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Lec 06 _ Introduction to Antennas and Propagation Models — Transcript

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0:14Good morning welcome to lecture 6, in today’s lecture the topics that we will be covering
0:19are a quick summary of the topics that we have covered in lecture 5 and basic introduction
0:28to antennas leading to link to the propagation using the breakpoint model and in the course
0:36of today’s lecture we will be having several examples which will highlight the concepts
0:40that we are discussing. The purpose of these topics is to work our way towards understanding
0:46the system design and two of the elements that will play a big important role in system
0:52design one is fading margin and another one is the noise figure.
0:57Now, all of these concepts together help us to understand the noise limited systems that
1:04is when I have a transmitter the further I go away from the transmitter the noise the
1:09signal level decreases noise level being constant we find that eventually the noise becomes
1:15a limiting factor, and that is what we refer to as a noise limited system a system in which
1:20only the additive Gaussian noise is the only impairment. Now on the other hand when it
1:27is a cellular system we find that the noise is in addition to the noise we also have an
1:33impairment in the form of interference. So, that will be our entry point into our discussion
1:38on the interference limited systems and of course, in the examples you see that there
1:43is one example that we would like to discuss that of the NASA deep space network which
1:49is probably one of the most powerful antennas built in wireless communications and it is
1:54a very interesting and useful example for us to study, that is something that we will
1:58be looking at in today’s lecture. So, first we begin with a quick review of
2:04what is happened in the previous lecture and for that I would like to look at the main
2:11concepts that we had discussed in the last lecture.
2:15The concepts are that we can have the uplink and the downlink in the form of 2 different,
2:23in 2 different frequencies, this would be referred to as frequency division duplex and
2:30at that transceiver we have a single antenna that is used both for the transmission and
2:35reception and to separate the 2 signals we have the use of a duplex filter. Several examples
2:41of the FDD system for example: the GSM system that we use in the second generation, the
2:47wideband CDMA in the third generation or examples of a frequency division duplex system. Now
2:52on the other hand if you were to look at a time division duplex system.
2:57Now, in this case also we have a single antenna that is that is used for both transmission
3:02and reception; however, on the transmission side we find that the uplink and the downlink
3:09happen on the same frequency, they are happening on different time slots. So, this helps us
3:15to use the same frequency for both uplink and downlink by suitably sharing between the
3:22uplink and the downlink. So, in such a system you find that the structure is different we
3:26do not need a duplex filter because the transmitter and reception are happening on the same frequency
3:31and this is what we see in a TDD system. Now in example of a TDD system is the wideband
3:41CDMA version which has a TDD version of it. So, given these 2 basic examples I thought
3:50it would be good for us to have a summary of the systems that we have discussed so far.
3:58So, let us go back to our notes and quickly pick up from there. So, we now have basic
4:05understanding of the FDD and the TDD systems.
4:08When a duplex filter is required and the spacing between the transmit frequency and the received
4:14frequency referred to as a duplex spacing. Now there is a fundamental difference or in
4:22terms of the duplex is what we call as the uplink downlink spacing versus multiple access
4:29sometimes this may cause some confusion. So, what I would like to do is just summarize
4:33what we had mentioned in the last lecture. So, we would like to look at 3 different systems
4:39one of them is the GSM system, the second is the CDMA system, and then the third would
4:45be the wideband CDMA system and the LTE system the fourth generation. So, we have the second
4:53generation, we have the third generation and the fourth generation.
4:59So, first I would like to address the aspect of multiple access; how is multiple access
5:05done in these different systems how are the different users sharing the common resources
5:11and in GSM it is done through time division multiple access each of the users gets a time
5:19slot on which they do the transmission and reception and therefore, that is a time division
5:23multiple access system. Now the CDMA system as the name suggests uses code division multiple
5:32access for sharing the resources between different users and the fourth generation the LTE system
5:40uses orthogonal frequency division multiplexing. So, 3 different multiple access methods in
5:45the second third and in the fourth generation, now we would like to take us next look at
5:52what is the frequency or the type of duplexing that is done.
5:56So, the type of duplexing that is done decides the design of the transceiver. So, the duplex
6:03method that is used for GSM, GSM uses an FDD’s duplexing frequency division duplexing, it
6:13the CDMA system uses a FDD system FDD also and the LTE the fourth generation can have
6:22an FDD or TDD system. So, the minute you have a TDD system you do not need a duplex filter.
6:31So, in this case we can straight away write down no duplex filter, no duplex filter on
6:40the other hand CDMA which is an FDD system it is continuously transmitting and receiving
6:46at the same time this requires a duplex filter, a CDMA system must have a duplex filter.
6:54Now, by the definition of being in FDD system you would expect that GSM also has a requirement
7:02of a duplex filter. So, let me ask the question would a GSM receiver necessarily have a duplex
7:09filter? The answer comes in the following fashion in the case of a GSM system, the transmit
7:17and receive time slots are offset in terms of their time positioning. So, if I transmit
7:23at a particular time slot the receive time slot is offset with respect to the transmit
7:28time. So, the transmission and reception though they are happening on different frequencies
7:34are not happening at the same time. So, essentially because of this spacing between transmitter
7:39and receiver timing wise this essentially says that we do not require a duplex filter.
7:46So, TDMA in the case of GSM a duplex filter is not needed for a single slot operation
7:54because you can have them separated in this fashion.
7:56So, the answer for this would be no for single slot for single slot operation. Now if you
8:05have a GSM transceiver which is using multiple time slots, then you may require the transmission
8:11reception may happen at the same time and there is a requirement that you would have
8:16a duplex filter. So, if you have a multi slot transmission then you will need to have a
8:22duplex filter. So, these are the basics that of the thing concepts that we have discussed
8:27and with this we would now like to dive into the topics of today’s lecture. So, the first
8:34topic that we would like to pick up is to have a basic understanding of antennas, now
8:41antennas is our ability to transmit or receive electromagnetic radiation and is a topic that
8:49is studied extensively in electromagnetic theory our interest is to understand it from
8:54a systems perspective.
8:55So, the first concept that we would like to introduce in the topic of antennas would be
9:02the topic of isotropic antenna. It is a concept that helps us understand the behavior and
9:10the property of antennas. So, this is our concept such a antenna does not exist physically,
9:16but it helps us to explain the basic behavior of antennas and also for us to understand
9:22the system impacts that these antennas will have.
9:25So, the isotropic radiator I would like to explain it in the context of a diagram and
9:34the isotropic radiator you can think of it as a point radiator it is a point radiator
9:41from which the radiations emanate in all directions in 3 dimensions. So, it is a 3 dimensional
9:48radiator and it is a point source it radiates in all direction. So, this is one of the basic
9:54concepts that we use in our understanding of antenna. So, an isotropic radiator.
10:01Now, how do we incorporate this into our design? Basic explanation of isotropic antenna, isotropic
10:13antenna; it is a 3 dimensional radiating element it radiates in all directions and it is one
10:27that we use for a explaining in a conceptual manner. So, in contrast to this we have a
10:34Omni directional antenna, Omni directional antenna now this antenna has got a property
10:43that is different from that of an. So, if you think of the isotropic radiator as a 3
10:50dimensional radiator in all directions, the isotropic antenna. So, basically the isotropic
10:58antenna radiates in all directions. The Omni directional antenna we think of it
11:02primarily in the context of a dipole antenna, typically the size of it would be of the order
11:11of lambda by 2 this would be a dipole antenna and this has a radiation pattern that is equal
11:23in it is in the horizontal plane in the azimuthal plane. So, the radiation pattern is equal
11:29now it is not equal in the vertical direction. So, we will come to describe that in a little
11:35bit. So, we have isotropic antenna Omni directional antenna isotropic and Omni directional are
11:43not the same. So, in the initial part of our lecture we are going to be looking at the
11:48isotropic antenna.
11:50Now, if we have a radiated power, radiated power of P watts. So indicate that as P subscript
12:02t transmit power this is the transmitted power. So, we define for this antenna for a isotropic
12:11radiator; isotropic antenna is our point topic of discussion now, the power flux density
12:26this is the total power that is radiated experienced at a unit area at a radius R. So, power flux
12:35density at a radius R radius from the point source radius R, this is denoted by phi subscript
12:44R and this is given by the transmit power divided by area of the sphere since we are
12:50radiating in all directions 4 pi R square and the units would be watts per meter square
12:59. Watts per meter square that is the power flux
13:03density and this tells us how much of is the intensity of the radiation when you are at
13:08a distance R the number of the flux intercepted per unit area. Now if we were to look at the
13:16other side of it from the transmit side to the receive side, the received signal power
13:23of course, the ideal scenario would be is if you could have a spherical antenna in enveloping
13:29the entire transmit entire power flux density, but such a receive antenna is not practical.
13:35So, the received signal power is proportional to the area that you can with which you can
13:41intercept the power flux density. So, Ae stands for the effective area, effective area of
13:50the antenna that area of the receive antenna effective area of the receive antenna.
13:56So, this is what tells us what portion of the transmitted signal power can be effectively
14:02intercepted. Now the effective area basically depends on some elements, it first of all
14:10depends on the size of the antenna the larger the size a larger it is effective or effective
14:17power that it can in the effective power that it can intercept. So, there is a proportionality
14:22to the size, it is also proportional to the cross sectional area cross sectional area;
14:30the cross section that will intercept the transmitted power that is a cross sectional
14:36area and of course, it is also dependent on the orientation of the antenna because you
14:47need something that is sufficiently large with a good cross section and pointed in the
14:53direction. So, that it can intercept the transmitted power.
14:58And these are 3 elements that will affect the effective in the or reception of a signal.
15:06Now we have the power flux density which is P t by 4 pi R squared, now the power flux
15:14density multiplied by the effective area of an of the receive antenna is going to be my
15:21received signal power. So, if we assume that we also have just like we had an isotropic
15:27transmitter we had an isotropic transmitter, that we also have a isotropic receiver a point
15:37source isotropic receiver. Now in the in the in the theory of antennas we associate with
15:47an isotropic receiver an effective area of an isotropic receiver as a lambda squared
15:54by 4 pi. You can think of it as a constant that that
15:58is used for understanding the basic operations. So, where lambda is the wavelength, wavelength
16:07in meters wavelength of the or that transmitted electromagnetic wave and a 4 pi is a constant
16:14dimensionally consistent because lambda squared would give you a meter squared and that corresponds
16:19to the effective cross sectional area. Now the received signal power in the case of an
16:27isotropic transmitter isotropic receiver would now can now be written as the transmitted
16:33power divided by 4 pi R squared as the effective as the power flux density multiplied by the
16:45effective area of a isotropic antenna which would be lambda squared by 4 pi and this can
16:52be written into a convenient form P t divided by 4 pi R by lambda whole square.
17:02Now, this is a very very useful result because this is what we referred to as free space
17:10propagation, where we have an isotropic transmitter and isotropic receiver and we have effectively
17:19line of sight transmission in free space between the transmitter and receiver. So, this is
17:24the free space model and this also gives us a very useful form with which we can work.
17:33So, let me recap the free space model free space propagation model it also called the
17:40free space path loss model referred to as a free space model. So, the transmitted signal
17:48power or the received signal power we initially we referred to the power flux density at a
17:54distance R and typically when we talk about transmission we do not talk about the radius
17:59of transmission. But more at distance of transmission from
18:02the transmitted receiver. So, we going to be make a change of variable from R to d,
18:10so that we can refer to it in terms of the distance between the transmitter and receiver.
18:15So, the received signal power at a distance d can be written as the transmitted signal
18:22power divided by 4 pi I have replaced r with d, d by lambda whole square and writing it
18:31in dB the received signal power in decibels at a distance d represented in decibels is
18:39given by the transmitted signal power in decibels and what is the difference between the received
18:47signal power and the transmitted signal power is the path loss we call that as PL stands
18:54for path loss at a distance d also expressed in dB.
18:59So, this quantity which is the path loss is the is referred to as the free space path
19:07loss free space path loss and it is given by 4 pi d by lambda whole squared and of course,
19:21this can be expressed in dB if you want to express the path loss in dB. So, this is a
19:28very good way for us to understand how the free space path loss a path loss works. So,
19:37what I would like to do is work do compute a simple worth example to help us consolidate
19:45whatever we have understood with the free space model and then we will try to build
19:50on this going forward.
19:52So, the first example would be that I have a system where the transmitted power transmit
20:01power is given as 100 milliwatts; 100 milliwatts is my transmitted power. Now I would like
20:10to convert this into dB scale. So, power if I want to convert into dB scale can be expressed
20:21as dBW or as dB watts or dBm which stands for dB milliwatts. So, the conversion this
20:29would be 0.1 watt. So, 100 mille watts divided by a thousand point one watt this can be written
20:39as minus 10 dBW or if I were to convert it in from dBW to dBm always have to add plus
20:4930 which can also be converted directly from 100 milliwatts converted into dB would be
20:5520 dBm. So, 100 milliwatts can be represented either as minus 10 dBW or s plus 20 dBm.
21:03Now, we are given that the carrier frequency for the electromagnetic transmission carrier
21:11frequency is 800 megahertz and this will help us compute the wave length. So, the wave length
21:21would be the velocity of light 3 into 10 power 8 divided by the carrier frequency 800 into
21:2910 power 6 that gives us 3 8 th of a meter; and we are told that the received signal power
21:40received signal power is minus 90 dBm received signal power is minus 90 dBm.
21:48Now if by any chance you made a mistake and you wrote down the received signal power as
21:55plus 90 dBm, the minute you see it you should definitely it should strike you because my
22:02plus 90 dBm would be a very large value of power, because it is its 10 power 9, 10 power
22:139 dB 10 power 9 milliwatts. So, which means it is a large number of amount of power. So,
22:19typically the received signal powers that we are working with of the order of minus
22:2390 dBm much much less than a milliwatt. Now the difference between the transmitted power
22:30and the received power is the path loss. So, that tells me that the path loss is equal
22:36to 20 dBm minus of minus 90 dBm this gives me 110 difference of 2 dBm quantities or the
22:50ratio of 2 powers is a dimensionless quantity. So, this is 110 dB please do not write it
22:56as dBm it is a 110 dB because it is a ratio of 2 powers.
23:01So, the path loss in dB is 110 and this effectively corresponds to on a linear scale 10 to the
23:09power of 11 which is a very large number. Now we would like to know what is this correspond
23:14to in terms of the distance. So, if you write down the expression for path loss, path loss
23:20at a distance d we are given as 10 power 11 like to recreate it to 4 pi d by lambda whole
23:30squared substitute for lambda where lambda is equal to 3 by 8 of a meter, you can verify
23:37that the distance that we can get will be approximately 9.4 kilometers. So, this is
23:45what the free space propagation loss tells us, and is a very useful measure for us to
23:51know that if there is a isotropic transmitter and isotropic receiver and only free space
23:57between the 2 of them 100 mille watts of transmission we could reach all the way up to 9.4 kilometers,
24:04and still give you a received signal power which is of the order of minus 90 dBm.
24:10It is also helpful for us to make a note that because of the exponential term in terms of
24:17the distance, a small change in the power does affect the distance also. So, if the
24:23transmitted power was reduced to 10 milliwatts you can redo the calculation and we find that
24:32the distance at which you can achieve minus 90 dBm is now only 2.98 kilometers, so the
24:42non-linear relationship between a tenfold in the decrease in power translating into
24:48a distance. Now an important point to mention which is helpful for us to keep in mind is
24:56the fact that free space we write it with an exponent of 2 this is writing of this as
25:05a path loss exponent n equal to 2 is for free space.
25:10Typically in a cellular system there is non-line of sight as opposed to free space propagation,
25:19free space propagation will be line of sight. So, cellular is non line of sight and as you
25:27would expect the losses are more in a non line of sight system. So, n is typically greater
25:34than 2 for a non line of sight system. So, if you have non line of sight system and typically
25:42it is in the range of 3 to 4 which tells us that the losses will be much more significant
25:49when you traverse a distance d. So, with the same transmit power you will be able to in
25:56a non line of sight environment you will be able to transmit the signal to a shorter range
26:02as compared to a line of sight system. Now, this also leads us to is the propagation
26:11model which is very helpful for us in terms of understanding the entire system.
26:18So, this would be the breakpoint model; breakpoint model the breakpoint model says that there
26:27are 2 regions in your signal propagation, this is a 2 two a single breakpoint model
26:38and it says that if I am below the breakpoint, then I can assume that it is free space propagation.
26:49Free space propagation and my exponent they receive signal power at a distance d is given
27:00by the transmit signal power by 4 pi d by lambda whole square.
27:08So, this would be my free space propagation before. So, if I were to sketch it plotting
27:16the received signal power P r at a distance d measured in dB, as a function of log of
27:25the distance, then up to a certain point which is the break point, I see a slope which is
27:36minus 2 or a path loss exponent which is n equal to 2. So, this is the basic model up
27:43to this is the d break, break point. Now once we go beyond the break point for d greater
27:53than d break the relationship is given in the following way, the received signal power
28:01at a distance d beyond d break is the received signal power at d break into d break by d
28:15raised to the power n where n is typically greater than 2 that is when you have the non
28:23line of sight component. Now, let us make a quick check to see if this
28:27is how this is working; if n had been equal to 2 then we would have still have had the
28:33same expression as free space propagation; however, when you have an path loss that is
28:40greater than 2 for d break, you find that beyond this break point you find that the
28:45path loss exponent causes the receive signal power to decay much faster. So, here we have
28:53a slope that is more negative than minus 2 or a path loss exponent that is greater than
29:002. So, basically we find that the received signal
29:04power decay is much faster as the distance increases. So, this is essentially the breakpoint
29:11model and typically the path loss exponent can be anywhere from 3 to 4 for non line of
29:20sight environments, and this is a way by which we can have a first quick prediction of what
29:26the received signal power is likely to be ok.
29:31Now, I would like to go back and quickly explain a few elements about the antennas and make
29:45before that let me also suggest that for this portion a good reference would be Molisch
29:55chapter 9, which has a good systems level description of antennas the different types
30:02of antennas and their construction and their basic properties different types, and it is
30:09very useful for us to we are only talked about a isotropic antenna and in a very very limited
30:16sense the Omni directional antenna. But there are several types of antennas let
30:22me just mention some of those. So, we have the isotropic that is our conceptual description
30:31of how the radiation occurs then we have the dipole antenna, then in Molisch chapter 9
30:39you can see that there are several other types of antennas which are used in our cellular
30:44systems something called a micro strip antenna micro strip antenna there is a planar inverted
30:54f antenna inverted f antenna very commonly used antenna.
30:59It also referred to as PIFA planar inverted f antenna and so there are several flavors
31:08of them which you can see and understand their properties, but very important for us to understand
31:15that at the end of the day what we are looking for is the effective cross sectional area
31:20we are looking for the ability for the antenna to transmit and receive signals in a suitable
31:28direction. So, let us quickly take a look at the isotropic the Omni directional antenna
31:36and then. So, if you were to look at a dipole antenna which would be a basis for discussion
31:44on Omni directional antennas.
31:47Now, if we were look at the radiation pattern ideally in the horizontal direction all 360
31:56degrees the antenna radiates equally, but the minute you look at the elevation side
32:03you find that the antenna is radiation pattern is very different it is not radiating like
32:10an isotropic radiator. So, for example, if we were to look at the radiation along the
32:15dipole axis at a point above that, you find that there is almost no radiation at this
32:22point similarly at a point below the dipole axis also you would find that there is no
32:26radiation. So, that is what is reflected in terms of the radiation as a function of elevation.
32:32If the angle is 0 measured with respect to the vertical the elevation angle being 0 size
32:38that the radiation is there is no radiation in the angle 0 no radiation at angle plus
32:43or minus 180 degrees, and the maximum radiation occurs in the plane that is perpendicular
32:49to the dipole axis and that is what you see in terms of the peak of the radiations.
32:55So, this is different from the isotropic radiator, but it has got a very useful property that
33:00in the azimuth you find that the radiation pattern is more or less equal. Now based on
33:06how the antenna is mounted you may find that this Omni directional pattern is sometimes
33:14disturbed and it is slightly deviates from the Omni directional behavior and that is
33:20something that is to be expected because this is this would mean that there is no other
33:25metallic objects or obstruction. So, that you get a perfect pattern, but the property
33:31that we use in most of our cell phone devices is that the antenna should have a Omni directional
33:38radiation pattern and the reason for that is first of all we do not know in which direction
33:44the base station is. So, irrespective of our orientation we want the signal to be able
33:49to reach the base station. So, an Omni directional radiation pattern is something that is very
33:54desirable in the handset of a cellular system. On the other hand when we talk about a base
34:01station as we saw in the last lecture, initially we concept we talked about base stations that
34:08could radiate in all directions, but when it comes to practical deployment we find that
34:12it the practical way to deploy a base station would be to have radiation in a sector, and
34:19typically the sectors that are used in a cellular system is about 120 degrees in terms of the.
34:25So, when we start talking about a directional antenna like a base station antenna, then
34:34we find that we can now talk about a radiation pattern. This is not a Omni directional antenna
34:42it has got a very specific radiation pattern it has got a direction of maximum gain and
34:48then as you move away in terms of angle from the point of on the direction of maximum gain.
34:53You find that the gain eventually starts to drop and we are very interested to find that
34:59angle at which the power drops to minus 3 dB. So, typically we would like to use that
35:07as the main direction of radiation and this region wherethe transmitted gain is between
35:15maximum and minus 3 dB of the maximum as what we refer to as the beam width or the half
35:21power beam width or the range in which they are that angle in which the antenna radiates.
35:27So, typically we would like this to be of the order of 120 degrees for a sectored cellular
35:33antenna. Now, we find that in any practical antenna
35:37you will find that the radiation will extend beyond the main beam width and of course,
35:44we would also there may be also be some side lobes, now radiation outside of the main beam
35:51width and the side lobes are undesirable and we would like to keep that as small as possible
35:57to get the maximum amount of radiation in the direction of interest.
36:01So, this is how the directional antennas are designed and the main difference between a
36:07base station antenna which is directional and a handset antenna which is Omni directional.
36:14So, with this we have a basic understanding of the antennas, what we would like to do
36:21is pick up and look at some additional examples in the context of the system that we are working
36:32with. So, let us go back to our examples and we will take up, we will take up an example
36:41that we can look at and understand how the system works. So, this would be a practical
36:47example that we would like to study.
36:50So, we are told that this is example 2 of our system we are told that we are using a
36:57QPSK modulation. QPSK where each bit or each symbol carries
37:042 bits of information. So, QPSK modulation at a rate of 400 kilo bits per second. Now
37:13QPSK at four hundred kilobits per second says that my symbol rate since I am transmitting
37:222 bits per symbol, symbol rate is 200 kilo symbols per second kilo symbols per second
37:31this is also referred to as the baud rate where the baud rate is 200 kilo baud. Now
37:41this is a basic definition of a system one thing that the baud rate always tells us is
37:46an estimate of what is the bandwidth that you would require to transmit the system.
37:51So, typically a 200 kilo baud or a 200 kilo symbol transmission would require a bandwidth
37:58of the order of 200 kilohertz. So, I have assuming that you have done a proper
38:04pulse shaping at the transmitter then you should be able to fit it in a bandwidth that
38:09is of the order of 200 kilohertz. Now some additional information that is given to us
38:15the received signal power is minus 102 dBm, where also told that the noise figure of the
38:25receiver is 8 dB. So, now, we would like to be able to understand how the system or the
38:34receiver is being designed. So, the question or what we are asked to compute is what is
38:42the received signal to noise ratio. So, in a practical system we would refer to the signal
38:49to noise ratio in terms of the signal power to the noise power. So, signal to noise ratio.
38:56So, what is the signal to noise ratio for such a system having already talked about
39:04the how the antennas work, now we want to understand the impact that it will have in
39:12terms of this system. So, the first step that we would like to do a recall what we have
39:19done in the last lecture would be the following calculation. So, the first step would be the
39:27noise floor trying to estimate what would be the noise floor. So, the noise floor for
39:33the receiver noise floor for the system is given by k times T where k is the Boltzmann’s
39:42constant Boltzmann’s constant as we have introduced in the last lecture Boltzmann Boltzmann’s
39:52constant. And it is given by one and it is value is
40:011.381 into 10 power minus 23 joules per Kelvin. Now we are talking about watts and hertz that
40:17is what we have in our input parameters. So, I want to be able to move from joules per
40:25Kelvin in to things at we can work with. So, in terms of just interpreting the units of
40:31the Boltzmann’s constant it is helpful for us to visualize joules per Kelvin in the following
40:37way, joules is energy which is the power into time. So, I can write joules as watts seconds
40:46per Kelvin, and this can also be written as seconds can be written hertz is one by second.
40:57So, this can be written as watts per Kelvin per hertz. So, now, if I multiply k by T then
41:09what we get is watts per Kelvin hertz multiplied by Kelvin. So, I get watts per hertz as the
41:17noise spectral density. So, noise spectral a noise spectral density noise spectral density
41:27which is k t is given by noise spectral density multiplies k times T this is a calculation
41:39that we you can substitute and obtain with a the 300 Kelvin if you take it at temperature
41:47to be 300 Kelvin then we can get minus 174 dBm per hertz, dBm per hertz basically you
41:59have converted it into a dB expression. Now, the second term that we would like to
42:06incorporate in order to estimate the noise floor would be bandwidth of the signal the
42:11bandwidth of the signal and since we are having a 200 kilowatt system and an equivalent bandwidth
42:18of 200 kilo hertz then we can take the bandwidth to be equal to 200 kilo hertz, which if you
42:26express it in dB will be 53 dB hertz and the third element is the noise figure noise figure
42:39which is 8 dB. So, the noise power that we will encounter in the receiver will be k T
42:51B times F, the noise power is given by k T B F, k T being in the noise spectral density,
42:59B being the bandwidth of the signal F being the noise figure.
43:02So, this is when you are doing it in a dB scale it will be a plus b plus c which is
43:11given by, estimate the compute the answer minus 113 dBm minus 113 dBm.
43:22So, if you now look at the received ratio of the received signal power to the noise
43:32power in dB, this will be we are told that the received signal power is minus 102 dBm
43:42minus 102 dBm minus of minus 113 dBm. So, this gives us a Ps by Pn of 11 dB again remember
43:55it is a ratio of power. So, the answer should be in dB.
44:01Now, a very interesting observation and an extension to what we can learn from this particular
44:08example is to translate Ps by Pn the signal to noise ratio into the parameter that we
44:16are very familiar with in digital communications which is Eb by n naught as we have discussed
44:22in the last lecture Eb by n naught is given by in terms of Ps by Pn is equal to Eb by
44:32naught is equal to Ps by Pn into Tb by Ts or the ratio of the bit duration to the symbol
44:42duration. Now for QPSK symbol duration is equal to 2 times the bit duration because
44:52each symbol are carries 2 bits of information. So, Tb by Ts is a factor of one by half. So,
45:00Eb by n naught is equal to Ps by Pn into one half and if I were to express it in dB, Eb
45:12by n naught dB is equal to Ps by Pn in dB minus 3 dB. So, in this particular example
45:27for QPSK, Eb by n naught would be corresponding to 8 dB and for this I am sure each of you
45:37can look up a standard textbook like Proakis and tell me what will be the bit error rate
45:45at Eb by n naught equal to 8 dB ok. Now, when we look at ber curves we always
45:53are interested in the context of Eb by n naught we are not talking about Ps by Pn, but in
45:57a practical system we measure the received signal power we know what is a noise power
46:02and therefore, what is easy for us to compute this Ps by Pn. Now extending it just a little
46:08bit more for to reinforce the concept now instead of QPSK if the system had been BPSK
46:17would anything have changed in terms of the Ps by Pn assuming that we still had the same
46:27expressions before us what we would find is that the relationship between Eb by n naught.
46:35Eb by n naught for BPSK would be Ps by Pn in dB. So, if you achieved 11 dB Ps by Pn
46:47it would translate to 11 dB, Eb by n naught. So, for a given Ps by Pn signal to noise ratio
46:56in terms of signal power to noise power you will find that the Eb by n naught for a QPSK
47:02system is 8 dB, for a BPSK system is 11 dB there is a difference between the two and
47:08therefore, you would find that the corresponding BPSK system would have a better BER.
47:14Now, this may seem a little counterintuitive because we are always told that QPSK and BPSK
47:19are the same performance keep in mind it is the same performance with respect to Eb by
47:24n naught. Now Eb by n naught translates differently to Ps by Pn when it is BPSK and then when
47:32it is QPSK. Now just for one more one more illustration if it had been 8 PSK. Eb by n
47:45naught in dB would have been Ps by Pn in dB the Tb by Ts now would be a factor of one
47:55by 3 and if you were to converted into dB scale this would translate into minus 4.77
48:05dB. So, keep in mind that this is the way that we would like to understand.
48:12Now, in addition to all of this we now also have to make a note of the following now we
48:19achieved a Ps by Pn of 11 dB that is what we have been able to achieve. Now if you make
48:31the following observations the following are observations regarding the factors that affect
48:40the value of Ps, now if the original system had been designed to have a gain of 0 dB for
48:52example, the gain of the receive antenna to be 0 dB and we achieved a Ps by Pn of 11 dB
48:59let us say the that was how it was designed. Now, if I am able to increase the gain of
49:05the receive antenna Gr dash to 2 dB, a gain of 2 dB on the receive antenna then this would
49:14directly mean that the Ps by Pn would be whatever was the original Ps by Pn the signal power
49:22whatever I received previously will now increase by a factor of 2 dB. So, this would become
49:2713 dB. So, there is a positive impact if you are able to increase the transmit gain or
49:35the receive gain in terms of the received signal to noise ratio. On the other hand if
49:41you would reduced the received gain to minus 2 dB of course, this would naturally be Ps
49:52by Pn would be instead of 11 dB now it would be 9 dB.
49:57Now, you may be wondering what would be a scenario where you would implement a gain
50:03of minus 2 degree, actually we do not deliberately intend to gain introduced a gain of minus
50:082 d B, but what typically happens is that if you have a cell phone and it is designed
50:15with let us say 0 dB antenna and the 0 dB gain antenna. So, when you hold it in such
50:24a fashion that your hands are covering the antenna then you could find that the tissue
50:31can actually cause a further loss of signal. So, effectively the gain of the antenna instead
50:37of being 0 dB becomes minus 2 dB. So, this is not by design it could be by because of
50:44the way it is used that the signal. So, we need to keep in mind that when we design the
50:50system we do have to keep a close watch on what is the gain that we are able to achieve,
50:58this gain can be in terms of the transmit antenna it can be in terms of the receive
51:03antenna and we also are very keen to monitor what is the path loss because what is between
51:10what is transmitted and what is received the path loss makes a big difference.
51:16Now, to conclude today’s lecture, I would like for you to look at the following example
51:24and maybe once you can have a chance to read it we will pick it up from there in the next
51:30lecture.
51:31So, this is also taken from Molisch chapter 3 this is example 3.2 and by the way the previous
51:45example that we had looked at was example 3.1 in Molisch the second one is 3.2 is the
51:52expression that we have. So, the problem that statement are that I
51:57have a signal that is at a carrier frequency of 900 megahertz, the bandwidth of the signal
52:06is 25 kilohertz and we are given that there is a transmit gain of 8 dB and a receive gain
52:19of minus 2 dB, and there are several losses in the system we call that as the losses in
52:33the system and that is of the order of 2 dB, the noise figure of my receiver is 7 dB and
52:44what we would like to achieve is a Ps by Pn of 18 dB we would like to achieve a signal
52:55to noise ratio and we are told that we can take the d break as a distance of 10 meters
53:05with a path loss exponent of 3.8.. Now, the question that we would like to answer
53:13which is a very important question is what is the transmit power under these assumptions
53:21that we have expressed or give explained in this example what will be the transmit power
53:28that will ensure that we achieve a Ps by Pn of 18 dB with these conditions that are mentioned
53:36here. Now to make it a little bit more realistic we would also like to look at what is the
53:44transmitted power that will achieve this with a margin with a fading margin of 10 dB.
53:56As we discussed the role of a fading margin is to make your systems more robust. So, when
54:02we have a fading margin incorporated into the design, we would have to transmit with
54:07higher power. So, we want to what would be this with a fading margin of 10 dB. So, again
54:14these are 2 questions that we would like to answer. So, what I would like to request you
54:19is do read Molisch chapter 9 to understand about antennas and chapter 3 where examples
54:303.1 examples 3 point worked examples 3.1 and examples 3.2 to understand link budgets system
54:39design fading margin and how all of these will work together to help us understand and
54:46build robust wireless systems. Thank you very much.

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