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Reduction of multiple subsystem [compatibility mode]

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  • 1. Reduction of Multiple Subsystems
  • 2. INTRODUCTIONComplicated system/multiple subsystem are represented by the interconnectionof many subsystems.Multiple subsystem are represented in two ways: as block diagrams and assignal-flow graphs.Block diagrams are usually used for frequency-domain analysis and design.Signal-flow graphs for state space analysis and design.Techniques to reduce multiple subsystem to a single transfer function:-1) Block diagram algebra –to reduce block diagrams2) Manson’s rule - to reduce signal-flow graphs.
  • 3. BLOCK DIAGRAMS Figure 1Components of a block diagram for a linear, time-invariant system
  • 4. A. CASCADE FORM Figure 2 a. Cascaded subsystems; b. equivalent transfer functionIntermediate signal values are shown at the output of each system.Each signal is derived from the product of the input times the transfer function.The equivalent transfer function, Ge(s), shown in Figure 1(b), is the outputLaplace transform divided by the input Laplace transform from Figure 1(a), or Ge ( s )  G3 ( s )G2 ( s )G1 ( s ) [1] which is the product of the subsystems’ transfer functions. Eq.[1] was derived under the assumption that interconnected subsystems do not load adjacent subsystems.
  • 5. Figure 3Loading in cascaded systems
  • 6. B. PARALLEL FORM Figure 4 a. Parallel subsystems; b. equivalent transfer function Parallel subsystems have a common input and an output formed by the algebraic sum of the outputs from all of the subsystems. The equivalent transfer function, Ge(s), is the output transform divided by the input transform from Figure 4(a) or Ge ( s )  G1 ( s )  G2 ( s )  G3 ( s ) [2] which is the algebraic sum of the subsystems’ transfer function; it appears in Figure 5(b).
  • 7. C. FEEDBACK FORM Figure 5 a. Feedback control system; b. simplified model; c. equivalent transfer function
  • 8. From Figure 5(b), E ( s)  R( s)  C ( s) H ( s) [3]But since C(s)=E(s)G(s), C ( s) [4] E ( s)  G( s)Substituting Eq. [4] into Eq.[3] and solving for the transfer function,Ge(s)=C(s)/R(s), we obtain the equivalent, or closed-loop, transfer functionshown in Figure 5(c), G( s) Ge ( s)  [5] 1  G( s) H ( s) The product, G(s)H(s), in Eq.[5] is called the open-loop transfer function, or loop gain.
  • 9. MOVING BLOCKS TO CREATE FAMILIAR FORMS Figure 6 Block diagram algebra for summing junctions—equivalent forms for moving a block a. to the left past a summing junction; b. to the right past a summing junction
  • 10. Figure 7Block diagram algebra for pickoff points— equivalent forms for moving a blocka. to the left past a pickoff point; b. to the right past a pickoff point
  • 11. Block diagram reduction via familiar formsExample 1: Reduce the block diagram shown in Figure 8 to a single transferfunction. Figure 8 Block diagram for Example 1
  • 12. Figure 9 Steps in solving Example 1:a. collapse summing junctions; b. form equivalent cascaded system in the forward path and equivalentparallel system in the feedback path; c. form equivalentfeedback system andmultiply by cascaded G1(s)
  • 13. Block diagram reduction by moving blocksExample 2: Reduce the system shown in Figure 10 to a single transfer function. Figure 10 Block diagram for Example 2
  • 14. Figure 11 Steps in theblock diagram reduction for Example 2
  • 15. Example 3: Find the equivalent transfer function, T(s)=C(s)/R(s), for the systemshown in Figure 12. Figure 12 Block diagram for Example 3
  • 16. Example 4: Reduce the block diagram shown in Figure 13 to a single transferfunction, T(s)=C(s)/R(s). Use the block diagram reduction. Figure 13
  • 17. Example 5: Reduce the block diagram shown in Figure 14 to a single transferfunction, T(s)=C(s)/R(s). Use the block diagram reduction. Figure 14
  • 18. ANALYSIS AND DESIGN OF FEEDBACK SYSTEMS Figure 15 Second-order feedback control systemThe transfer function K/s(s+a), can model the amplifiers, motor, load and gears.From Eq.[5], the closed-loop transfer function, T(s), for this system is K T ( s)  2 [6] s  as  Kwhere K models the amplifier gain, that is, the ratio of the output voltage to theinput voltage. As K varies, the poles move through three ranges of operation ofa second-order system: overdamped, critically damped, and underdamped.
  • 19. For K between 0 and a2/4, the poles of the system are real and are located at a a 2  4K [7] s1, 2   2 2As K increases, the poles move along the real axis, and the system remainsoverdamped until K=a2/4. At that gain, or amplification, both poles are real andequal, and the system is critically damped.For gain above a2/4, the system is underdamped, with complex poles located at a 4K  a 2 [8] s1, 2   j 2 2As K increases, the real part remains constant and the imaginary part increases.Thus, the peak time decreases and the percent overshoot increases, while thesettling time remains constant.
  • 20. Example 4: For the system shown in Figure 14, find the peak time, percentovershoot, and settling time. Figure 16 Feedback system for Example 4 The closed-loop transfer function Substituting Eq.[10] into Eq.[11], found from Eq.[6] is   0.5 [12] 25 T (s)  2 [9] Using the values of  and  n s  5s  25  From Eq.[9], Tp   0.726 s [13] n 1   2 n  25  5 [10] %OS  e  / 1 2 100  16.303 [14] 2 n  5 [11] Ts  4  1 .6 s [15]  n
  • 21. Example 5: Design the value of gain, K, for the feedback control system ofFigure 15 so that the system will respond with a 10% overshoot. Figure 15 Feedback system for Example 5The closed-loop transfer function of From Eq.[17] and Eq.[18],the system is 5 K [16]   [19]T (s)  2 2 K s  5s  K A 10% overshoot implies thatFrom Eq.[16],  / 1 2 %OS  e 100  10% 2 n  5 [17]   0.591 [20] n  K [18] Substituting   0.591 Eq. [19] K  17.9 [21]
  • 22. Example 6: For a unity feedback control system with a forward-path transferfunction G(s)=16/s(s+a), design the value of a to yield a closed-loop stepresponse that has 5% overshoot.
  • 23. SIGNAL-FLOW GRAPHSA signal-flow graph consists only of branches, which is represent systems, andnodes, which represent signals.A system is represented by a line with an arrow showing the direction of signalflow through the system.Adjacent to the line we write the transfer function.A signal is a node with the signal’s name written adjacent to the node. Figure 17 Signal-flow graph components: a. system; b. signal;c. interconnection of systems and signals
  • 24. Each signal is the sum of signals flowing into it. From Figure 16c, example for the signal, V ( s )  R1 ( s )G1 ( s )  R2 ( s )G2 ( s )  R3 ( s )G3 ( s ) [1] C2 ( s )  V ( s )G5 ( s )  R1 ( s )G1 ( s )G5 ( s )  R2 ( s )G2 ( s )G5 ( s )  R3 ( s )G3 ( s )G5 ( s ) [2]C3 ( s )  V ( s )G6 ( s )   R1 ( s )G1 ( s )G6 ( s )  R2 ( s )G2 ( s )G6 ( s )  R3 ( s )G3 ( s )G6 ( s ) [3] Notice that in summing negative signals we associate the negative sign with the system and not with a summing junction, as in the case of block diagrams.
  • 25. Converting common block diagrams to signal-flow graphsExample 1: Convert the cascaded, parallel, and feedback forms of the blockdiagrams in Figure 2a, 4a and 5b, respectively, into signal-flow graphs. Figure 18 Building signal-flow graphs: Figure 2(a) a. cascaded system nodes (from Figure 2(a)); b. cascaded system signal-flow graph;
  • 26. c. parallel systemnodes (from Figure 4(a));d. parallel systemsignal-flow graph; Figure 4(a)
  • 27. e. Feedback system nodes(Figure 5(b))f. feedback systemsignal-flow graph Figure 5(b)
  • 28. Converting a block diagram to a signal-flow graphExample 2: Convert the block diagram of Figure 10 to a signal-flow graph. Figure 10
  • 29. Figure 19 Signal-flow graph development: a. signal nodes; b. signal-flow graph;c. simplified signal-flow graph
  • 30. Example 3: Convert the block diagram of Figure 12 to a signal-flow graph. Figure 12
  • 31. MASON’S RULEMason’s signal – a technique for reducing signal-flow graphs to single transferfunctions that relate the output of a system to its input.Definitions[1]Loop gainThe product of branch gains found by traversing a path that starts at a node andends at the same node, following the direction of signal flow, without passingthrough an other node more than once.There are four loop gains (Refer Figure 20): (1) G2 ( s) H1 ( s) [4a] (2) G4 ( s) H 2 ( s) [4b] (3) G4 ( s )G5 ( s ) H 3 ( s ) [4c] (4) G ( s )G6 ( s ) H 3 ( s ) [4d] 4
  • 32. Figure 20 Signal-flow graph for demonstrating Mason’s rule[2]Forward-path gainThe product of gains found by traversing a path from the input node to theoutput node of the signal-flow graph in the direction of signal flow.These are two forward-path gains:(1) G1 ( s )G2 ( s )G3 ( s )G4 ( s )G5 ( s )G7 ( s ) [5a](2) G1 ( s )G2 ( s )G3 ( s )G4 ( s )G6 ( s )G7 ( s ) [5b]
  • 33. [3]Nontouching loopsLoops that do not have any nodes in common. In Figure 20, loop G2(s)H1(s)does not touch loops G4(s)H2(s), G4(s)G5(s)H3(s), and G4(s)G6(s)H3(s).[4]Nontouching-loop gainThe product of loop gains from nontouching loops taken two, three, four, ormore at a time. In Figure 20 the product of loop gain G2(s)H1(s) and loop gainG4(s)H2(s) is a nontouching-loop gain taken two at a time. In summary, all threeof the nontouching-loop gains taken two at a time are (1) G2 ( s) H1 ( s)G4 ( s) H 2 ( s) [6a] (2) G2 ( s) H1 ( s)G4 ( s)G5 ( s) H 3 ( s) [6b] (3) G2 ( s) H1 ( s)G4 ( s)G6 ( s) H 3 ( s) [6c]
  • 34. Mason’s RuleThe transfer function, C(s)/R(s), of a system represented by a signal-flow graphsis C ( s )  k Tk  k G (s)   [7] R( s) where k Number of forward paths Tk  The kth forward-path gain  1-  loop gain +  nontouching-loop gains taken two at a time -  nontouching-loop gains taken three at a time +  nontouching-loop gains taken four at a time - …  k   -  gain term in  that touch the kth forward path. In loop other words,  k is formed by eliminating from  those loop gains that touch the kth forward path.
  • 35. Transfer function via Mason’s ruleExample 1: Find the transfer function, C(s)/R(s), for the signal-flow graph inFigure 21. Figure 21 Signal-flow graph
  • 36. Step 1: Identify the forward-path gains. In this case, there is only one. G1 ( s )G2 ( s )G3 ( s )G4 ( s )G5 ( s ) [8]Step 2: Identify the loop gains. There are four, as follow: (1) G2 ( s ) H1 ( s ) [9.1] (2) G4 ( s ) H 2 ( s ) [9.2] (3) G7 ( s ) H 4 ( s ) [9.3] (4) G2 ( s )G3 ( s )G4 ( s )G5 ( s )G6 ( s )G7 ( s )G8 ( s ) [9.4]Step 3: Identify the nontouching loops taken two at a time. From Eq.[9] and Figure 21, loop 1 does not touch loop 2, loop 1 does not touch loop 3 and loop 2 does not touch loop 3. Notice that loops 1, 2, and 3 all touch loop 4. Thus, the combinations of nontouching loops taken two at a time are as follows:
  • 37. G2 ( s ) H1 ( s )G4 ( s ) H 2 ( s ) [10.1] Loop 1 and loop 2: Loop 1 and loop 3: G2 ( s ) H1 ( s )G7 ( s ) H 4 ( s ) [10.2] Loop 2 and loop 3: G4 ( s ) H 2 ( s )G7 ( s ) H 4 ( s ) [10.3] Finally, the nontouching loops taken three at a time as follows: Loop 1, 2 and 3: G2 ( s ) H 1 ( s )G4 ( s ) H 2 ( s )G7 ( s ) H 4 ( s ) [11]Step 4: From Eq.[7] and its definitions, we form  and  k . Hence,   1  [G2 ( s ) H 1 ( s )  G4 ( s ) H 2 ( s )  G7 ( s ) H 4 ( s ) [12]  G2 ( s )G3 ( s )G4 ( s )G5 ( s )G6 ( s )G7 ( s )G8 ( s )]  [G2 ( s ) H 1 ( s )G4 ( s ) H 2 ( s )  G2 ( s ) H 1 ( s )G7 ( s ) H 4 ( s )  G4 ( s ) H 2 ( s )G7 ( s ) H 4 ( s )]  [G2 ( s ) H 1 ( s )G4 ( s ) H 2 ( s )G7 ( s ) H 4 ( s )]
  • 38. We form  k by eliminating from  the loop gains that touch the kth forwardpath: 1  1  G7 ( s) H 4 ( s) [13]Step 5: Expressions [8], [12] and [13] are now substituted into Eq.[7], yieldingthe transfer function: T11 [G1 ( s)G2 ( s)G3 ( s)G4 ( s)G5 ( s)][1  G7 H 4 ( s)] [14] G( s)    Since there is only one forward path, G(s) consists of only one term, rather thana sum of terms, each coming from a forward path.
  • 39. Example 2: Use Mason’s rule to find the transfer function of the signal-flowdiagram shown in Figure 22. Figure 22 Signal-flow graph
  • 40. Example 3: Use Mason’s rule to find the transfer function of the signal-flowdiagram shown in Figure 23. Figure 23
  • 41. SIGNAL-FLOW GRAPHS OF STATE EQUATIONSConsider the following state and output equations: x1  2 x1  5 x2  3x3  2r  [15a] x2  6 x1  2 x2  2 x3  5r [15b]  x3  x1  3x2  4 x3  7r  [15c] y  4 x1  6 x2  9 x3 [15d]Step 1: Identify three nodes to be the three state variables, x1, x2 and x3; alsoidentify three nodes, placed to the left of each respective state variable, to bederivatives of the state variables. Also identify a node as the input, r, and anothernode as the output, y.Step 2: Next interconnect the state variables and their derivatives with thedefining integration, 1/s.
  • 42. Step 3: Then using Eqs.[15], feed to each node the indicated signals. Example: Eq.[15a] - x1 receives 2 x1  5 x2  3x3  2r (Figure 24c)  Eq.[15b] - x2 receives 6 x1  2 x2  2 x3  5r (Figure 24d)  Eq.[15c] - x3 receives x1  3x2  4 x3  7r (Figure 24e)  Eq.[15d] – the output,  4 x1  6 x2  9 x3 (Figure 24f) Figure 24f – the final phase-variable representation, where the state variables are the outputs of the integrators.
  • 43. Figure 24 Stages of development of asignal-flow graph for the system of Eqs.15: a. place nodes; b. interconnectstate variables and derivatives; c. form dx1/dt ; d. form dx2/dt(figure continues)
  • 44. Figure 24 (continued)e. form dx3 /dt; f. form output
  • 45. Example 1: Draw a signal-flow graph for the following state and outputequations:  2 1 0 0  x  0    3 1  x  0 r     3  4  5   1    y  0 1 0 x

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