Question

For the two port network shown below, the [Y]-parameters is given as [Y] = (1/100) * [[2 -1], [-1 4/3]] S. The value of load impedance Z_L, in ohms, for maximum power transfer will be _____________ (rounded off to the nearest integer).

Hint:

With $Y$-parameters, terminating the other port by an admittance $Y_T$ gives $Y_{\text{seen}}=Y_{22}-\dfrac{Y_{21}Y_{12}}{Y_{11}+Y_T}$. Set $Z_L=1/Y_{\text{seen}}$ for maximum power (real case).

Answer 1

With the source killed, the $120$ V source becomes a short, so port~1 is terminated by its series resistance $R_S=10\,\Omega \Rightarrow Y_S=\frac{1}{R_S}=0.1\,\text{S}$.
For a two–port with $Y$-parameters and port~1 terminated by $Y_S$, the driving-point admittance seen at port~2 is
\[ Y_{\text{out}}=Y_{22}-\frac{Y_{21}Y_{12}}{Y_{11}+Y_S}. \] Given $Y_{11}=\frac{2}{100}=0.02$, $Y_{12}=Y_{21}=-\frac{1}{100}=-0.01$, $Y_{22}=\frac{4/3}{100}=0.013\overline{3}\ \text{S}$.
Compute: $Y_{11}+Y_S=0.02+0.1=0.12$, $Y_{21}Y_{12}=(-0.01)(-0.01)=10^{-4}$.
Hence \[ Y_{\text{out}}=0.013\overline{3}-\frac{10^{-4}}{0.12} =0.013\overline{3}-0.000\overline{83}=0.0125\ \text{S}. \] Therefore $Z_{\text{th}}=\dfrac{1}{Y_{\text{out}}}=80\,\Omega$.
For maximum power transfer, $Z_L=Z_{\text{th}}$ (all-real network).
\[ \boxed{Z_L = 80~\Omega} \]