UNIT 2
Block Diagrams and Signal flow graphs
It is the ratio of Laplace transform of the output to Laplace transform of input with all initial conditions zero.
Fig 1 System with Transfer Function G(s)
Poles and zero of a Transfer Function:
The transfer function can be represented by the ratio of two polynomials
G (S) = a0sn+a1 sn1+an/b0sm+b1sm1++bn
a0—an  constants
G(S) = K(s+z1) (s+z2) (as2+bs+c)/(S+A) (s+p2) (As2+Bs +c)
K= a0/b0 ( Gain of system)
For poles –They are the values of s for which
G(S)
(S+p1)(S+p2)(A
S= p1, p2 , B±B24Ac/2A
For ZEROS – They are the values of s for which
G(S)
S=z1, z2 , b±b24ac/2a
 Location of poles and zeros in s place determines the reliability of the system
 There can be multiple poles and zeros
 The numerator of transfer function when equalized to zero gives zero of system
 The denominator of transfer function which equalized to zero gives poles of system.
Key takeaways
 The numerator of transfer function when equalized to zero gives zero of system.
 The denominator of transfer function which equalized to zero gives poles of system.
Advantages of Block diagram reduction technique:
 Very simple to Construct the Block diagram of complicated electrical & mechanical systems.
 The function of individual element can be visualized form block diagram
 Individual as well as overall performance of the system can be studied by the figure shown in Block diag.
 Overall CLTF can be easily calculated by Block diagram reduction rules.
Disadvantages of Block diagram reduction technique:
It does not include any information above physical construct of system (completely mathematical approach).
 Source of energy in generally not shown in the block diagram so w.g diff. block diagram can be drawn for the same function
CLTf: ve feedback
C(s)/R(s)= G(s)/1+G(s)H(s)
CLTF:+vefeedback
C(S)/R(S) = G(S)/1G(s)H(S)
Rules of Block diagram Algebra:
Block in cascade
Moving summing point after a block
+
Moving summing point ahead of block
Moving take off point after a block

Moving take off point ahead a block

Eliminating a feedback Loop

Q. 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)/1G1,G2(G3+G4) H2/1+G1G2H1 = G1G2(G3+G4)/1+G1G2H1G1G2H2(G3+G4) =G1G2(G3+G4)/1+(H1H2)(G1G2) (G3+G4) C(s)/R(S) = G1G2(G3+G4)/1+(H1H2(G3+G4)) G1 G2

Reduce the Block diagram



C(s)/R(s)= G1(G3+G2)/(1G1G3X1) (1G2X2) H1 = G(G3+G2)/(1G3G1H1) (1G2H2) + G1H1(G3+G2) = G1(G3+G2)/1G3G1H1G2H2+G1H1(G3+G2H1 =G1(G3+G2)/1G3H2+G1G2H1(1+G3H2)


Block Diagram

SFG
Q: Draw SFG for the following block diagram?

Sol:

Ra+cb =c
c/R= a/1b
RULES:
1) The signal travels along a branch in the direction of an arrow.
2) The input signal is multiplied by the transmittance to obtain the o/p.
3) I/p signal at a node is sum of all the signals entering at that node.
4) A node transmits signal at all branches leaving that node.
Q. The SFG shown has forward path and singles isolated loop determine overall transmittance relating X3 and X1
Soln:
X1 I/p node
X2Intenmediale 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 (1bc) X1 = x3 (1bc)/ab Ab/(1bc) = x3/x1 T= p/1L 

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/1cb x2 = ax1/(1cb) xy = db x2 + f x6/g xy = db [ax1]/1cb + f xb/g x5 = c db ( ax1)/1cb + efxb/g xb = gx5 = gedb (ax1)/1cb + g efxb/g Xb = gx5 gedb (ax1)/1cb + g efxb/g (1 gef/g) xb = gedb ax1/1ab Xb/x1 = gedb a/ (1 ef – bc + beef Xb/x1 = p/ 1 (L1+L2) + L1 L2 for isolated loops

Key takeaways
 The function of individual element can be visualized from block diagram
 The signal travels along a branch in the direction of an arrow.
 The input signal is multiplied by the transmittance to obtain the o/p.
Reference
1. “Modern Control Engineering “, K. Ogata, Pearson Education Asia/
PHI, 4th Edition, 2002.
2. “Automatic Control Systems”, Benjamin C. Kuo, John Wiley India
Pvt. Ltd., 8th Edition, 2008.
3. “Feedback and Control System”, Joseph J Distefano III et al.,
Schaum’s Outlines, TMH, 2nd Edition 2007.
4. J. Nagarath and M.Gopal, “Control Systems Engineering”, New Age
International (P) Limited, Publishers, Fourth edition – 2005