UNIT-1
Introduction to Control System
Q1)Find SFG for the block diagram below?
Sol:
Ra+cb =c
c/R= a/1-b
Q2) The SFG shown has forward path and singles isolated loop determine overall transmittance relating X3 and X1?
Sol:-
X1- I/p node
X2-Intenmediale node
X3- o/p node
ab- forward path (p)
bc- 1 loop (L)
At node XQ:
X2 = x1a + x3c [Add i/p signals at node]
At node x3:
x2b =x3
(x1a+x3c) b = x3
X1ab = x3 (1-bc)
X1 = x3 (1-bc)/ab
Ab/(1-bc) = x3/x1
T= p/1-L
X1:- I/p node x2, x3,x4,x5,Qnlexmedili node
X0:- o/p node abdeg:- forward path
bc, ef :- Loop [isolated]
x2 = ax1+c x3
x3= bx2
x4 = d x3+f x5
x5 = e x4
x6= g x5
x6 = g(e x4) = ge [dx3+ e f x5]
xb = ge [d (bx2) + f (e x4)]
xb = ge [ db (ax1+cx3) + fe (dx3+ fx5)]
xb = ge [db (ax1+cb (ax1+x3) +fe[cdbx2]+
f( e [db (ax1+ cx3)
x2 = ax1 + cb (x2) x4 = d bx2 + f exq
x2 = ax1 + cbx2 = db (d4) + fe/1-cb
x2 = ax1/(1-cb) xy = db x2 + f x6/g
xy = db [ax1]/1-cb + f xb/g
x5 = c db ( ax1)/1-cb + efxb/g
xb = gx5
= gedb (ax1)/1-cb + g efxb/g
Xb = gx5
gedb (ax1)/1-cb + g efxb/g
(1- gef/g) xb = gedb ax1/1-ab
Xb/x1 = gedb a/ (1- ef – bc + beef
Xb/x1 = p/ 1- (L1+L2) + L1 L2 for isolated loops
Q3) Reduce given B.D to canonical (simple form) and hence obtain the equivalent Tf = c(s)/ R(S) ?
Sol:
C(S)/R(S) = (G1G2) (G3+G4)/1+G1G2H1)/1-G1,G2(G3+G4) H2/1+G1G2H1
= G1G2(G3+G4)/1+G1G2H1-G1G2H2(G3+G4)
=G1G2(G3+G4)/1+(H1-H2)(G1G2) (G3+G4)
C(s)/R(S) = G1G2(G3+G4)/1+(H1-H2(G3+G4)) G1 G2
Reduce the Block diagram
C(s)/R(s)= G1(G3+G2)/(1-G1G3X1) (1-G2X2) H1
= G(G3+G2)/(1-G3G1H1) (1-G2H2) + G1H1(G3+G2)
= G1(G3+G2)/1-63G1H1-G2H2+G1H1(G3+G2H1
=G1(G3+G2)/1-G3H2+G1G2H1(1+G3H2)
Q4) Reduce using Masons gain formula
Sol:
P1= G1 p2 =G2 Delta1 =1
L1= -G1 H1
= 1-(-G1H1)
= 1+G1H1
T= G1+G2/1+G1H1
Q5) Determine overall gain reliably x5 and x1 also draw SFG
Sol:
X2 = ax1+ f x2
X3= bx2 +exy
X4 = cx3+hx5
X5 =dx4 + gx2
P1 = abcd p2 = ag
L1 = f L2 = ce, L3= dh
1 = 1
2= 1-ce
= 1-[L2+L2+L3] + [L1 L2 +L1 L3]
= 1-[f+ le = dh] + [fce +fdh]
T= abcd+ ag (1-ce)/1-[ftce + dh ] + (fce + fdh)
Q6) For the given electrical system below draw the analogous system find V0(s)/V1(s)
Sol:
Let Z1 = R2 11 1/c2
=R2*1/c2s/R2+1/c2s
Z1= R2/1+R2c2s
Let Z2 = R1+1/c1s
Z2= 1+R1c1s/c1s
V0(s)/vi(s) = z2/z1+z2
= 1+R1c1s/c1s/R2/1+R2c2s+1+R1c1s/c1s
V0(s)/v0(s) =(c1+R1c1s) (1+R2c2s)/R2c1s+1+sR1c1s2R1R2c1c2
Q7) Difference between open and closed-loop system?
Sol:
Open Loop System | Closed-Loop system |
1.It does Not have any feedback. | 1. This system comprises a feedback |
2.As no feedback so easier to build. | 2.As it has feedback so difficult to build |
3.Theaccuracy of this system depends on the calibration of input. | 3.They are accurate because of the feedback. |
4.Open Loop systems are more stable. | 4.In closed Loop system stability depends on system components. |
5.optimization is not possible | 5. optimization is possible |
6.These systems are not reliable. | They are more reliable |
Q8) Explain Mason’s gain formula?
Sol: T= 
The overall transmittance Coverall gain can be determined by the Masks formula.
Explanation:-
Pk
forward path transmittance of k+n path from a specified i/p node to n o/p nods
While calculating ipnode to n o/p nods.
While calculating ip no node should be encountered (used) more than ones.
it is the graphics determined which involve transmittances and multiple increases b/w non-touching loops.
= 1- [sum of all individual loop transmitting]
+[ sum of loop transmittance product of all possible non-touching loops]
-[sum of loop transmittance of all possible triples of non-touching loops]
path factor associated with concered path & involves all an in the graphic which is isolated from the forward path under consideration.
The path factor 
for kthis equal to graph determinant of SFG which effect after erasing the kth path from the graph
Q9) Write rules for drawing SFG?
Sol:
1) The signal travels along a branch in the direction of an arrow.
2) The lip signal is multiplied by the transmittance to obtain the o/p.
3) I/p signal at a node is the sum of all the signals entering at that node.
4) A node transmits a signal at all branches leaving that node.
Q10) List advantages of block diagram reduction technique?
Sol:
(1) Very simple to Construct the Block diagram of complicated electrical & mechanical sys.
(2) The function of the individual elements can be visualized form a block diagram
(3) Individual, as well as the overall performance of the system, can be studied by the if shown in Block diag.
(4) Overall CLTf can be easily calculated by Block diagram reduction rules.
