Burn in hell i watched your stupid video and i still could not get the answer

A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel

Hatshy [7]

Answer:

The load reflection coefficient, $\Gamma =0.62\angle 82.875^{\circ} \Omega$

Reflection coefficient at input, $\Gamma = 0.62\angle - 147.518^{\circ} \Omega$

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, $Z_{c} = 75 \Omega$

Length of the cable, L = 2.0 cm = 0.02 m

$Z_{Load} = 37.5 + j75 \Omega$

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = $3.0 \times 10^{9} Hz$

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance $Z_{i}$ is given by:

$Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}$ (1)

Now, we calculate the value of $\beta$ :

$\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}$

(since, $\lambda' = \farc{c}{f\sqrt{K}}$ )

$\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53$

Now, Substituting the value in eqn (1):

$Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega$

Now, the load reflection coefficient is given by:

$\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}$

Thus

$\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega$

Similarly,

Reflection coefficient at input:

$\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}$

$\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega$

Now, the SWR is given by:

SWR, Standing Wave Ratio = $\frac{1 +|\Gamma|}{1 - |\Gamma|}$

SWR = $\frac{1 +|0.62|}{1 - |0.62|} = 4.26$

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Answer:

peak-hour volume = 1890 veh/h

Explanation:

<u>Determine the peak-hour Volume </u>

Applying the equation below

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where :

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v ( peak - hour volume ) = ?

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N = 2 lanes per direction

Fg ( grade adjustment for rolling terrain ) = 0.99 ≈ 1

Fdp = 0.90

<u>Back to equation 1 </u>

v = Vp ( PHF * N * Fg * Fdp )

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