3.3 Transfer Function Solution : The s-domain network is shown in the Fig. 3.1. Applying KVL to the two loops, R I (s) I (s) sL I (s) sL V(s)1 1 1 2 = 0 i.e. I (s) [R sL] I (s) sL1 1 2 = V(s) ... (1) I (s) R 1 sC I (s) sL I (s) sL I (s)2 2 2 2 1 = 0 I (s)1 = sL s LC 2 2 2 2 2 2 s LC sL2 2 2 1 I (s) V(s) 2 = s LC s LCR s R R C R s L C s L R C sL s L C 2 3 2 2 2 s LC (R R ) s (R R C L) R 2 TECHNICAL PUBLICATIONS - An up thrust for knowledge TM(3 - 1) 3 Transfer Function and Impulse Response 1(s) R2R1 sL Q.2 Solution : The s-domian network is shown in the Fig. 3.2 (a). T.F. = E (s) E (s) Z Z Z (R R )+ sR R (C C ) 2 1 1 1 2 1 2 1 2
Principles of Control Systems 3 - 2 Transfer Function and Impulse Response TECHNICAL PUBLICATIONS - An up thrust for knowledge TM E (s)i 5.3 Rules for Block Diagram Reduction Q.11 Solution : Shifting take off point after G 2 and seperating feedback paths we get, TECHNICAL PUBLICATIONS - An up thrust for knowledge TM(5 - 1) 5 Block Diagram Representation of Control Systems G1 G2 2 2 1 2 1 G1 G (1 + G ) 1 + G H + G H + G G H H + G G H + G G G H H 3 2 3 3 2 2 2 3 2 3 1 2 1 1 2 3 1 3 R(s) C(s) Series https://gioumeh.com/product/principle-of-control-system-solution/ https://gioumeh.com/product/principle-of-control-system-solution/ Q.12 Solution : No series, parallel combination and no minor feedback loop exists. So shifting take off point before the block of s s + 10
. Principles of Control Systems 5 - 2 Block Diagram Representation of Control Systems TECHNICAL PUBLICATIONS - An up thrust for knowledge TM – – + (s + 11) (s 3) (s 4) + s(2s + 7) (s + 8)3 C(s) R(s) = 2 Q.4 Solution : As there are two inputs, consider each input separately. Consider R(s), assuming Y(s) = 0. C(s) R(s) = 1 2 1 i.e. C(s) = R(s) G G 1 2 1 Now consider Y(s) acting with R(s) = 0. Now sign of signal obtained from H1 is negative which must be carried forward, though summing point at R(s) is removed, as R(s) = 0, so we get, Key Point While finding equivalent G, trace forward path from input summing point to output in direction of signal. While finding equivalent H, trace the feedback path from output to input summing point in the direction of signal. Now equivalent G = G 2 , tracing forward path from input summing point to output. Equivalent H = G H1 1 tracing feedback path from output to input summing point. While sign of the final feedback is positive at the input summing point. Principles of Control Systems 5 - 3 Block Diagram Representation of Control Systems TECHNICAL PUBLICATIONS - An up thrust for knowledge TM – + + C(s) G2 3_1 5_1 6_1 7_1 8_1 10_1 11_1 12_1 14_1