---
tags: Communication System Design, ss, ncu
author: N0-Ball
title: HW2
GA: UA-208228992-1
---

# Problem

A line-of-sight microwave link operates at 25.0 GHz using dish antennas with a
diameter of 0.5 m. The aperture efficiency of the antennas is 65%. The transmitter
output power at the transmitting antenna input port is 3.0 W to transmit a signal with a
bandwidth of 50 MHz. The noise temperature of the receiver system is 1000 K in a
noise bandwidth of 50 MHz, referred to the receiving antenna output port. The length
of the link is 25 km and in clear air conditions, atmospheric loss on the link is 6.0 dB.
Using decibel units throughout, calculate the following parameters for this link. Give
all answers to a precision of ± 0.1 dB. Note: You may or may not need all the
parameters.

:::info
- P~t~ = 3.0 W $\approx$ 4.7 dBW
- A~e~ = 65% $\times \pi (0.5)^2$
- R = 25 km
- L~a~ = 6.0 dB
:::

# a.
:::info
Calculate the path loss for the link at 25.0 GHz.
:::

**Since**
$$
L_p = (\frac{4 \pi R}{\lambda})^2
$$

and $\lambda = \frac{c}{f}$

$\Rightarrow L_p = (\frac{4 \pi \times 25 \times 10^3 \times 25 \times 10 ^{9}}{3.0 \times 10 ^{8}})^2 \approx \underline{6.9 \times 10 ^ {14} \approx 148.39\ dB}$

## Ans.

$$
148.39\ dB
$$

# b.
:::info
Calculate the gain of the transmit and receive antennas in dB. Use the exact formula.
:::

**Since**

$$
\begin{split}
G &= \frac{4 \pi A_e}{\lambda ^2} = \frac{4 \pi \eta_e A f^2}{c ^2} \\[1em]
&= \frac{4 \pi \times 0.65 \times \pi \times (0.25)^2 \times (25 \times 10 ^ 9)^2}{(3.0 \times 10^8)^2} \approx 4.46 \times 10^4 = 10 \times log(4.46 \times 10^4) \approx \underline{40.46\ dB}
\end{split}
$$

## Ans.
$$
40.46\ dB
$$

# c.

:::info
Calculate the received power at the output of the receiving antenna, in dBW
:::

**Since**
$$
\begin{split}
P_r &= G_t + G_r + P_t - L_a - L_p \\[1em]
& = 2 \times 40.46 + 4.7 - 6.0 - 148.39 = -56.70\ dB
\end{split}
$$

## Ans.

$$
-68.77\ dB
$$