Anches, Harris Joe T. MSEE 1

Tutorial 1

Q1.

i.) Compared with traditional wireless networks, what are the physical limitations restricting

the deployment of sensor networks?

Ans: Compared with traditional wireless networks, WSN can be deployed without the limitation

of an accessible location while traditional wireless should be installed on accessible locations so

extra effort must needed.

ii.) Explain what is meant by energy-aware design.

Ans: One of the most important requirement for the development and implementation of

wireless sensor networks (WSN) is the energy consumption. Hardware components, operations

of the sensors, the communication protocols, the application algorithms, and the application

duty cycle are some of the many design aspects affecting the energy consumption. A full design

is therefore required to estimate the contribution to energy consumption of all of these factors,

and significantly decrease the effort and time spent to choose the right architecture that fits

best to a particular application. In order to redesign to a lowest possible energy consumption

design so that energy sources can last as long as possible to maintain network operation.

iii.) How can system performance be traded off for energy efficiency?

Ans: In a Wireless Sensor Network, communication plays an important role in a system to

work properly and efficiently. But in order to have an energy efficient system, we need to limit

communication between nodes and try to compute instead. Also adjusting the clock frequency

and or the voltage supply can also be traded to have an efficient energy consumption. This is

called Dynamic Voltage Scaling and Dynamic Frequency Scaling.

Q2.

i.) The Transmeta Crusoe processor can be scaled down from 700 MHz at 1.65 V down to 200

MHz at 1.1 V. By how much (in percentage terms) does this scaling reduce power

consumption? What is the corresponding reduction in the energy required per instruction?

Ans: @ f = 700Mhz V = 1.65V P ( fV2 ) (700)(1.65)2 = 1905.75

@ f = 200Mhz V = 1.1V P ( fV2 ) (200)(1.1)2 = 242

1 P1/P0 = 1-(242)/(1905.75) = 0.873 x 100% = 87.3%

Reduced to as much as 87.3% of power consumption.

Po = 1905.75 P1 = 242

Instructionso = 1905.75 * (1 instruction / 1 nJ) * (1 / 700MHz) = 2,722.5 M instructions

Instructions1 = 242 * (1 instruction / 1 nJ) * (1 / 200MHz) = 1,210 M instructions

%reduced = (2,722.5 1,210)/2,722.5 * 100% = 55.56%

Reduced to as much as 55.56% energy per instruction

Q3.

i.) On Slide 37 of Lecture Notes 1 there is an example showing the requirements for the

processing power of a microprocessor embedded in a smart dust. If the processing power

of a low power processor is 100 mW, and the other assumptions are the same as those on

Slide 45, what is the size of the resulting mote (assuming that the resulting lithium battery

takes all the space)?

Ans: @ 100mW processing power (assuming a single day operational lifetime)

Energy = 100mW x 86,400s = 8,640 Ws = 8,640 Joules

for primary battery;

8640 J / (2880 J/cm3 ) = 3 cm3

for secondary battery;

8640 J / (1080 J/cm3) = 8 cm3