Tutorial Wireless Basics
Revision of dB, dBW, dBm, dBV, etc.
Q. Define the terms dB, dBW, dBm, dBi, and Signal to Noise Ratio (SNR).
Q. Why do we use the dB scale? When do we use dB, dBW or dB?
Q. Find the gain in dB if a 10 mW signal is inserted into a transmission line and the measured power some distance away
is 5 mW? Repeat if the output is 20mW. Answer -3dB, +3dB
NdB=10 Log(5/10)= 10(-0.3)= -3 dB
Generally doubling power means 3 dB and halving power is -3 dB
Q. A network has an input of 5mW and a gain of +9dB. What power level in mW would be expect to measure at the
output port? Answer 40mW
Given that each 3dB is doubling power then 9dB corresponds to 2 * 2 * 2 = 8 times
So if the input is 5mW then output = 8 * 5 = 40 mW.
Alternatively using the formulae 9 dB = 10 log10( Output/ 5 mW)
Q. Fill the following table of approximate power ratios for various dB levels.
Decibels
±1
±2
±3
±4
±5
Losses
0.5
0.8
0.63
0.4
0.32
Gains
2
1.25
1.6
2.5
3.2
±6
0.25
4.0
±7
0.2
5.0
±8
0.16
6.3
±9
0.125
8.0
±10
0.1
10
You can use the calculator to work out the numbers in the table using
dB = 10 log(Power Gain ) where Power Gain = Output Power / Input Power
For + 2dB Power Gain = 10 ^( 2 /10) = 1.59
For -2 dB Power Gain = 10^(-2/10) = 0.63
Q. An antenna has an output of 10 W. What is its output in dBW and dBm? Repeat for 20W. Ans: 10 dBW and 40 dBm
Answer Power (dBW) = 10 log (Power/1W) = 10 log10 = 10 dBW = 10 +30 dBm = 40 dBm
Alternatively Power (dBm) = 10 log (Power in mW/1mW)= 10 log (10000mW/1mW)= 40 dBm
Q. Prove that Power in dBm = Power in dBW + 30
Power (dBm) = 10 log (Power in mW/1mW)
= 10 log (1000 * Power in W/1) using log (x * y) = log (x) + log (y) high school math
= 10 log (1000) + 10 log (power in W/1)
= 30 + Power in dBW
Q. Fill following table with dBW and dBm values
Decibels
1W
100mW
10mW
dBW
-20
0
-10
dBm
10
30
20
1mW
-30dBW
0dBm
0.1mW
-40
-10
Using the formulae 100mW in dBW = 10 log10( 0.1 W/ 1W) = - 10 dBW
Using the formulae 100mW in dBm = 10 log10( 100 mW/ 1mW) = 20 dBm
Q. Fill in the missing values in the figures below
0.01mW
-50
-20
0.001mW
-60
-30
0.0001mW
-70
-40
Tutorial Wireless Basics
Exercise 1a: Output = - 6 dBm + 19 dB = 13 dBm = 10 ^ (13/10) = 20 mW
Alternatively – 6Bm = - 0.25 W
Gain of 19 dB = 10 * 2 * 2 * 2 = 80 power ratio
Output = 0.25 * 80 = 20 mW
Exercise 1b: Overall gain = -2 - 4 + 35 = 29 dB and Input = 1W = 0 dBW
Output = 0 dBW + 29 dB = 29 dBW
Alternatively: Gain of 29 dB = 10 * 10 * 2 * 2 * 2 = 800 power ratio
Output = 1W * 800 = 800 W = 10 log (800/1) = 29 dBW
Exercise 1c: Overall gain = -28 + 12 = -16 dB and Output = 40 mW = 16 dBm
Input = 16 dBW – (-16 dB) = 32 dBm
Exercise 1c: Overall gain = 32 - 6 = 26 dB and Output = -19 dBm
Input = -19 dBm – (26 dB) = -45 dBm = -75 dBW = 10 ^ (-75/10) = 31.6 * 10 ^ (-9) Watts
Q. Voltage Ratio Gain: If an antenna has a 20 dBi voltage ratio gain, what voltage ratio does the gain represent?
Answer: For a voltage ratio,
NdB = 20 = 20 log(V2/V1) = V2/V1 = 1020/20 = 101 =10
Q. Fill following table with Power Ratio values
SNR
+10dB
+9dB
Power Ratio
10
8.0
+7dB
5
+6dB
4
+2dB
1.6
Q. What is the output of the ideal combiner (same as an adder)? Answer: +7.76 dBm
Answer: 6 dBm = 10 ^ 0.6 mW = 4 mW
and 3 dBm = 2 mW
Combiner adds the signal power = 4 + 2 = 6 mW = 7.76 dBm
+1 dB
1.25
= 10
Tutorial Wireless Basics
Shannon and Nyquist Theorem
Shannon Channel Capacity C=B log2(1+SNR) where B is Bandwidth, SNR is Signal to Noise as a Power Ratio
Nyquist Bandwidth C=2B log2 M where M= No. of discrete signal or voltage levels.
Hint: logxa = logya / logyx
Q. The spectrum of a channel is between 30 MHz and 40 MHz and SNRdB=24 dB. Find the maximum channel capacity.
Based on Nyquist’s formula how many signalling levels are required? Answer: 80Mbps, M=16
Bandwidth = 40MHz - 30 MHz = 10 MHz
SNRdB=24dB=10 log10(SNR) => SNR=251 (Power Raio)
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Using Shannon C=B log2(1+SNR)= 10 *log2(1+251)=80 Mbps
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7
Now C=2B log2M => 8*10 =2*(10 )*log2M => 4=log2M => M=16
Q. Sydney University and UTS are linked by a microwave radio system with the following parameters.
Analogue bandwidth = 12 MHz
Signal to Noise Ratio = 100,000
Length of the channel = 4.6 km
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-1
Velocity of propagation of the signal = 2.9 x 10 ms
(a). What is the maximum possible data rate that the channel will support? Answer=200Mbps
The maximum possible data rate is provided by the Shannon Limit:
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Bmax = Wlog2(1 + S/N) = (12x10 )log2(1+100,000) ≈ 200 Mb/s (199.3 in fact!)
Source Coding
Q. Explain how digitized speech using PCM has a data rate of 64 kbps. Is this data rate used in all telecommunication
networks?
Answer
The bandwidth of human voice has been standardized at 4 kHz by the ITU.
An analogue speech signal has to be sampled, quantized, and digitized to convert it to a digital signal.
Nyquist Theorem states that a signal needs to be sampled at two times the highest frequency to be able to
reproduce at the receiver. So speech has to be sampled at 2*4000 = 8000 times a second.
Representing voice samples by an 8 bit pattern as in PCM provides sufficient quality for conversation.
Thus data rate = 8000 samples/sec * 8 bits/sample = 64 kbps.
In PSTN, a data rate of 64 kbps is the standard speech data rate.
In mobile communication, to efficiently use the available bandwidth, 64 kpbs
Q. Explain how digitized speech using PCM has a data rate of 64 kbps.
Answer
Q. A digitized TV broadcast is to be transmitted at 30 picture frames per second. The source uses a matrix of 480 x 600
pixels, where each pixel is a combination of 3 colours (red, green, blue) and each colour can have 10 intensity values.
a) How many possible values can each pixel take and how many bits are required to represent each pixel.
b) Find the source rate R in bps.
c) Can this TV broadcast be transmitted over a wireless channel with 4.5 MHz bandwidth and a 35-dB signal-to-noise
ratio.
Answer
a. Each pixel can take on (10*10*10=) 1000 values and can therefore be represented by 10 bits
2^5=32
2^6=64:
2^8=256
2^10= 1024
b Source Rate = (30 pictures/s) (480 * 600 pixels/picture)* 10 bits/pixel = 86.4*106 bps
c. We use the formula: C = B log2 (1 + SNR)
B = 4.5 * 106 MHz = bandwidth, and SNRdB = 35 = 10 log10 (SNR), hence SNR = 1035/10 = 103.5,
C = 4.5 * 106 log2 (1 + 103.5) = 4.5 * 106 * log2 (3163) = (4.5 * 106 * 11.63) = 52.335 * 106 bps
NO, It cannot be transmitted.
Protocols
Q. What is protocol layering? Draw a diagram to illustrate this concept.
Tutorial Wireless Basics
Q. Protocols are often implemented in layers or stacks. Why is this so?
Q. List the major disadvantages with the layered approach to protocols.
Q. What is the 7-layer ISO Reference Model? List the names of each layer (in order from highest layer to lowest layer)
and give a single sentence explanation of the function of each layer
Fragmentation and Reassembly
Q. What is fragmentation in an IP network? Why is it performed? Where is it performed? How is it performed?
Q. What are the pros and cons of intermediate reassembly of an internet fragmented datagram versus reassembly at the
final destination?
Transmission Basics
Q. Distinguish between the terms transmission time, propagation delay and processing delay?
Q. Two wireless nodes connected by a 3km point-to-point link and a data rate of 10 Mbps.
(a) How long after one node begins sending a 1000 byte packet will it be completely received by the other?
(b) What is the round trip delay if the size of the ACK packet is 100 bytes. Draw a timing diagram to illustrate the situation.
Q. Find the round-trip delay of data sent between a satellite and the earth for LEO, MEO, and GEO satellites assuming
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the speed of light is 3 × 10 m/s. If the maximum acceptable delay for a voice system is 30 milliseconds, which of these
satellite systems would be acceptable for two-way voice communication? Assume the following: GEO: geostationary
orbit, ca. 36000 km above earth surface; LEO (Low Earth Orbit): ca. 500 - 1500 km; MEO (Medium Earth Orbit) or ICO
(Intermediate Circular Orbit): ca. 6000 - 20000 km.