Block Diagrams of Systems

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Sources:

  1. B. P. Lathi & Roger Green. (2018). Chapter 4: The Laplace Transform. Signal Processing and Linear Systems (3rd ed., pp. 386-401). Oxford University Press.

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.

Figure 4.18a

Large systems can be represented by suitably interconnected subsystems, each of which can be readily analyzed. Each subsystem can be characterized in terms of its input–output relationships.

Subsystems may be interconnected by using:

  1. cascade,
  2. parallel, and
  3. feedback

interconnections.

The cascade case

When transfer functions appear in cascade, as depicted in Fig. 4.18b, the transfer function of the overall system is the product of the two transfer functions. \[ \frac{Y(s)}{X(s)}=\frac{W(s)}{X(s)} \frac{Y(s)}{W(s)}=H_1(s) H_2(s) \] Figure 4.18b

The parallel case

Similarly, when two transfer functions, \(H_1(s)\) and \(H_2(s)\), appear in parallel, as illustrated in Fig. 4.18c, the overall transfer function is given by \(H_1(s)+H_2(s)\).

Figure 4.18c

The feadback case

Figure 4.18d

When the output is fed back to the input, as shown in Fig. 4.18d, the overall transfer function \(T(s)\) is

\[ T(s)=\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s) H(s)} \]

The feedback system is also called the close-loop system, as shown in Fig. 4.33b, whereas the system without feedback is called the open-loop system, as shown in Fig. 4.33a.

Fig. 4.33a

Derivation of transfer function

The inputs to the adder are \(X(s)\) and \(-H(s) Y(s)\). Therefore, \(E(s)\), the output of the adder, is: \[ E(s)=X(s)-H(s) Y(s) \]

Since:

\[ \begin{aligned} Y(s) & =G(s) E(s) \\ & =G(s)[X(s)-H(s) Y(s)] \end{aligned} \]

Rearranging terms gives:

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

Thus, the transfer function \(\frac{Y(s)}{X(s)}\) is:

\[ \frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s) H(s)} \] 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 \[ \begin{aligned} Y(s) & =G(s) E(s) \\ & =G(s)[X(s)-H(s) Y(s)] \end{aligned} \]

Therefore, \[ Y(s)[1+G(s) H(s)]=G(s) X(s) \] so that \[ \frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s) H(s)} \]