Question

SORU 1: C(s) R(S) + K Gp(s) Kh 100 Yukarıda verilen sistemde plant transfer fonksiyonu: G, (s) = olarak verilmiştir. s +10s +100 C(s) a) Kapalı çevrim transfer fonksiyonu M(s) ifadesini bulunuz. R(S)

Plant transfer function in the system given above: p   2 100 G h s 10s 100    It is given as.
a) Closed loop transfer function       C s M s = R s Find the expression.
b) According to the statement you found in (a), ss c for unit step entry Please describe.
c) you obtained in (b) ss c the sensitivity of K a and Khar individually according to parameters write.
D) K a and Khar parameters have uncertainties of 10% and 5%, respectively. ss c , at rest for less than 1% change K K K  a h According to the testimony K define the range of and this interpret whether the condition will be met.

Please can you help.

Plant transfer function in the system given above:Gp(s) =100/s^2+10s+100
a) Closed loop transfer function M s = Cs/Rs Find the expression.
b) According to the statement you found in (a), ss c for unit step entry Please describe.
c) you obtained in (b) ss c the sensitivity of K a and Khar individually according to parameters write.
D)Ka and Kh
parameters have uncertainties of 10% and 5%, respectively. Css
, at rest
for less than 1% change
K=KaKh
According to the testimony
K
define the range of and this
interpret whether the condition will be met.

Answer 1

loo RIS) C) | Gp (8) given GP (6) Ra 8² + 108tloo SO I kh OLTF = Ka. Gp (8) = a) CLIF = C(s) RS) 100 ka 8² + lost 100 Feedback = kh CLTF= looka OLT F TOLTFX Feedback looka 82 + lost 100 It looka 8² +10st 100 RA 8² + lost 100 Cltkah) Х ANS. Hence the CLTF is looka ANS 8²+ lost 100 (1+kakh) 6) We need Steady State (value; We can use final value theorem as SS.C= lim & C(8) given R(8) = input = step= 1 sto lim 100 ka lim looka Sto S 8? +108 + 100 CI+Rakh) sto g²+108 + 100 (Itka Rh) SS.C= ka This is the steady state value It Rakh of c. c) Sensitivity of c wart parameters Ka and Kh Part & i) wert barometer Ka. S. Jc dc Ka (trakh) 1 - Rakh) ya dka С (Itkakht Ya dka It halth) Ka This is s.s. Songitivity of C wort Ra It Rakh x ka 1 re ANS

partii Sensitivity wit Ih. Sue де akh kh C с dkh kh Ka.(-1) (ka) (1+ Kakhe kh ka (Ithakh) - Kakh It Rakh This is the sensitivity of Cwest Kh ANS. .05 d) incertamity Ka => + 10% = Kh = + 25% = We need to find change or uncertainity in Steady State C change or un certainity in C = Sengitivity of C wit kax Change in Kat Х X.05 2 20 2-k 20 Sensitivity of C wort khx Change in Kh .lt (- Rakh It Kalb It Rakh Kakh (2-kakh) (It Rakh 2o 20 Clt ka kh) ANS. For Steady State c less than 1% and Rakh taken as k. C.s.s. is less than 1% CARCI+R) that is 12-kl 20 (HR) 100 for <2 2-K 10-5k LHK GR> 9R>1.5 1+R 5 thus Range 1.5 to 2 ANS K-2 et 5K - 10 <lHk 4K 411 k <2.75 HK this Range 2 to 2.75 Thus Possible Range of his 1.5 to 2 to 2.75 OR 1.5 to 2.75 ANS, This Range is definately Possible with proper values of Ra and kh. > for K 472 1-2 233
The whole problem is about basics of transfer function and it's dependency on forward path and feedback variables.

Hope you get it easily,

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