Lecture 1: Introduction to IoT & Edge Computing

Contents

• Self-Test
  • Question 1:
  • Question 2:
  • Question 3:
  • Question 4:
  • Question 5:
  • Question 6:
  • Question 7:
  • Question 10:
  • Question 12:
  • Question 13:
  • Question 14:
  • Question 15:

Self-Test

Question 1:

What are the key challenges of the cloud based architecture and why migration towards edge can help?

Solution: Refer to Key Challenges with Cloud Based Architecture section.

Question 2:

What is the difference between *application edge and network edge?

Solution:

Network Edge refers to the boundary of a network where end devices (users, IoT sensors, base stations, routers, etc.) connect to the broader internet or backbone.

Example: The point where your phone connects to a 5G base station, or where a home router connects to an ISP. Focus: Traffic routing, access control, and connectivity. Used in telecom/ISP contexts (e.g., "edge of the mobile network").

Application Edge refers to the deployment of applications/services close to end users, often leveraging edge computing resources.

Example: Running an AI inference engine at a nearby edge server instead of a distant cloud data center. Focus: Low-latency service delivery, computation, and user experience. Used in cloud/edge computing contexts (e.g., "edge AI applications").

Also refer to diagram in Slide 2 of A Journey in to IoT.

Question 3:

What are cyber-physical system and what are their key applications?

Solution Refer to the definition and key characteristics of CPS in Lec 1, Part 1.

Question 4:

What are various layers of CPS systems?

Solution Refer to CPS section of Lec 1, Part 1.

Question 5:

What is the difference between CPS and IoT?

Solution Refer to the diagram for the IoT, CPS and SoS.

Question 6:

What are Digital twins and what are key attributes associated?

Solution Refer to the DT section of Lec. 1.

Question 7:

Setup. Two IoT-enabled streetlights are connected in series (the system fails if either streetlight fails).

Component reliabilities:

$$R_1(k) = (0.98)^k,\qquad R_2(k) = (0.96)^k$$

Find the system reliability after $k = 7$ days.

Solution

For components in series, the system reliability is the product of component reliabilities:

$$R_{\text{sys}}(k) = R_1(k) \cdot R_2(k).$$

Substitute the given expressions:

$$R_{\text{sys}}(k) = (0.98)^k \cdot (0.96)^k.$$

Combine common exponent $k$:

$$R_{\text{sys}}(k) = \big(0.98\cdot 0.96\big)^k = (0.9408)^k.$$

Evaluate at $k=7$:

$$R_{\text{sys}}(7) = (0.9408)^7 \approx 0.6523507426.$$

Final answer:

$$R_{\text{sys}}(7) \approx 0.6523507426.$$

Question 8: What are core 5G service classes?

Solution Check 5G and 6G discussion in Lec 1.

Question 9: Describe types of signals produced by sensor including example of sensors which can yield these signals?

Solution Check Sensor Type section in Lec 1, Part 2.

Question 10:

A smart thermostat measures room temperature. The signal contains frequency components up to 40 Hz due to fast fluctuations in the environment.

Tasks:

1. Determine the Nyquist rate for this signal.
2. Find the minimum sampling interval (in milliseconds).
3. If the system samples at 150 samples/second, check whether aliasing will occur.

Solution

1. The Nyquist rate is twice the maximum signal frequency, so

$$f_{\text{Nyquist}} = 2 f_{\max} = 2\times 40\ \text{Hz} = 80\ \text{Hz}.$$

2. The minimum sampling interval $T_s$ (period between samples) corresponding to the Nyquist rate is

$$T_s = \frac{1}{f_{\text{Nyquist}}} = \frac{1}{80}\ \text{seconds} = 0.0125\ \text{s} = 12.5\ \text{ms}.$$

3. If the system samples at $f_s = 150\ \text{samples/s} = 150\ \text{Hz}$, compare with the Nyquist rate:

$$150\ \text{Hz} > 80\ \text{Hz},$$

so the sampling frequency exceeds the Nyquist rate and aliasing will not occur for frequency components up to 40 Hz.
There is margin — the designer should still consider anti-aliasing filters and practical margins.

Question 11: What is Aliasing and when do we encounter it?

Solution Check Aliasing discussion in Part 2, Lec 1.

Question 12:

An IoT acoustic sensor records sound signals with amplitudes normalized between $(-1)$ and $(+1)$. The signal is digitised using a uniform quantiser with a bit depth of 8 bits.

Tasks:

1. Determine the number of quantisation levels.
2. Calculate the quantisation step size ($\Delta$).
3. Using the approximate formula for signal-to-quantisation-noise ratio (SQNR), estimate the SQNR in dB.
4. Repeat the SQNR calculation for a bit depth of 12 bits and comment on the improvement.

Solution

1. The number of quantisation levels is

$$L = 2^N = 2^8 = 256.$$

2. The dynamic range of the signal is 2 (from −1 to +1). The quantisation step size is

$$\Delta = \frac{2}{L} = \frac{2}{256} = 0.0078125.$$

3. The approximate SQNR formula for a uniform quantiser is

$$\text{SQNR}_{\text{dB}} \approx 6.02N + 1.76.$$

For $N = 8$:

$$\text{SQNR}_{\text{dB}} \approx 6.02\times 8 + 1.76 = 48.16 + 1.76 = 49.92\ \text{dB}.$$

4. For $N = 12$:

$$\text{SQNR}_{\text{dB}} \approx 6.02\times 12 + 1.76 = 72.24 + 1.76 = 74.0\ \text{dB}.$$

Comment: Increasing the bit depth from 8 to 12 improves the SQNR by about 24 dB, which corresponds to a substantial reduction in quantisation noise and better signal fidelity — useful for high-quality acoustic sensing in IoT applications.

Question 13:

What is regression?

Solution Refer to Part 4 of Lec 1.

Question 14:

What is type of signals are frequently generated by IoT devices?

Solution Refere to Part 4, Lec 1.

Question 15:

A small IoT temperature sensor records the following data over 5 hours:

$$\begin{array}{c|c} \text{Hour }(x) & \text{Temperature }(^\circ\text{C}) (y) \\ \hline 1 & 15 \\ 2 & 17 \\ 3 & 18 \\ 4 & 21 \\ 5 & 22 \\ \end{array}$$

Task:

1. Fit a simple linear regression model of the form $y = mx + c$ by hand.
2. Determine the slope $m$ and intercept $c$.
3. Predict the temperature at hour 6.

Solution

1. Compute the means:

$$\bar{x} = \frac{1+2+3+4+5}{5} = 3, \quad \bar{y} = \frac{15+17+18+21+22}{5} = 18.6$$

2. Compute the slope $m$:

$$m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} = \frac{(-2)(-3.6)+(-1)(-1.6)+(0)(-0.6)+(1)(2.4)+(2)(3.4)}{(-2)^2+(-1)^2+0^2+1^2+2^2}$$ $$m = \frac{7.2 + 1.6 + 0 + 2.4 + 6.8}{4+1+0+1+4} = \frac{18}{10} = 1.8$$

3. Compute the intercept $c$:

$$c = \bar{y} - m \bar{x} = 18.6 - (1.8)(3) = 18.6 - 5.4 = 13.2$$

4. Regression equation:

$$y = 1.8 x + 13.2$$

5. Predict temperature at hour 6:

$$y = 1.8(6) + 13.2 = 10.8 + 13.2 = 24.0^\circ\text{C}$$