Question

2. An effective control design concerns itself with more than the transfer function between the reference input, r(t), and the system output, y(t). For well-posedness and eventually internal stability (see (DFT) Sections 3.1 and 3.2), all nine transfer functions between the input signals R[S], D[s], and N[s] and the outputs of the summing junctions e1[s], e2[S], and e3 [S], must exist and be stable, respectively. For this problem, find the following four transfer functions in terms of G1 [s], G2 [s], and H[s]: (i) T, [s] = "18, (ii) T2 (iii) [s] = RIS T4[S] = . Because we are dealing with linear systems and superposition holds, you may set R[S] to zero the inputs that are not associated with the transfer function in question. For example, when solving for Tz [s] = Yig, you may set D[s] = 0 and N[s] = 0. is , and (iv) D(A) Controller Plant R(e) + Gel Glo u & G, (0] MAC el Ghitans eeld Glosto Feed back Gain R(o)= Reference Input Y(0) = Output D(0] = Disturbance Input N[0] = Noise Input a(o)= Input Signal

Answer 1

The Given Control Design is shown in the figure below-

D(59 Controller Plant Ru Do al un šo como y (5) [H(s)k eg(s) x N(S) Feedback Gain

Now We have to find out T1(s), T2(S), T3(s).

Let's find out all one by one----

R (S) (i) finding Ti(): Ti(s)= y(s) when N(5)=0 and DCs?=0. N(5)=0 € D (5) = O makes the control design as shown in below tigure. R69 _ 6ix) {GI (59 ) {G12657F1 Y ) - H(s) This design can again be modified as, RCS) 100 19) 6126591T +rs) I H (S) H e() R(S) – Y(S) H(S) again, Y(5) = eics) Gi (5) G2(5) :-) Y (5)= TR (S) - Y(S) H(S)] Gi (5) G2(s) • Y(s) - R() G ($) G2 (3) - Y(s) (5) GSDG2(S) » Y() Fit Gy(5) G2(5) H(S)] = R(S) Gi(5) G215) Y(s) Gi (s) G2 (s) R (5) 1+ G, (S) G12 (5) H (5) → Ti(s) = Gi(3) G218) 4+ G (6) G2(S) HS. This is the answer of question (;) ?

This is time to find out T2(S),

(ii) Finding Tz (s) : : T2 (5) = yes when R(S)=o and N(5)= 0 R(3)=0 and Nis)=0 makes the control design as shown in the below figure D(S) D(59 019-6115) 2 0 0171 Yes) HB) K EKO simplied to--the figure as shown below - This can be again Pl? {6265) Poris) - G1 (5) 4(3) 12 Thus, we can write, &(52= D(S)- Gi(S) H (5) Y(S) Y(S)= ez (5) G12() Ore Y151 = TD (5244)G510 (5) +(59) G12 (5) Or, y(s)= 015) G2(s) – Y15) Gi () # (s) . op, Y(S) Tit Gi (s) 4() 1 - 0 (5) G2(5 or Y(S)/0 = G2(S) / 1+ G1 (59 H (5) on, T315) = _G2(8) 1 + G11(5) H(5) + Answer for the question in duostas ©

Now, let's find out the last transfer function T3(S),

N (S) (tii) finding T3(5): T3 (39= y 2 when R65)=0 and D65)=0 RCS)=0 and D(52-0 makes the control design as shown in the below figure op 61 62 482677 pre H() L ek N(s) This can be again Simplified to the figure as shown - N(S) -6,44f-H(s) GI(C) G2(8) 71 VIS) R3(s)= N(5) +Y(s) ? YISDE 23659 FHC) Gu(5) 4726577 => Y(S)- TN (S) + y(s)] F-HIS) Gı(5) G2(S)] > Y(52= - N(3) H (3). G11 (5) G2 (s) - Y(5) H(5) G. (5) G2 (5) => -4(5) TI+ HCS) Gi(5) G2 (s)] = N(S) H(5) G, (3) G2 (5) => Y(s) HCS) Guils) G12 (5) _ 16 HCS) G (5) G2(s) - NIS) = Ħ() GI(S) G2(5) TH(S) GI(S) G2 (5) > Answer for question the m

In this way, We have found out all the transfer functions. Look at the answers again-

Thus we have found out, Tils), Tals) and 73(5), Tils) = GI(S) G2 (6) 1+ G. (5) G2 (5) H(5) Tz (S) == G2(S) 1+ Gils) 8 (5) and T3 (s) = - - H(S) GI(S) G2 (6) + H (5) Gi($) G2 (s)