---
# System prepended metadata

title: HW3
tags: [' ss', Communication System Design, ' ncu']

---

---
tags: Communication System Design, ss, ncu
author: N0-Ball
title: HW3
GA: UA-208228992-1
---

# Question #1 
:::info
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.
:::

## Preknowledge

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

Ans. $148.39\ dB$

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

Ans. $40.46\ dB$

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

Ans. $-68.77\ dBW$

## d.
:::info
Hence find the C/N ratio, in dB, at the output of the receiving antenna.
:::

$$
P_n = kT_sB_n = 10 \times \log(1.38 \times 10^{-23} * 1000 * 50 \times 10^6) \approx -121.6115\ dbW \\[1em]
C/N = \frac{P_r}{P_n} = P_r - P_n = \underline{53.0\ dB}
$$

### Ans.

$$
53.0\ dB
$$

## e.
:::info
Heavy rain will cause high attenuation on this link at 25.0 GHz. If the threshold C/N ratio at the 
receiver is 22.0 dB, what is the rain attenuation margin on the link? What is the maximum rain 
attenuation in dB/km that can be tolerated before the link is in outage with uniform rain along the 
entire link? (Ignore any noise temperature increase at the receiver caused by the rain.)
:::

### Rain attenuation margin
$$
C/N - (C/N)_t = 53 - 22 = \underline{31\ dB}
$$

### Maximum rain attenuation

$$
\frac{35}{25} = \underline{1.2 \ dB/km}
$$

# Question #2
:::info
- Transponder and satellite characteristics
    - Transponder bandwidth 25 MHz
    - Transponder saturated output power 90 W
    - Downlink frequency 11.5 GHz
    - Uplink frequency 14.7 GHz
    - Downlink and uplink antenna gain, on axis 32.0 dB
    - Atmospheric clear air loss on downlink 0.5 dB
    - All other losses on downlink 0.5 dB
- Ground station parameters
    - Antenna diameter 0.9 m
    - Aperture efficiency 70%
    - Antenna noise temperature (clear air) 40 K
    - Receiver noise temperature 90 K
    - Receiver noise bandwidth 15 MHz
    - Minimum overall C/N ratio (threshold) 8.0 dB
- Downlink parameters
The satellite transmits a continuous TDM data stream that can be received in a 15 MHz bandwidth. 
The transponder is operated with 2 dB output back-off. (This means that the transponder output power 
is set to 2 dB below the maximum available output power level of 90 W.) The earth station, equipped 
with a high-gain low noise amplifier (LNA), is at a distance of 36,000 km from the satellite, and is on 
the -3 dB contour of the satellite antenna beam.
:::

## a.
:::info
Find the path loss, in dB, for the downlink from the satellite to the receiving earth station.
:::

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

$$
L_p = (\frac{4 \pi \times 3.6 \times 10 ^ 7 \times 11.5 \times 10 ^ 9}{3 \times 10 ^ 8})^2 \approx 3.0 \times 10 ^ {14} \approx \underline{204.78\ dB}
$$

### Ans.

$$
204.78\ dB
$$

## b.
:::info
Calculate the gain of the receiving antenna in dB, and the received power, in dBW, at the input to 
the earth station’s LNA. Set out your calculation as a link budget.
:::

$A_e$
:    $A_e \times A = 0.7 \times \pi \times \left( \frac{0.9}{2} \right)^2 \approx 0.445$

$G_r$
:    $10 \times \log \left( \frac{4 \pi A_e}{\lambda ^ 2} \right) = 10 \times \log \left( \frac{4 \pi \times 0.445 \times (11.5 \times 10 ^ 9)^2}{(3 \times 10^8)^2} \right) \approx 39.14\ dB$

$P_t$
:    $10 \times \log(90) - 2 \approx 17.54\ dbW$

$G_t$
:    32.0 dB

$L_p$
:    204.78 dB

$L$
:    0.5+0.5 = 1dB

**Therefore**
$$
P_r = P_t + G_r + G_t - L_p - L = 17.54 + 39.14 + 32.0 - 204.78 - 1 - 3 = \underline{-120.1\ dBW}
$$

### Ans.

$$
-120.1\ dBW
$$

## c.
:::info
Calculate the noise power, in dBW, referred to the input of the earth station receiver LNA. Hence 
find the downlink C/N ratio in decibels. Ignore any noise radiated by the satellite.
:::

$T-s$
:    T_A + T_R = 90 + 40 = 130 K

$B_n$
:    15 MHz

$P_n$
:    $kT_sB_n = 10\times \log (1.38 \times 10 ^ {-23} \times 130 \times 15) \approx -135.7 dB$

**Therefore**
$$
C/N = P_r - P_n = -120.1 - (-135.7) = \underline{15.6\ dB}
$$

### Ans.

$$
15.6\ dB
$$

## d.
:::info
Heavy rain affects the downlink from the satellite causing 6 dB of rain attenuation. Ignore the 
increase in antenna temperature caused by the rain. What is the new value for the downlink C/N in 
the earth station receiver? Give your answers in dB.
:::

$$
15.6 - 6 = \underline{9.6\ dB}
$$

### Ans.

$$
9.6\ dB
$$

## e.
:::info
The threshold C/N ratio for this link is 8.0 dB. What is the rain attenuation margin 
for this downlink?
:::

$$
15.6 - 8 = \underline{7.6\ dB}
$$

### Ans.

$$
7.6\ dB
$$