Question 19.16: Find the z parameters for the circuit in Fig. 19.52 at ω = 1...

Notice that we used dc analysis in Example 19.15 because the circuit in Fig. 19.49 is purely resistive. Here, we use ac analysis at ƒ = ω/2π = 0.15915 MHz, because L and C are frequency dependent.
In Eq. (19.3), we defined the z parameters as

(19.3):           $\pmb{z _{11}}=\left.\pmb{\frac{ V _1}{ I _1}}\right|_{ \pmb{I _2}=0}, \quad \pmb{z _{12}}=\left.\pmb{\frac{ V _1}{ I _2}}\right|_{ \pmb{I _1}=0} \\ \pmb{z _{21}}=\left.\pmb{\frac{ V _2}{ I _1}}\right|_{ \pmb{I _2}=0}, \quad \pmb{z _{22}}=\left.\pmb{\frac{ V _2}{ I _2}}\right|_{ \pmb{I _1}=0}$

$\pmb{z _{11}}=\left.\pmb{\frac{ V _1}{ I _1}}\right|_{\pmb{ I _2}=0}, \quad \pmb{z _{21}}=\left.\pmb{\frac{ V _2}{ I _1}}\right|_{ \pmb{I _2}=0}$

This suggests that if we let $\pmb{I_1} = 1  A$ and open-circuit the output port so that $\pmb{I_2} = 0$, then we obtain

$\pmb{z _{11}}=\frac{\pmb{ V _1}}{1} \quad \text { and } \quad \pmb{z _{21}}=\frac{ \pmb{V _2}}{1}$

We realize this with the schematic in Fig. 19.53(a). We insert a 1-A ac current source IAC at the input terminal of the circuit and two VPRINT1 pseudocomponents to obtain $\pmb{V_1} \text{ and } \pmb{V_2}$. The attributes of each VPRINT1 are set as AC = yes, MAG = yes, and PHASE = yes to print the magnitude and phase values of the voltages. We select Analysis/Setup/AC Sweep and enter 1 as Total Pts, 0.1519MEG as Start Freq, and 0.1519MEG as Final Freq in the AC Sweep and Noise Analysis dialog box. After saving the schematic, we select Analysis/Simulate to simulate it. We obtain $\pmb{V_1} \text{ and } \pmb{V_2}$ from the output file. Thus,

$\pmb{z _{11}}=\frac{ \pmb{V _1}}{1}=19.70 \underline{/175.7^{\circ}}  \Omega, \quad \pmb{z _{21}}=\frac{ \pmb{V _2}}{1}=19.79 \underline{/170.2^{\circ}}  \Omega$

In a similar manner, from Eq. (19.3),

$\pmb{z _{12}}=\left.\pmb{\frac{ V _1}{ I _2}}\right|_{ \pmb{I _1}=0}, \quad \pmb{z _{22}}=\left.\pmb{\frac{ V _2}{ I _2}}\right|_{ \pmb{I _1}=0}$

suggesting that if we let $\pmb{I_2} = 1  A$ and open-circuit the input port,

$\pmb{z _{12}}=\frac{ \pmb{V _1}}{1} \quad \text { and } \quad \pmb{z _{22}}=\frac{ \pmb{V _2}}{1}$

This leads to the schematic in Fig. 19.53(b). The only difference between this schematic and the one in Fig. 19.53(a) is that the 1-A ac current source IA C is no w at the output terminal. We run the schematic in Fig. 19.53(b) and obtain $\pmb{V_1} \text{ and } \pmb{V_2}$ from the output file. Thus,

$\pmb{z _{12}}=\frac{ \pmb{V _1}}{1}=19.70 \underline{/175.7^{\circ}}  \Omega, \quad \pmb{z _{22}}=\frac{ \pmb{V _2}}{1}=19.56 \underline{/175.7^{\circ}}  \Omega$