptotes and sketch them in the s-plane. b) Find the angle of departure from the complex poles. c) Determine the gain when two roots lie on the imaginary axis. d) Sketch the root locus.

          2) Suppose this unity feedback system for the mirror segments of a telescope has the loop transfer
function, (8 marks)
$L(s) = G_c(s)G(s) = \frac{K}{s(s^2 + 2s + 5)}$
a) Find the asymptotes and sketch them in the s-plane.
b) Find the angle of departure from the complex poles.
c) Determine the gain when two roots lie on the imaginary axis.
d) Sketch the root locus.
        
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2) Suppose this unity feedback system for the mirror segments of a telescope has the loop transfer
function, (8 marks)
L(s) = Gc(s)G(s) = (K)/(s(s^2 + 2s + 5))
a) Find the asymptotes and sketch them in the s-plane.
b) Find the angle of departure from the complex poles.
c) Determine the gain when two roots lie on the imaginary axis.
d) Sketch the root locus.

Added by Christopher H.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Course Name: Control Systems Please solve with all steps in detail. Fast urgent!!! 2) Suppose this unity feedback system for the mirror segments of a telescope has the loop transfer function: (8 marks) L(s) = K * G(s) * G(s) / (s^2 + 2s + 5) a) Find the asymptotes and sketch them in the s-plane. b) Find the angle of departure from the complex poles. c) Determine the gain when two roots lie on the imaginary axis. d) Sketch the root locus.
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Transcript

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00:01 Hello students in this question we have an open loop transfer function of a unity feedback system so this is given by this equation k times g of s is equal to k times s minus 1 times s minus s minus 2 times s minus 1 divided by s into s plus 1 first we need to find the asymptotes as well as the breakaway and break -in points so we can write down the characteristic equation so this is the equation 1 plus k of g of s is equal to 0 so let's substitute the value of k of g of s in here so this is the substitution and you can just equate this equation and you get a square plus s plus k a square k times s square minus 2s minus s plus 2 is equal to 0 and you simplify this to this form so a square we have 1 plus k so that will be the a coefficient 1 plus k and b is equal to k that is s power of s will be equal to power of s will be equal to so this is the actually b so this is actually b and c we have the third element which is a constant is this value so this is c okay so we got the coefficients now what we can do is that to find the angle conditions for the asymptotes is given by this condition right sigma delta theta is equal to 2k plus 1 times pi by n so this is this you can substitute the values of n pi n here so that is 3 so this will be the the condition for condition for theta value now coming to the b part for the system we have the characteristic equation right there so we can substitute s equal to i omega in here so in so the characteristic equation now will become i omega square or j omega square j omega square j omega plus k times j omega the whole square minus 3 k j omega plus 2 k and once you solve that equation you will get so j omega let's do that so j omega will be equal to what j omega square will…
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