4.5 BLOCK [DIAGRAMS](#page-10-0)

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4.5 BLOCK DIAGRAMS

Large systems may consist of an enormous number of components or elements. As anyone who has seen the circuit diagram of a radio or a television receiver can appreciate, analyzing such systems all at once could be next to impossible. In such cases, it is convenient to represent a system by suitably interconnected subsystems, each of which can be readily analyzed. Each subsystem can be characterized in terms of its input–output relationships. A linear system can be characterized by its transfer function H(s). Figure 4.18a shows a block diagram of a system with a transfer function H(s) and its input and output X(s) and Y(s), respectively.

Subsystems may be interconnected by using cascade, parallel, and feedback interconnections (Figs. 4.18b, 4.18c, 4.18d), the three elementary types. When transfer functions appear in cascade, as depicted in Fig. 4.18b, then, as shown earlier, the transfer function of the overall system is the product of the two transfer functions. This result can also be proved by observing that in Fig. 4.18b

$$\frac{Y(s)}{X(s)} = \frac{W(s)}{X(s)} \frac{Y(s)}{W(s)} = H_1(s)H_2(s)$$

Figure 4.18 Elementary connections of blocks and their equivalents.

We can extend this result to any number of transfer functions in cascade. It follows from this discussion that the subsystems in cascade can be interchanged without affecting the overall transfer function. This commutation property of LTI systems follows directly from the commutative (and associative) property of convolution. We have already proved this property in Sec. 2.4-3. Every possible ordering of the subsystems yields the same overall transfer function. However, there may be practical consequences (such as sensitivity to parameter variation) affecting the behavior of different ordering.

Similarly, when two transfer functions, H1(s) and H2(s), appear in parallel, as illustrated in Fig. 4.18c, the overall transfer function is given by H1(s) + H2(s), the sum of the two transfer functions. The proof is trivial. This result can be extended to any number of systems in parallel.

When the output is fed back to the input, as shown in Fig. 4.18d, the overall transfer function Y(s)/X(s) can be computed as follows. The inputs to the adder are X(s) and −H(s)Y(s). Therefore, E(s), the output of the adder, is

$$E(s) = X(s) - H(s)Y(s)$$

But

$$Y(s) = G(s)E(s)$$

= $G(s)[X(s) - H(s)Y(s)]$

Therefore,

$$Y(s)[1 + G(s)H(s)] = G(s)X(s)$$

so that

$$\frac{Y(s)}{X(s)} = \frac{G(s)}{1 + G(s)H(s)}$$

(4.35)

Therefore, the feedback loop can be replaced by a single block with the transfer function shown in Eq. (4.35) (see Fig. 4.18d).

In deriving these equations, we implicitly assume that when the output of one subsystem is connected to the input of another subsystem, the latter does not load the former. For example, the transfer function H1(s) in Fig. 4.18b is computed by assuming that the second subsystem H2(s) was not connected. This is the same as assuming that H2(s) does not load H1(s). In other words, the input–output relationship of H1(s) will remain unchanged regardless of whether H2(s) is connected. Many modern circuits use op amps with high input impedances, so this assumption is justified. When such an assumption is not valid, H1(s) must be computed under operating conditions [i.e., with H2(s) connected].

EXAMPLE 4.21 Transfer Functions of Feedback Systems Using MATLAB

Consider the feedback system of Fig. 4.18d with G(s) = K/(s(s + 8)) and H(s) = 1. Use MATLAB to determine the transfer function for each of the following cases: (a) K = 7, (b) K = 16, and (c) K = 80.

We solve these cases using the control system toolbox function feedback.

(a)
>> H = tf(1,1); K = 7; G = tf([0 0 K],[1 8 0]); TFa = feedback(G,H)
    Ha =
            7
      -------------
      s^2 + 8 s + 7
Thus, Ha(s) = 7/(s2 +8s+7).
(b)
>> H = tf(1,1); K = 16; G = tf([0 0 K],[1 8 0]); TFb = feedback(G,H)
    Hb =
            16
      --------------
      s^2 + 8 s + 16
Thus, Hb(s) = 16/(s2 +8s+16).
(c)
>> H = tf(1,1); K = 80; G = tf([0 0 K],[1 8 0]); TFc = feedback(G,H)
    Hc =
            80
      --------------
      s^2 + 8 s + 80
Thus, Hc(s) = 80/(s2 +8s+80).