secord.doc - Second-order Systems Definition the transfer function contains terms in s2(and often but not necessarily s as well but none in s3 or above

secord.doc - Second-order Systems Definition the transfer...

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Second-order Systems Definition -- the transfer function contains terms in s 2 (and often, but not necessarily, s as well!) but none in s 3 or above. What would it be in terms of the differential equation? To see what the problems might be, let us determine how systems having the following transfer functions will respond to a unit step input. 1. 10/(s 2 + 5s + 4) 2. 10/(s 2 + 2s + 4) 3. 10/(s 2 + 4s + 4) System 1 The transform of the unit step is 1/s, so the transform of the output will be: 10 ---------------- s(s 2 + 5s + 4) We have, unfortunately, no direct conversion for this one in the tables. We will need to use Partial Fractions. Fortunately, we note that (s 2 + 5s + 4) will factorise as (s + 1)(s + 4). So we have: A B C ----- + ------ + -- s + 1 s + 4 s We put it all over its common denominator as usual: As(s + 4) + Bs(s + 1) + C(s 2 + 5s + 4) ----------------------------------------------------------------------- s(s + 1)(s + 4) The numerator of the fraction turns out to be: s 2 (A + B + C) + s(4A + B + 5C) + (4C) and it all has to equal the 10 (+ 0s + 0s 2 ) we started out with. So 4C = 10, and C = 2.5. The s2 terms then give us A + B + 2.5 = 0, whilst the s terms give us 4A + B + 12.5 = 0. If we subtract the first equation from the second, we obtain: 3A + 10 = 0, so A = -10/3 = -3.333. Putting this back into A + B + 2.5 = 0, we obtain: -3.333 + B + 2.5 = 0, giving B = 0.833. The Laplace Transform of the output is therefore: 2.5/s - 3.333/(s + 1) + 0.833/(s + 4) and the time response becomes: 2.5 - 3.333e -t + 0.833e -4t .
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We will sketch the graph in the lecture but it is particularly noteworthy that the response does not overshoot the steady-state value. System 2 On the face of it, this one looks like a repeat performance - but there are actually important differences! The output transform is 10/[s(s 2 + 2s + 4)]. Unfortunately, s 2 + 2s + 4 does not factorise ... we will have to adopt another approach. The tables do give us the following guidance: w/[(s + a) 2 + w 2 ] = e -at sin(wt) (s + a)/[(s + a) 2 + w 2 ] = e -at cos(wt) so we will have to resort to partial fractions again.
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