Question 10.1: A series–parallel circuit is shown in Fig. 10.4. Calculate t......
A series–parallel circuit is shown in Fig. 10.4. Calculate the current gain \frac{I_{0}(\omega)}{I_{i(\omega)}} and its zeros and poles.ω
The output current can be determined as,
I_{0}(\omega)=I_{i}(\omega)\times\frac{(1+j2\omega)}{1+j2\omega+2+\frac{1}{j0.5\omega}} (10.12)
\frac{I_{0}(\omega)}{I_{i}(\omega)}=\frac{j0.5\omega(1+j2\omega)}{(1+j2\omega+2)j0.5\omega+1} (10.13)
Considering s = jω, Eq. (10.13) can be modified as,
\frac{I_{0}(\omega)}{I_{i}(\omega)}=\frac{0.5s(1+2s)}{(1+2s+2)0.5s+1} (10.14)
{\frac{I_{0}(\omega)}{I_{i}(\omega)}}={\frac{s(s+0.5)}{s^{2}+1.5s+1}} (10.15)
Setting the numerator of Eq. (10.15) equal to zero yields,
s(s+0.5)=0 (10.16)
z_{1}=0,\;z_{2}=-0.5 (10.17)
Setting the denominator of Eq. (10.15) equal to zero yields,
s^{2}+1.5s+1=0 (10.18)
s_{1,2}={\frac{-1.5\pm{\sqrt{1.5^{2}-4}}}{2}} (10.19)
p_{1}=-0.75+j0.66,\;p_{2}=-0.75-j0.66 (10.20)