0
358views
Explain $Y$ parameter in terms in $Z$ parameter
1 Answer
0
6views

Solution:

$ \begin{aligned} \\ &\text { Equations of } z \text {-parameter are given by }\\\\ &V_1(s)=Z_{11} I_1(S)+Z_1 2 \cdot I_2(S)...(1)\\\\ &V_2(s)=Z_{21} I_1(s)+Z_{22} \cdot I_2(s)...(2)\\\\ &\Delta Z=\left|\begin{array}{ll} Z_{11} & Z_{12} \\\\ Z_{21} & Z_{22} \\ \end{array}\right|=Z_{11} \cdot Z_{22}-Z_{12} \cdot Z_{21}...(3)\\\\ &\Delta_1=\left|\begin{array}{ll} V_1(s) & Z_{12} \\\\ V_2(s) & Z_{22} \\ \end{array}\right|=Z_{22} V_1(s)-Z_{12} V_2(s)...(4) \\ \end{aligned} \\ $

$ \therefore I_1(S)=\frac{\Delta 1}{\Delta Z}=\frac{Z_{22} V_1(S)-Z_{12} V_2(S)}{\Delta Z} ...(5)\\ $

$ \begin{aligned} \\ &I_1(s)=\frac{Z_{22}}{\Delta z} V_1(s)-\frac{Z_{12}}{\Delta z} V_2(s)...(6) \\\\ &I_2(s)=\frac{\Delta z}{\Delta}=\frac{-Z_{21} V_1(s)+Z_{11} V_2(s)}{\Delta z} \\\\ &I_2(s)=-\frac{Z_{21}}{\Delta z} V_1(s)+\frac{Z_{11}}{\Delta z} V_2(s)....(7) \\\\ &\text { from } e q-n(6) \&(7), y-\text { parameters are } \\\\ &{\left[\begin{array}{ll} Y_{11} & Y_{12} \\\\ Y_{21} & Y_{22} \\ \end{array}\right]=\left[\begin{array}{cc} \frac{Z_{22}}{\Delta z} & -\frac{Z_{12}}{\Delta z} \\\\ \frac{Z_{21}}{\Delta z} & \frac{Z_{11}}{\Delta z} \\ \end{array}\right]} \\ \end{aligned} \\ $

Please log in to add an answer.