Problem 24

Subject: 5. Reduction of Multiple Subsystems.


 Given the system below, draw a signal-flow graph and represent the system in state space in the following forms:
  1. Phase-variable form
  2. Cascade form

(Page 310)


    Part a: 1. Phase variables
    2. State-space representation
    3. Signal-flow graph
    Part b: 1. State space: cascade form representation
    2. Converting block diagrams to signal flow graphs


  1. Part a:
    Find the associated differential equation. 
  2. Select Phase variables and write the state and output equations. 
  3. Draw a signal-flow graph of the state and output equations. 
  4. Part b:
    Draw a block diagram of the transfer function in cascade form.
  5. Draw a signal-flow graph for each first-order system.
  6. Draw a signal-flow graph for the entire system by combining the signal-flow graphs of the subsystems. Write the state and output equations. 

Detailed Solution

  1.  Part a:
    We need to represent the system eqimages/q61_2.png  in phase-variable form and draw a signal-flow graph. 
  2.  First, we find the associated differential equation. 
  3.  The corresponding differential equation is found by taking the inverse Laplace transform, assuming zero initial conditions:
  4.  We choose the output and its two derivatives as phase variables :
  5.  Differentiating both sides of (4) and using the differential equation (3), we obtain the state equations: 
  6.  The output equation is simply:
  7.  Rewriting equations 5 and 6 in vector-matrix form, we obtain:
  8.  Now, we can draw a signal-flow graph of the state equations (5). First, we identify three nodes corresponding to the three state variables, three nodes corresponding to the derivatives of the state variables, the input node, eqimages/q61_9.png , and the output node, eqimages/q61_10.png , as shown in Fig. 24-1:
    (Figure 24-1)
  9.  Now we connect the state variables with their derivatives with the defining integration, eqimages/q61_11.png , as shown in Fig. 24-2:
    (Figure 24-2)
    Note that we place each "derivative" node to the left of the corresponding "state" node.
  10.  Finally, using the state and output equations (5 and 6), we feed to each node the appropriate signal as shown in Fig. 24-3:
    (Figure 24-3)
  11.  Part b:
    We need to represent the system eqimages/q61_12.png in cascade form, and draw a signal-flow graph.
  12.  Cascading each term of the transfer function eqimages/q61_13.png , we obtain the block diagram shown in Fig. 24-4. The output of each block corresponding to a first-order system is labeled as a state variable. 
    (Figure 24-4)
  13.  In order to draw a signal-flow graph of the system, we first find the signal-flow graph for each first-order system. For example, cross-multiplying the transfer function eqimages/q61_14.png , yields:
  14.  Taking the inverse Laplace transform, we have: eqimages/q61_16.png  and solving for eqimages/q61_17.png  yields:
    eqimages/q61_18.png .
  15.  Assuming a node for eqimages/q61_19.png , a node for eqimages/q61_20.png  and a node for the input eqimages/q61_21.png , we obtain the signal-flow graph for eqimages/q61_22.png :
    (Figure 24-5)
  16.  Similarly, we find the signal-flow graphs for eqimages/q61_23.png  and eqimages/q61_24.png , shown in Fig. 24-6 and Fig. 24-7, respectively:
    (Figure 24-6)
    (Figure 24-7)
  17.  Cascading the transfer functions shown in Figures 24-7, 24-6, and 24-5, we obtain the system representation shown in Fig. 24-8:
    (Figure 24-8)
  18.   From the signal-flow graph depicted in Fig. 24-8 we can write the state and output equations: 
  19.  Rewriting (18) in vector-matrix form yields:

Final Answer

a. eqimages/q61_29.png  
b. eqimages/q61_30.png