Question

The input impedance of a \(\frac{\lambda }{8}\) section of a lossless transmission line of characteristic impedance 50 Ω is found to be real when the other end is terminated by a load Z_L (= R + jX)Ω. If X is 30Ω, the value of R (in Ω) is ______

Answer 1

Concept:

The impedance of a transmission line at a distance ‘l’ from the load is given as:

\(Z\left( \ell \right) = {Z_0}\left( {\frac{{{Z_L} + j{Z_0}\tan \beta \ell }}{{{Z_0} + j{Z_L}\tan \beta \ell }}} \right)\) 

Z₀ = characteristic impedance of the transmission line

Z_L = Load impedance

β = phase length defined as:

\(\beta = \frac{{2\pi }}{\lambda }\) 

Analysis:

With \(\ell = \frac{\lambda }{8}\)

\(\beta \ell = \frac{{2\pi }}{\lambda } \times \frac{\lambda }{8}\) 

\(\beta \ell = \frac{\pi }{4}\) 

The impedance of the transmission line will be:

\(Z\left( {\frac{\lambda }{8}} \right) = {Z_0}\left( {\frac{{{Z_L}\; +\; j{Z_0}\tan \left( {\frac{\pi }{4}} \right)}}{{{Z_0}\; +\; j{Z_L}\tan \left( {\frac{\pi }{4}} \right)}}} \right)\) 

\(Z\left( {\frac{\lambda }{8}} \right) = {Z_0}\left( {\frac{{{Z_L} \;+ \;j{Z_0}}}{{{Z_0}\; +\; j{Z_L}}}} \right)\) 

With Z_L = (R + j30)Ω and Z₀ = 50Ω, we get:

\(Z\left( {\frac{\lambda }{8}} \right) = 50\left( {\frac{{R\; + \;j30\; +\; j50}}{{50\; +\; j\left( {R\; +\; j30} \right)}}} \right)\) 

\(= \frac{{50\left( {R\; + \;j80} \right)}}{{20\; + \;jR}}\) 

\( = \frac{{50\left( {R\; + \;j80} \right)\left( {20\; -\; jR} \right)}}{{\left( {20\; + \;jR} \right)\left( {20\; - \;jR} \right)}}\) 

\(= \frac{{50\left( {R\; +\; j80} \right)\left( {20\; -\; jR} \right)}}{{400\; -\; {R^2}}}\) 

\(= \frac{{50}}{{400\; + \;{R^2}}}\left( {20R + 80R + 1600j - j{R^2}} \right)\) 

For Z_in to be real, the imaginary part must be equal to 0, i.e.

1600 j – jR² = 0

1600 = R²

R = 40 Ω