Question 19.18: Design the LC ladder network terminated with 1 - Ω a resisto......

The denominator shows that this is a third-order network, so that the LC ladder network is shown in Fig. 19.63(a), with two inductors and one capacitor. Our goal is to determine the values of the inductors and capacitor. To achieve this, we group the terms in the denominator into odd or even parts:

D(s)  =  (s³  +   2s)  +  (2s²  +  1 )

so that

H(s) = \frac{1}{(s³  +  2s)  +  (2s²  +  1)}

Divide the numerator and denominator by the odd part of the denominator to get

H(s) = \frac{\frac{1}{s³  +  2s}}{1  +  \frac{2s²  +  1}{s³  +  2s}}              (19.18.1)

From Eq. (19.82), when Y_{L} = 1 ,

H(s) = – \frac{y_{21} /Y_{L}}{1  +  y_{22} / Y_{L}}                (19.82)

H(s) = \frac{-  y_{21}}{1  +   y_{22}}                      (19.18.2)

Comparing Eqs. (19.19.1) and (19.19.2), we obtain

y_{21}  = –  \frac{1}{s^{3}  +   2s}  ,        y_{22}  = \frac{ 2s^{2}  +   1}{s^{3}  +   2s}

Any realization of  y_{22} will automatically realize  y_{21} since y_{22}   is the output driving-point admittance, that is, the output admittance of the network with the input port short-circuited. We determine the values of L and C in Fig. 19.63(a) that will give us  y_{22} .  Recall that y_{22}   is the short-circuit output admittance. So we short-circuit the input port as shown in Fig. 19.63(b). First we get L_{3}   by letting

Z_{A} = \frac{1}{y_{22}} = \frac{s^{3}  +  2s}{2s^{2}  +  1} = sL_{3}  +  Z_{B}                         (19.18.3)

By long division,

Z_{A}   =  0.5 s  +  \frac{1.5s}{2s^{2}  +  1}                     (19.18.4)

Comparing Eqs. (19.18.3) and (19.18.4) shows that

L_{3}   =  0.5 H    ,           Z_{B}  = \frac{1.5s}{2s^{2}  +  1}

Next, we seek to get  C_{2} as in Fig. 19.63(c) and let

Y_{B}  = \frac{1}{Z_{B}} = \frac{2s^{2}  +  1}{1.5 s}  =  1.333 s  + \frac{1}{1.5 s}  =  s C_{2}  +  Y_{C}

from which  C_{2} = 1.33 F   and

Y_{C}  =  \frac{1}{1.5s} =  \frac{1}{sL_{1}}        ⇒     L_{1}  =  1.5  H

Thus, the LC ladder network in Fig. 19.63(a) with L_{1}  =  1.5  H  and  C_{2} = 1.33 F and L_{3} = 0.5  H    has been synthesized to provide the given transfer function H(s) .  This result can be confirmed by finding H(s) = V_{2} /V_{1}    in Fig. 19.63(a) or by confirming the required y_{21}    .