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  f = ω/2π = 0.15915 MHz,   because L and C are frequency dependent. In Eq. (19.3), we defined the z parameters as

\begin{matrix}z_{11}  =  \frac{V_{1}}{I_{1}} \mid_{I_{2} = 0}  ,        z_{12}  =  \frac{V_{1}}{I_{2}} \mid_{I_{1} = 0} \\\\ z_{21} =  \frac{V_{2}}{I_{1}} \mid_{I_{2} = 0} ,    z_{22}  =  \frac{V_{2}}{I_{2}} \mid_{I_{1} = 0}\end{matrix}\quad (19.3) \\

z_{11} = \frac{V_1}{I_1}|_{I_2=0} ,\quad z_{21} = \frac{V_2}{I_1}|_{I_2=0}

This suggests that if we let I_{1} = 1  A   and open-circuit the output port so that  I_{2} = 0  , then we obtain

z_{11}  =  \frac{V_{1}}{1}        and          z_{21}  =  \frac{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 V_{1}  and  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 V_{1}  and  V_{2}   from the output file. Thus,

z_{11}  = \frac{V_{1}}{1}  =  19.70 \underline{/175.7°} Ω ,       z_{21}  =  \frac{V_{2}}{1}  =  19.79 \underline{/170.2° }  Ω

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

z_{12} =  \frac{V_{1}}{I_{2}} \mid_{I_{1} = 0} ,    z_{22}  =  \frac{V_{2}}{I_{2}} \mid_{I_{1} = 0}

suggesting that if we let  I_{2} = 1 A and open-circuit the input port,

z_{12}  = \frac{V_{1}}{1}       and      z_{22}  =  \frac{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 IAC is now at the output terminal. We run the schematic in Fig. 19.53(b) and obtain V_{1}   and  V_{2}   from the output file. Thus,

z_{12}  = \frac{V_{1}}{1}  =  19.70 \underline{/175.7°} Ω ,       z_{22}  =  \frac{V_{2}}{1}  =  19.56 \underline{/175.7° }  Ω