Application of The Laplace Transform to the Network Synthesis

Electrical Circuit Analysis > The Laplace Transform

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Network synthesis may be regarded as the process of obtaining an appropriate network to represent a given transfer function. Network synthesis is easier in the $ s $ domain than in the time domain.

In network analysis, we find the transfer function of a given network. In network synthesis, we reverse the approach: given a transfer function, we are required to find a suitable network.

Network synthesis is finding a network that represents a given transfer function.

Keep in mind that in synthesis, there may be many different answers-or possibly no answers - because there are many circuits that can be used to represent the same transfer function; in network analysis, there is only one answer.

Network synthesis is an exciting field of prime engineering importance. Being able to look at a transfer function and come up with the type of circuit it represents is a great asset to a circuit designer. Although network synthesis constitutes a whole course by itself and requires some experience, the following examples are meant to whet your appetite.

Example 1: Given the transfer function

$$H(s)=\frac{V_{o}(s)}{V_{i}(s)}=\frac{10}{s^{2}+3 s+10}$$

realize the function using the circuit in Fig. 1(a).
(a) Select $ R=5 \Omega $, and find $ L $ and $ C $.
(b) Select $ R=1 \Omega $, and find $ L $ and $ C $.

Fig. 1: Example 1.

Solution: The $ s $-domain equivalent of the circuit in Fig. 1(a) is shown in Fig. 1(b). The parallel combination of $ R $ and $ C $ gives

$$R \| \frac{1}{s C}=\frac{R / s C}{R+1 / s C}=\frac{R}{1+s R C}$$

Using the voltage division principle,

$$V_{o}=\frac{R /(1+s R C)}{s L+R /(1+s R C)} V_{i}=\frac{R}{s L(1+s R C)+R} V_{i}$$

$$\frac{V_{o}}{V_{i}}=\frac{R}{s^{2} R L C+s L+R}=\frac{1 / L C}{s^{2}+s / R C+1 / L C}$$

Comparing this with the given transfer function $ H(s) $ reveals that

$$\frac{1}{L C}=10, \quad \frac{1}{R C}=3$$

There are several values of $ R, L $, and $ C $ that satisfy these requirements. This is the reason for specifying one element value so that others can be determined.

(a) If we select $ R=5 \Omega $, then

$$C=\frac{1}{3 R}=66.67 \mathrm{mF}, \quad L=\frac{1}{10 C}=1.5 \mathrm{H}$$

(b) If we select $ R=1 \Omega $, then

$$C=\frac{1}{3 R}=0.333 \mathrm{~F}, \quad L=\frac{1}{10 C}=0.3 \mathrm{H}$$

Making $ R=1 \Omega $ can be regarded as normalizing the design.

In this example we have used passive elements to realize the given transfer function. The same goal can be achieved by using active elements, as the next example demonstrates.

Example 2: Synthesize the function

$$T(s)=\frac{V_{o}(s)}{V_{s}(s)}=\frac{10^{6}}{s^{2}+100 s+10^{6}}$$

using the topology in Fig. 2.

Fig. 2: Example 2.

Solution:
We apply nodal analysis to nodes 1 and 2 . At node 1 ,

$$\left(V_{s}-V_{1}\right) Y_{1}=\left(V_{1}-V_{o}\right) Y_{2}+\left(V_{1}-V_{2}\right) Y_{3}$$

At node 2,

$$\left(V_{1}-V_{2}\right) Y_{3}=\left(V_{2}-0\right) Y_{4}$$

But $ V_{2}=V_{o} $, so Eq. (15.26.1) becomes

$$Y_{1} V_{s}=\left(Y_{1}+Y_{2}+Y_{3}\right) V_{1}-\left(Y_{2}+Y_{3}\right) V_{o}$$

and Eq. (15.26.2) becomes

$$V_{1} Y_{3}=\left(Y_{3}+Y_{4}\right) V_{o}$$

$$V_{1}=\frac{1}{Y_{3}}\left(Y_{3}+Y_{4}\right) V_{o}$$

Substituting Eq. (15.26.4) into Eq. (15.26.3) gives

$$Y_{1} V_{s}=\left(Y_{1}+Y_{2}+Y_{3}\right) \frac{1}{Y_{3}}\left(Y_{3}+Y_{4}\right) V_{o}-\left(Y_{2}+Y_{3}\right) V_{o}$$

$$Y_{1} Y_{3} V_{s}=\left[Y_{1} Y_{3}+Y_{4}\left(Y_{1}+Y_{2}+Y_{3}\right)\right] V_{o}$$

Thus,

$$\frac{V_{o}}{V_{s}}=\frac{Y_{1} Y_{3}}{Y_{1} Y_{3}+Y_{4}\left(Y_{1}+Y_{2}+Y_{3}\right)}$$

To synthesize the given transfer function $ T(s) $, compare it with the one in Eq. (15.26.5). Notice two things: (1) $ Y_{1} Y_{3} $ must not involve $ s $ because the numerator of $ T(s) $ is constant; (2) the given transfer function is second-order, which implies that we must have two capacitors. Therefore, we must make $ Y_{1} $ and $ Y_{3} $ resistive, while $ Y_{2} $ and $ Y_{4} $ are capacitive. So we select

$$Y_{1}=\frac{1}{R_{1}}, \quad Y_{2}=s C_{1}, \quad Y_{3}=\frac{1}{R_{2}}, \quad Y_{4}=sC_2$$

Substituting Eq. (15.26.6) into Eq. (15.26 .5) gives

$$\begin{split} \frac {V_{0}}{V_{s}}&=\frac {1 / \left(R_{1} R_{2} \right)}{1 /\left(R_{1} R_{2}\right)+s C_{2}\left(1 / R_{1}+1 / R_{2}+s C_{1}\right)} \\ &=\frac{1 /\left(R_{1} R_{2} C_{1} C_{2}\right)}{s^{2}+s\left(R_{1}+R_{2}\right) /\left(R_{1} R_{2} C_{1}\right)+1 /\left(R_{1} R_{2} C_{1} C_{2}\right)} \end{split}$$

Comparing this with the given transfer function $ T(s) $, we notice that

$$\frac{1}{R_{1} R_{2} C_{1} C_{2}}=10^{6}, \quad \frac{R_{1}+R_{2}}{R_{1} R_{2} C_{1}}=100$$

If we select $ R_{1}=R_{2}=10 \mathrm{k} \Omega $, then

$$\begin{array}{c}C_{1}=\frac{R_{1}+R_{2}}{100 R_{1} R_{2}}=\frac{20 \times 10^{3}}{100 \times 100 \times 10^{6}}=2 \mu \mathrm{F} \\C_{2}=\frac{10^{-6}}{R_{1} R_{2} C_{1}}=\frac{10^{-6}}{100 \times 10^{6} \times 2 \times 10^{-6}}=5 \mathrm{nF}\end{array}$$

Thus, the given transfer function is realized using the circuit shown in Fig. 3.

Fig. 3: For Example 1.

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