Question

For a Mechanical Engineering System Dynamics class

1. 4 C(s) 1 S A system has a block diagram as shown. The input is R(s) and the output is C(s). a) Using only the block diagram reduction method*, R(s) find the transfer function of the system. b) Determine the characteristic function and the order of the system. c) Find the characteristic roots of the system. d) Find the natural frequency of the system. Find the damped natural frequency of the system. 8 * NOTE: All stages of block diagram reduction must be clearly shown until you arrive at the last simple closed loop diagram and then the final open-loop equivalent with the transfer function in question. In addition, the final transfer function must be simplified, i.e., no fraction(s) over fraction(s).

Answer 1

SOLUTION:

- Given: 144 c(s) RCS t 8 K Using Block Reduction Method Step 1) Simplifying the feedback R(S) + 사 R(5) ts ***** 1+ 4(12 uts

After 1st Reduction : C(s) R(S) Colt 4+S sk Step 2) Simplifying the blocks in Series R(S) CCS) LAS 18 6,75 TL S 45+5? After 2nd Reduction, cs

RE CES) I 45+82 18k - The above is the simple closed loop diagram Step 3) Reduction of feedback loop To get the open loop cuith Transfer function (IF) (CS) usts? sk 4 Res) CCS) 4sts? 1+ 8 (452) usts? [es]

-R(S) (5) S²+ 45+8 TF= 52+ 45+8 Formula : De --- G 7 GH H 2) xail yazt Raiga by characteristic function and order of the system, Comparing the TF cwith Standard form, I s2+ 45+8 S²+25unstown? una

) Natural Frequency From TE= S2+ 45+8 Com paring with the standard form cun? S²+2 scunst aan?. ce get, asun=4 and ain²=8. cun= √8 Con=22 rad S rad e Damped natural frequency cwd-on Ji-s 2 - from step (a) we have Con=22 2 x 5 = 4 un ។ 2x252 3= . √2 Ts=0nol wod=252 I-conoruz od=2 rad Is

ANSWERS:

A) TRANSFER FUNCTION (TF) = 1 / ($s^{2}$ + 4s + 8)

B) CHARACTERISTIC EQN = $s^{2}$ + 4s + 8 , WITH ORDER OF 2

C) CHARACTERISTIC ROOTS ARE (-2 + 2i ) AND (-2 - 2i )

D) NATURAL FREQUENCY (wn) = $2\sqrt{2}$ rad / s

E) DAMPED NATURAL FREQUENCY (wd) = 2 rad / s