Assignment 1

1. Assume a transmitter power of | Watt at 60 GHz is fed into the transmitter antenna. The
antennas used at the transmitter and receiver have a gain of 20 dB.

(a) Calculate the fre

Question

Assignment 1

1. Assume a transmitter power of | Watt at 60 GHz is fed into the transmitter antenna. The
antennas used at the transmitter and receiver have a gain of 20 dB.

(a) Calculate the free spoce path los a 1 m, 100 m, and 1000 m.
(b) Calculate the received signal power at these distances.

2. Signal strength field measurements were taken inside a building. Post processing revealed
that the measured data it a distant-dependent mean power law model having a log-nommal
distribution about the mean. The mean received power law was found tbe (d) xd". A
signal strength of | mW was measured at d, = | m from the transmitter. Ata distance of 10
m, 10% of the measurements were stronger than - 20 dBm. Determine the standard deviation,
or forthe path loss model.

3. Ifthe received power at a reference distance d = | kis | microwat, find the received
powers at distances of 2 km, 5 km. 10 km, and 20 km from the same transmiter or the
following path loss models: (a) Free space: (b) a channel with pathloss exponent 1 = 3; (6)
two-ray pathloss model,

4. Assume an SNR of 25 dB s desired at the receiver. The bandwidth of the receiver is 200
Kiz, The room temperature is T

VIPSolutions · Accepted Answer

# Assignment 1 Breakdown

Here is structured overview and of each question from the. This will help you understand the requirements and how to approach each problem.

---

## 1. Free Space Path Loss and Received Signal Power

**Given:**
- Transmitter power = 1 W
- Frequency = 60 GHz
- Antenna gain = 20 dB (for both transmitter and receiver)

**Tasks:**
- **(a)** Calculate free space path loss at 1 m, 100 m, and 100 m.
- **(b)** Calculate the received signal power at these distances.

---

## 2. Log-Normal Shadowing Path Loss Model

**Given:**
 Mean received power law: $ S(d) \propto d^{-3} $
- At $ d_ = 1 $ m, received power = 1 mW
- At $ d = 10 $ m, 10% of measurements > -20 dBm

**Task:**
- Determine **standard deviation, σ**, for the path loss model.

---

## 3. Received Power with Different Path Loss Models

**Given:**
- Reference distance $ d_ = 1 $ km, received power = 1 μW
- Find received power at 2 km, 5 km, 10 km, and 20 km

**Path Loss Models:**
- (a) Free space
- (b) Path loss exponent $ n = 3 $
- (c) Two-ray path loss model

---

## 4. Signal-to-Noise Ratio (SNR) at the Receiver

**Given:**
- Desired SNR = 25 dB
- Bandwidth = 200 kHz
- Temperature $ T = 290 $ K
- Transmitter power = 100 W
- Carrier frequency = 900 MHz
- Receiver antenna gain =  dB
- Noise figure = 10 dB

**Tasks:**
- (a) Find the **noise floor** for the receiver.
- (b) Find required **signal level** at 10 km for desired SNR.
- (c) Find received power at $ d_ = 1 $ km (free space model).
- (d) For $ n = 4, \sigma = 8 $, find the **probability** that desired SNR is achieved at 10 km.

---

## 5. Path Loss Exponent and Shadowing

**Given:**
- Received powers at 100 m, 200 m, 1 km, 2 km:  dBm, -25 dBm, -35 dBm, -38 dBm
- Path loss model:  
  $$
  L(d) = L(d_) - 10n \log(d/d_) + X_\sigma
  $$
  where $ d_ = 100 $ m, $ X_\sigma $ represents shadowing

**Tasks:**
- (a) Find **minimum mean square error estimate** for path loss exponent $ n $
- (b) Calculate **standard deviation** of shadowing
- (c) Estimate received power at $ d = 2 $ km
- (d) Find probability that received signal at 2 km is greater than -35 dBm (use Gaussian tail function Q).

---

### **Tips for Solving**

- **Free Space Path Loss Formula:**
  $$
  PL_{dB} = 20\log_{10}(d) + 20\log_{10}(f) - 147.55
  $$
  where $ d $ in meters, $ f $ in Hz.

- **Received Power Calculation:**
  $$
  P_r = P_t + G_t + G_r - PL
  $$
  (all in dB or dBm).

- **Log-Normal Shadowing:**
  $$
  PL(d) = PL(d_) + 10n \log_{10}(d/d_) + X_\sigma
  $$
  where $ X_\sigma $ is Gaussian with  mean and std dev $ \sigma $.

- **Noise Floor:**
  $$
  N = kTB \times 10^{NF/10}
  $$
  where $ k = 1.38 \times 10^{-23} $, $ T $ in Kelvin, $ B $ in Hz, $ NF $ is noise figure in dB.

- **SNR:**
  $$
  SNR_{dB} = P_{signal, dBm} - N_{dBm}
  $$

If you need step-by-step solutions for any specific question, please specify which one you'd like to start with!

VIPSolutions · Answer

# Assignment 1 Solutions

This document provides step-by-step solutions to each question in the assignment.

---

## 1. Free Space Path Loss and Received Signal Power

### Given:
- Transmitter Power, $ P_t = 1 \, \text{W} = 30 \, \text{dBm} $
- Frequency, $ f = 60 \, \text{GHz} = 60 \times 10^9 \, \text{Hz} $
- Gain of both antennas, $ G_t = G_r = 20 \, \text{dB} $

### (a) Free Space Path Loss (FSPL)

The Free Space Path Loss (in dB) can be calculated using the formula:
$$
FSPL(d) = 20\log_{10}(d) + 20\log_{10}(f) - 147.55
$$

**Calculations:**
1. **At 1 m:**
   $$
   FSPL(1) = 20\log_{10}(1) + 20\log_{10}(60 \times 10^9) - 147.55 = 0 + 20 \times 8.778 - 147.55 \approx -4.55 \, \text{dB}
   $$

2. **At 100 m:**
   $$
   FSPL(100) = 20\log_{10}(100) + 20\log_{10}(60 \times 10^9) - 147.55 = 40 + 20 \times 8.778 - 147.55 \approx 35.45 \, \text{dB}
   $$

3. **At 1000 m:**
   $$
   FSPL(1000) = 20\log_{10}(1000) + 20\log_{10}(60 \times 10^9) - 147.55 = 60 + 20 \times 8.778 - 147.55 \approx 100.45 \, \text{dB}
   $$

### (b) Received Signal Power

Using the formula:
$$
P_r = P_t + G_t + G_r - FSPL
$$

1. **At 1 m:**
   $$
   P_r(1) = 30 + 20 + 20 - (-4.55) = 74.55 \, \text{dBm}
   $$

2. **At 100 m:**
   $$
   P_r(100) = 30 + 20 + 20 - 35.45 = 34.55 \, \text{dBm}
   $$

3. **At 1000 m:**
   $$
   P_r(1000) = 30 + 20 + 20 - 100.45 = -30.45 \, \text{dBm}
   $$

---

## 2. Log-Normal Shadowing Path Loss Model

### Given:
- Power law model: $ S(d) = k \cdot d^{-3} $
- At $ d_1 = 1 $ m, $ S(1) = 1 \, \text{mW} = 0 \, \text{dBm} $
- At $ d = 10 $ m, 10% of measurements > -20 dBm

### Task: Determine Standard Deviation, $ \sigma $

Using the log-normal shadowing model:
$$
P(d) = P(d_1) - 10n \log_{10}(d/d_1) + X_\sigma
$$

Where $ X_\sigma $ follows a Gaussian distribution. Given $ n = 3 $:
$$
P(10) = 0 - 10 \times 3 \log_{10}(10) + X_\sigma = -30 + X_\sigma
$$
Setting $ P(10) = -20 $ dBm for the top 10%:
$$
-20 = -30 + X_\sigma \implies X_\sigma = 10 \, \text{dBm}
$$
Using the Q-function:
$$
Q\left(\frac{-20 - (-30)}{\sigma}\right) = 0.1
$$
This provides:
$$
\frac{10}{\sigma} \approx 1.2816 \implies \sigma \approx \frac{10}{1.2816} \approx 7.8 \, \text{dBm}
$$

---

## 3. Received Power with Different Path Loss Models

### Given:
- Reference distance $ d_1 = 1 $ km, received power $ P(1) = 1 \, \mu\text{W} = -30 \, \text{dBm} $

### Task: Find received power at 2 km, 5 km, 10 km, and 20 km

**(a) Free Space:**
Using FSPL:
$$
P(d) = P(1) - 10n \log_{10}(d/d_1) \quad (n=2)
$$

1. **At 2 km:**
   $$
   P(2) = -30 - 10 \cdot 2 \log_{10}(2) = -30 - 6.02 = -36.02 \, \text{dBm}
   $$

2. **At 5 km:**
   $$
   P(5) = -30 - 10 \cdot 2 \log_{10}(5) = -30 - 13.01 = -43.01 \, \text{dBm}
   $$

3. **At 10 km:**
   $$
   P(10) = -30 - 10 \cdot 2 \log_{10}(10) = -30 - 20 = -50 \, \text{dBm}
   $$

4. **At 20 km:**
   $$
   P(20) = -30 - 10 \cdot 2 \log_{10}(20) = -30 - 26.02 = -56.02 \, \text{dBm}
   $$

**(b) Path Loss Exponent $ n = 3 $:**

1. **At 2 km:**
   $$
   P(2) = -30 - 10 \cdot 3 \log_{10}(2) = -30 - 9.03 = -39.03 \, \text{dBm}
   $$

2. **At 5 km:**
   $$
   P(5) = -30 - 10 \cdot 3 \log_{10}(5) = -30 - 17.01 = -47.01 \, \text{dBm}
   $$

3. **At 10 km:**
   $$
   P(10) = -30 - 10 \cdot 3 \log_{10}(10) = -30 - 30 = -60 \, \text{dBm}
   $$

4. **At 20 km:**
   $$
   P(20) = -30 - 10 \cdot 3 \log_{10}(20) = -30 - 39.02 = -69.02 \, \text{dBm}
   $$

**(c) Two-Ray Path Loss Model:**
For the two-ray model, the received power is formulated differently. However, typically, you will find it to be less straightforward, often involving the heights of antennas.

---

## 4. Signal-to-Noise Ratio (SNR) at the Receiver

### Given:
- Desired SNR = 25 dB
- Bandwidth = 200 kHz
- Temperature $ T = 290 $ K
- Transmitter power = 100 W = 50 dBm
- Carrier frequency = 900 MHz
- Receiver gain = 0 dB
- Noise figure = 10 dB

### (a) Noise Floor Calculation

Using:
$$
N = kTB \times 10^{NF/10}
$$

Where $ k = 1.38 \times 10^{-23} \, \text{W/K} $:
$$
N = 1.38 \times 10^{-23} \times 290 \times 200 \times 10^3 \approx 8.06 \times 10^{-12} \, \text{W} = -107.9 \, \text{dBm}
$$

### (b) Required Signal Level at 10 km

Using:
$$
SNR_{dB} = P_{signal, dBm} - N_{dBm}
$$
$$
P_{signal} = SNR + N = 25 - 107.9 = -82.9 \, \text{dBm}
$$

### (c) Received Power at 1 km (Free Space Model)

Using:
$$
P_r(1 \text{km}) = P_t + G_t + G_r - FSPL
$$
Where $ FSPL $ can be calculated as shown previously.

### (d) Probability of Achieving Desired SNR (n=4, σ=8)

Using:
$$
Q\left(\frac{SNR_{min} - \mu}{\sigma}\right)
$$
Where $ \mu = -30 \, \text{dBm} $ (used previously):
- Calculate $ Q $-function for $ P_{signal} = -82.9 \, \text{dBm} $.

---

## 5. Path Loss Exponent and Shadowing

### Given:
- Measured powers: 0 dBm, -25 dBm, -35 dBm, -38 dBm at distances 100 m, 200 m, 1 km, 2 km respectively.

### (a) Minimum Mean Square Error Estimate for Path Loss Exponent $ n $

Using:
$$
L(d) = L(d_1) - 10n \log_{10}(d/d_1) + X_\sigma
$$

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