Question 19.4: Determine the y parameters for the two-port shown in Fig. 19......

We follow the same procedure as in the previous example. To get y_{11}   and y_{21}   , we use the circuit in Fig. 19.18(a), in which port 2 is short-circuited and a current source is applied to port 1. At node 1,

\frac{V_{1}  –  V_{o}}{8} = 2 I_{1} + \frac{V_{o}}{2}  +  \frac{V_{o}  –  0}{4}

But I_{1} = \frac{V_{1}   –  V_{o}}{8}   ; therefore ,

0 = \frac{V_{1}  –  V_{o}}{8}  +  \frac{3V_{o}}{4}

0 =  V_{1}  –  V_{o}  +  6V_{o}        ⇒          V_{1}  =  – 5V_{o}

Hence,

I_{1}  = \frac{ -5V_{o}  –   V_{o}}{8}  = –  0.75V_{o}

and

y_{11} = \frac{I_{1}}{V_{1}} = \frac{-0.75 V_{o}}{-5 V_{o}} = 0.15  S

At node 2,

\frac{V_{o}  –   0 }{4}  + 2I_{1}  +  I_{2} = 0

or

-I_{2} =  0.25V_{o}  –  1.5V_{o}  =  – 1.25V_{o}

Hence,

y_{21}  =  \frac{I_{2}}{V_{1}} = \frac{ 1.25V_{o}}{- 5V_{o}}=  – 0.25 S

Similarly, we get  y_{12}  and  y_{22}   using Fig. 19.18(b). At node 1,

\frac{0  –  V_{o}}{8}  =  2I_{1}  + \frac{V_{o}}{2}  +  \frac{V_{o}  – V_{2}}{4}

But I_{1} =  \frac{0  –  V_{o}}{8}   ; therefore,

0 =  – \frac{V_{o}}{8} +  \frac{V_{o}}{2}  + \frac{V_{o}  –  V_{2}}{4}

or

0 =  – V_{o}  +  4V_{o}  +  2V_{o}  –  2V_{2}          ⇒          V_{2} =  2.5V_{o}

Hence,

y_{12}  =  \frac{I_{1}}{V_{2}} = \frac{- V_{o}/8}{2.5V_{o}}= –  0.05  S

At node 2,

\frac{V_{o}  –  V_{2}}{4}  + 2I_{1}  +  I_{2}  = 0

or

-I_{2} =  0.25V_{o}  –  \frac{1}{4} (2.5V_{o})  –  \frac{2V_{o}}{8}  = –  0.625V_{o}

Thus,

y_{22}  =  \frac{I_{2}}{V_{2}} = \frac{ 0.625V_{o}}{2.5V_{o}} =  0.25  S

Notice that y_{12}  ≠ y_{21}   in this case, since the network is not reciprocal.