Respuesta :
Answer:
The load reflection coefficient, [tex]\Gamma =0.62\angle 82.875^{\circ} \Omega[/tex]
Reflection coefficient at input, [tex]\Gamma = 0.62\angle - 147.518^{\circ} \Omega[/tex]
SWR = 4.26
Given:
Characteristic impedance of the co-axial cable, [tex]Z_{c} = 75 \Omega[/tex]
Length of the cable, L = 2.0 cm = 0.02 m
[tex]Z_{Load} = 37.5 + j75 \Omega[/tex]
Dielectric constant, K = 2.56
frequency, f = 3.0 GHz = [tex]3.0 \times 10^{9} Hz[/tex]
Explanation:
In order to calculate the reflection coefficient at load, we first calculate these:
The line input impedance [tex]Z_{i}[/tex] is given by:
[tex]Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}[/tex] (1)
Now, we calculate the value of [tex]\beta[/tex]:
[tex]\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}[/tex]
(since, [tex]\lambda' = \farc{c}{f\sqrt{K}}[/tex])
[tex]\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53[/tex]
Now, Substituting the value in eqn (1):
[tex]Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega[/tex]
Now, the load reflection coefficient is given by:
[tex]\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}[/tex]
Thus
[tex]\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega[/tex]
Similarly,
Reflection coefficient at input:
[tex]\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}[/tex]
[tex]\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega[/tex]
Now, the SWR is given by:
SWR, Standing Wave Ratio = [tex]\frac{1 +|\Gamma|}{1 - |\Gamma|}[/tex]
SWR = [tex]\frac{1 +|0.62|}{1 - |0.62|} = 4.26[/tex]
