Question 6.3.5: The circuit shown in Figure 6.3.9a consists of two series RC......
The circuit shown in Figure 6.3.9a consists of two series RC circuits wired so that the output voltage of the first circuit is the input voltage to an isolation amplifier. The output voltage of the amplifier is the input voltage to the second RC circuit. The amplifier has a voltage gain G; that is, v_{2}(t)\,\,=\,G v_{1}(t). Derive the transfer function V_{o}(s)/V_{s}(s) for this circuit, and for the case G = 1 compare it with the transfer function of the circuit shown in Figure 6.3.1.
The amplifier isolates the first RC loop from the effects of the second loop; that is, the amplifier prevents the voltage v_{1} from being affected by the second RC loop. This in effect creates two separate loops with an intermediate voltage source v_{2}=G v_{1}, as shown in Figure 6.3.9b. Thus, for the left-hand loop we obtain
{\frac{V_{1}(s)}{V_{s}(s)}}={\frac{1}{R C s+1}}For the right-hand RC loop,
\frac{V_{o}(s)}{V_{2}(s)}=\frac{1}{R C s+1}For the amplifier with gain G,
V_{2}(s)=G\,V_{1}(s)To obtain the transfer function V_{o}(s)/V_{s}(s), eliminate the variables V_{\mathrm{1}}(s) and V_{\mathrm{2}}(s) from these equations as follows:
\frac{V_{o}(s)}{V_{s}(s)}=\frac{V_{o}(s)}{V_{2}(s)}\frac{V_{2}(s)}{V_{1}(s)}\frac{V_{1}(s)}{V_{s}(s)}=\frac{1}{R C s+1}G\frac{1}{R C s+1}=\frac{G}{R^{2}C^{2}s^{2}+2R C s+1} (1)
This procedure is described graphically by the block diagram shown in Figure 6.3.10a. The three blocks can be combined into one block as shown in part (b) of the figure.
The transfer function of the circuit shown in Figure 6.3.1 was derived in Example 6.3.2. It is
\frac{V_{o}(s)}{V_{s}(s)}=\frac{1}{R^{2}C^{2}s^{2}+3R C s+1} (2)
Note that it is not the same as the transfer function given by equation (1) with G = 1.
A common mistake is to obtain the transfer function of loops connected end-to-end by multiplying their transfer functions. This is equivalent to treating them as independent loops, which they are not, because each loop “loads” the adjacent loops and thus changes the currents and voltages in those loops. An isolation amplifier prevents a loop from loading an adjacent loop, and when such amplifiers are used, we can multiply the loop transfer functions to obtain the overall transfer function.
This mistake is sometimes made when drawing block diagrams. The circuit of Figure 6.3.9 can be represented by the block diagram of Figure 6.3.10, where the transfer function of each block can be multiplied to obtain the overall transfer function V_{o}(s)/V_{i}(s). However, the circuit of Figure 6.3.1 cannot be represented by a simple series of blocks because the output voltage \mathbf{}v_{o} affects the voltage v_{1}\,. To show this effect requires a feedback loop, as shown previously in Figure 6.3.2.
In general, even though elements are physically connected end-to-end, we cannot represent them by a series of blocks if the output of one element affects its input or the behavior of any preceding elements.
