Two Port Network And Network Functions – GATE EE Electric CircuitsPractice Questions & PYQs

From $Z_{p}$ matrix

Convert Z-parameters in to ABCD parameters

From equation (ii),

Sub equation (iii) in equation (i),

From equation (iii) and (iv),

Convert $Y$ -parameters to $\mathrm{ABCD}$ parameters From equation (vi),

Sub equation (vii) in equation ( $\mathrm{v}$ ),

From equation (vii) and equation (viii),

$Y =[Y]_{A}+[Y]_{B}$

Put $I_{2}=0$ and $V_{1}=20 \mathrm{~V}$ $0=-\frac{8.5}{3} \times 20+\frac{4}{3} V_{2}$

Equivalent circuit $I=\frac{42.5}{\frac{3}{4}+6}=6.296 \mathrm{~A}$

By KCL, $\frac{V_a-1}{1}+\frac{V_a}{1}+\frac{V_a+2I_1}{1}=0$ $3V_a+2I_1=1\;\;...(i)$ $I_1=\frac{1-V_a}{1}\;\;...(ii)$ Substitute equation (ii) in equation (i) [/latex] $V_a=-1$ $I_1=\frac{1-V_a}{1}=\frac{1-(-1)}{1}=2$ $h_{11}=\frac{V_1}{I_1}=\frac{1}{2}=0.5\Omega$

For two port network, we can write,

$A=A_1A_2+B_1C_2\;\;...(i)$ $B=A_1B_2+B_1D_2\;\;...(ii)$ $V_1=AV_2-BI_2\;\;...(iii)$ To get, $V_T(I_2=0)$ ; from equation (iii) $V_2=V_T=\frac{V_1}{A}=\frac{V_1}{A_1A_2+B_1C_2}$ To get $Z_T(V_1=0)$ ; from equation (iii) $V_1=AV_2-BI_2$ $0=AV_2-BI_2$ $Z_T=\frac{V_2}{I_2}=\frac{B}{A}=\frac{A_1B_2+B_1D_2}{A_1A_2+B_1C_2}$

Given, $Z_{11}=40\Omega ,Z_{12}=60\Omega$ $Z_{21}=80\Omega ,Z_{22}=100\Omega$ From the figure, $V_2=-20I_2\;\;...(i)$ $V_1=40I_1+60I_2\;\;...(ii)$ $V_2=80I_1+100I_2\;\;...(iii)$ From equation (i) and (iii), we get $-20I_2=80I_1+100I_2$ $\Rightarrow \; I_2=-\frac{2}{3}I_1\;\;...(iv)$ Using equation (ii) and (iv), we get $V_1=40I_1+60I_2$ $\;\;=40I_1+60\left ( \frac{-2}{3} I_1\right )$ $V_1=0$ From the figure, $20=10I_1+V_1$ Since, $V_1=0$ So, $I_1=2A$ $I_2=\frac{-4}{3}A$ Power dissipated in $P_{R_L}=I_2^2R_L=\left ( \frac{4}{3} \right )^2 \times 20$ $\;\;=\frac{16}{9} \times 20=35.55W$

To find impedance seen by $V_s$ $Z_s=\frac{V_s}{I_s}$ $V_1=2I_s$ Applying KCL at node A, $I_s=4V_1=\frac{V_A}{3}+\frac{V_A}{6}$ $V_A=V_s-V_1$ and $V_1=2I_s$ So, $I_s+8I_s=\frac{V_s-2I_s}{3}+\frac{V_s-2I_s}{6}$ $\Rightarrow \; 54I_s=2V_s-4I_s+V_s-2I_s$ $3V_s=60I_s$ $\frac{V_s}{I_s}=20\Omega$

Driving point impedance, Z(s) is , $Z(s)=s+\left ( \frac{\left ( s+\frac{1}{s} \right )\times \frac{1}{s}}{s+\frac{1}{s}+\frac{1}{s}} \right )$ $\;\;=s+\left ( \frac{s^{2+}1}{s^2} \right )\times \frac{s}{s^{2+}2}$ $\;\;=s+\frac{s^{2+}1}{s(s^{2+}2)}$ $\;\;=\frac{s^2(s^{2+}2)+^{2+}1}{s^{3+}2s}$ $Z(s)=\frac{s^{4+}3s^{2+}1}{s^{3+}2s}$

$(i)$ $V_1=10V,V_2=3V, I_2=-3A$ $V_1=AV_2-BI_2$ $10=3A+3B$ $(ii) V_2=5V, I_2=-2A$ $10=5A+2B$ $A=\frac{10}{9}$ $B=\frac{20}{9}$ Given, $V_1=8V$ $(V_2)_{OC}=?$ $I_2=0$ $V_1=AV_2-BI_2$ $8=A(V_2)_{OC}-0$ $(V_2)_{OC}=\frac{8}{A}=\frac{8}{10/9}=7.2V$

$V_1^{ab}=Z_{11}I_1+Z_{12}I_2$ $V_2=Z_{21}I_1+Z_{22}I_2$ As, $1\Omega$ resistor is connected in series with the network as port-1. $V_2$ does not get affected, $V_1^{ef}=V_1^{ab}+I_1 \times 1$ $\;\;=Z_{11}I_1+Z_{12}I_2+I_1$ $\;\;=(Z_{11}+1)I_1+Z_{12}I_2$ Modified Z-parameter

$I_1=g_{11}V_1+g_{12}I_2$ $V_2=g_{21}V_1+g_{22}I_2$ Since port-1 is open circuit, $I_1=0$ Port-2 is short-circuit, $V_2=0$

So, g-parameters,

Output voltage $V_0=AV_i$ $V_0=10^6 \times 1 \times 10^{-6}=1V$ To calculate input impedance, $V_{dc}$ source is connected at input port Input impedance $Z_i=\frac{V_{dc}}{I_i}$ as loop is not closed, $I_i=0$ So, $Z_i=\frac{V_{dc}}{0}=\infty$ To calculate output impedance, $V_{dc}$ source is connected at output port. Output impedance, $Z_o=\frac{V_{dc}}{I_o}=\frac{I_oR_o+AV_i}{I_o}$ As $V_i=0 Z_o=R_o=10\Omega$

$V_1=AV_2+BI_2$ $I_1=CV_2+DI_2$ $V_2=Z_2(I_1+I_2)\;\;...(i)$ $C=\frac{I_1}{V_2}|_{I_2=0}$ Putting $I_2=0$ in equation (i) $V_2=Z_2I_1$ $Z_2=\frac{V_2}{I_1}|_{I_2=0}$ $\;\;=\frac{1}{C}=\frac{1}{0.025\angle 45^{\circ}}$ $Z_2=40\angle -45^{\circ}\Omega$

$v_1=(i_1+i_2)Z_1 =Z_1i_1+Z_1i_2$ $v_2=Z_2i_1+Z_1(i_1+i_2) =Z_1i_1+(Z_1+Z_2)i_2$ From above, we can write,

$y_{22}=1.8$

$h_{12}=\frac{E_1}{E_2}|_{I_1=0}$ $h_{12}$ is ratio of $E_1 \; to \; E_2$ for the input open-circuited condition. Assuming $I_1=0$ $I_2=\frac{E_2}{2+(2+4)||4}=\frac{E_2}{4}$ $I_x=\frac{I_2}{(2+2)+4 }\times 4 =\frac{I_2}{2}=\frac{1}{2}\left ( \frac{E_2}{4} \right )=\frac{E_2}{8}$ $E_1=2I_x=2\frac{E_2}{8}=\frac{E_2}{4}$ $\Rightarrow \;\; \frac{E_1}{E_2}=0.25$

Using KVL, $E_1=2I_1+2(I_1+I_2)$ Again using KVL, $E_2=2I_2+2(I_1+I_2)$

$[y]=[z]^{-1}$

For a passive two port network, output powe can never be grater than input power.