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So next we are going to discuss about 60 gigahertz
channel and presence of noise So from the previous slide what we have seen that because
of this higher frequency we have very high attenuation due to path loss and it is 28
dB higher compared to 2 point 4 gigahertz And if I compared this value because of attenuation
due to atmosphere that is only 10 to 12 dB per kilometre so much higher compared to that
value So channel performance at 60 gigahertz the
first problem we will be facing is high path loss value then multipath fading effect we
have human shadowing effect Human shadowing effect it can be as high as 20 dB so if let
us say there is a link and any human is now present between the transmit and receive antennas
then the received power it can be it can be it can degrade by even as high as 20 dB so
that we have to keep in mind Then non line of sight propagation that is due to multipath
non negligible Doppler shift even at pedestrian velocities because we know Delta f change
in velocity Doppler shift change in frequency that is Delta V by c into f 0 and also we
have effect of noise So this noise it becomes an important issue we will discuss in details
about this noise there are many sources of noise thermal noise it can be shot noise Flicker
noise blackbody radiation So what is blackbody radiation Because of
the temperature of different objects they are constantly radiating we can model them
by blackbody radiation curve Plancs law And antennas receiving antenna it will collect
this radiation as well and it is a undesired signal and noise for the system Then theoretical
maximum rate at which error free digits can be transmitted over a bandwidth limited channel
in the presence of noise it can be given by Shannons law it is channel capacity C it is
equal to bandwidth of the channel multiplied by Log you note down here base is 2 log of
1 plus received power P rx divided by bandwidth multiplied by n 0 n 0 this is noise per hertz
so total receive noise is bandwidth multiplied by the n 0 value small n 0 Approximately it
is equal to 1 point 44 divided by receive power by total receive noise power here we
are considering only the effect of thermal noise so N 0 it represents thermal noise We can also call them this P rx by capital
N 0 this term SNR of the receiver signal to noise ratio at the receiver So this formulation
when we assume that bandwidth is sufficiently high fence to infinity So SNR it becomes one
very important quantity channel capacity it depends on this value and already we learned
how to calculate SNR in presence of noise so ratio of the signal power to the noise
power at the receiver in dB is SNR this is P t G (tra tra) G of transmitter G of receiver
minus now the path loss we are dividing it into 2 components PL 0 and PL r why PL 0 it
represents the path loss at 1 meter and then after that we have a different coefficient
L for path loss calculation we call it relative path loss at armature So for the first 1 meter
we will consider n equal to 2 in that path loss formulation that means 20 log 10 4 Pie
r by Lambda And for the next part after 1 meter we have
to experimentally determine what is the value of L usually varies between 3 and 5 which
we discussed earlier So now why first for first 1 meter we are using Friis equation
directly If we have any transmitter it can be placed uhh on top any on top of any pole
outside or inside even any room let us say it is placed at any one corner of the room
so at least for first 1 meter we can consider it is arm obstructed and we can use the Friis
So after that we will be considering the effect of all these multipath effect fading effect
human shadowing effect and we will consider a different value of L so that is why we are
dividing this path loss components into 2 parts First part due to the first 1 meter
from the antenna and second part after 1 meter Next is the IL we call it the implementation
loss so it represents all the losses in cables antennas et cetera Then we can calculate the
noise contribution we are calculating SNR so we are subtracting now the noise component
now there are different sources of noise for any given device If we have antenna antenna
it has metallic part dielectric part it will add some noise if we have cable again inside
cable it will add some noise if we have amplifier it will also add some noise so the noise contribution
and how to calculate the overall noise we will discuss after this in detail So sometimes
some devices it comes with a noise factor so noise factor is nothing but it is a ratio
of SNR at the input terminal to SNR at the output terminal of the device if the noise
factor is given we can directly put the value here So now the thermal noise it depends on the
effective noise temperature of the system and the bandwidth If we increase the bandwidth
of the system we will pick up more thermal noise so we have to then subtract these 2
components one due to thermal noise and another one which is coming from the active devices
which is given by the noise figure of the device So then in the expression we have minus
10 Log 10 KT B we are using decibel scale minus noise figure we are assuming noise figure
is given in decibels Once we have this information we can calculate then SNR signal to noise
power at the receiver and we can calculate then what is the channel (capashit) capacity
of that particular given channel Let us take many medical example let us say
transmit power is given as 10 dBm so 10 dBm in milliwatts it is 10 milliwatts Noise figure
this overall noise figure is already calculated and given 6 dB implementation loss IL is given
6 dB thermal noise you can calculate K into T it comes 174 dBm per hertz Bandwidth of
the system 1 point 5 gigahertz and distance between the transmit and receive antennas
20 meter and the path loss at 1 meter which is PL 0 if I put the values at 60 gigahertz
and consider N equal to 2 in that expression of path loss it comes 57 point 5 dB Now here
is the plot of channel capacity in bits per second versus distance so for the specifications
we will consider are equal to variable and we are changing it from let us say 1 meter
to 20 meter and we are considering different scenarios for indoor models Inside office for line of sight communication
n that path loss index in Friis equation it was n equal to 2 Here inside office we are
considering n equal to 1 point 77 and for Non line of sight communication it can be
via reflection n equal to 3 point 85 we have higher loss so these are some experimental
values we are considering them indoor communication inside let us say office building Now let
us see how the channel capacity to varies with distance for this given specification
So this black line solid line it represents free loss according to Friis equation And
we see if I increase distance then nice power it does not change but receives power it changes
it decreases So accordingly channel capacity it decreases so at 20 meters a day somewhere
in between 10 to the power 8 to 10 to the power 9 bits per second even though bandwidth
is given 1 point 5 gigahertz And you see for line of sight communication
indoor communication n equal to 1 point 77 more or less it following the curve for n
equal to 2 but for non line of sight communication when n equal to 3 point 85 if I follow this
blue line it decreases to 10 to the power 6 to 7 somewhat in in between this value at
20 meter so you see the effect of noise and distance If we have higher n then SNR decreases
as a result channel capacity also decreases Now we are considering another scenario here
we are plotting channel capacity article channel capacity versus antenna gain so what is the
minimum antenna gain required to maintain a given channel capacity of 5 GBPS So 5 GBPS
is my target so for that we need to maintain a typical SNR value and we are considering
let us say the separation between the transmit and receive antenna is 20 meters So then SNR if I consider the specifications
given above only thing is that will change the receive and transmit antenna gain and
we are assuming receive and transmit antenna they have the similar gain the same same gain
value So if we need to maintain now this 5 Gbps data rate at 20 meter we have to we have
to maintain the gain the receive or transmitter gain value the typical gain value given by
these curves For example let us say n equal to 1 point 77 this black daughter line so
for 5GBPS channel capacity antenna gain required at least a somewhat this is 15 30 so 23 dBi
both for transmit and receive antenna Now if n equal to 3 point 85 for non line of sight
communication minimum antenna gain it increases to 37 point 5 dBi if we need to maintain that
5 GBPS data rate at 20 meter So here we are considering that we do not have any control
over transmit power and other things only thing we can change the gain of the antennas
And this plot shows then what should be the minimum gain of the antennas to maintain this
5 GBPS data rate at 20 meter separation A typical 60 gigahertz measurement setup it
can be anything for this example we are considering let us say we are going to characterise one
antenna which is uhh separated from the transmitting antenna at different distances We are considering
transmitted power is 0 dBm that means 1 milliwatts Attenuation loss of the coalition cable 6
point 2 dB per meter at 60 gigahertz conversion loss of the mixer so you see in a typical
system at millimetre wave frequencies we know that at millimetre wave frequencies the component
cost is very high so for any given system then what we do we try as soon as possible
to down convert the frequency So just after antenna it may be a band pass filter or a
low noise amplifier then we use a mixer to down convert the frequency If you look at
any practical instrument for example let us say spectrum analyser inside the first stage
it might use one mixer to down convert the frequency So once it is down converted to lower frequency
then we can use cheaper low frequency component So in this example also we are using a sub
harmonic mixer for down conversion and mixer it comes with some conversion loss so if you
feed some power to mixer which is millimetre wave power when it will down convert to microwave
frequency or RF frequency (down) then the power available for that RF or microwave signal
it will be much less compared to that millimetre waves frequency So for this one the loss or
because of this change in frequency (the con) which you call the conversion loss of the
mixer it is 40 dB so RF power available it is 40 dB down compared to the millimetre wave
power at the left hand side So this is a typical value for a passive sub
harmonic mixer we have some different categories of mixer for them sometimes it can be improved
to even let us say 0 dB noise figure since conversion loss is 40 dB noise figure is also
40 dB and VSWR of the mixer is given by 2 point 6 is to 1 and noise floor of this spectrum
analyser is (130) minus 130 dBm so this spectrum analyser can sense power till minus 130 dBm
at the rate of 60 gigahertz Then what is the maximum and minimum power we can measure by
using this system The minimum power it depends on SNR again we have to calculate SNR for
the system and since the device under test or antenna under test it is connected to the
sub harmonic mixer so we cannot change this one so only control we have only transmit
antenna gain or maybe we have we can increase the transmit power so we will see then for
different total antenna gain what is the minimum power we can sense by using this spectrum
analyser at different distances And what limits the maximum power Maximum
power it is limited by nonlinearity of the device when we design any system it can be
microwave system or millimetre wave system even a low frequency amplifier we consider
it as a linear device so that is why a low frequency amplifier sometimes we call small
signal equivalent circuit or small signal amplifier Now if I increase the power input
power in that case it can move to its non linear region and then we have to model it
in different ways and it will be associated with other high frequency components So maximum
power would be then limited by the nonlinearity of the devices are being used in the system
Here we are considering the minimum power and maximum power difference so it is calculate
it for this data set so dynamic range for this measurement setup then it varies with
antenna gain and distance because of the different SNR value So at 10 meter you see when the antenna gain
is less than let us say 6 or 7 dBi it cannot sense anything you do not have any dynamic
range To sense anything it starts from approximately 7 dBi and now if I keep on increasing the
antenna gain so in that case this dynamic range will increase so maximum to minimum
power that can be dynamic range it is the difference between the maximum and minimum
power that can be sensed by this system Now obviously we decrease the distance let us
say to 0 point 5 meter in that case we have some dynamic range even 0 dBi gain and if
I increase the antenna gain then dynamic range increases So when we are going for any measurement or
any antenna or any other things that time also we have to keep in mind what about the
what is the SNR and because the dynamic range it also depends on SNR of the receiver So
here you see for non line of sight communication and extra 30 dB fade margin is suggested to
compensate severe fading effects because of multipath human shadowing et cetera for this
plot reconsider line of sight communication Next link budget typical we are considering
for indoor model So link budget it is basically signal power
plan under the given condition You do not have any control over noise figure and the
noise contribution then depending on the receiver to transmitter maximum separation what should
be the minimum antenna gain value what should be the minimum transmit power so all these
are calculated using this link budget So the calculation let us say we are considering
at 60 gigahertz it starts with the path loss calculation Now for the first 1 meter we will
consider we will consider the path loss it is 20 log 10 4 Pie f c by C so if I put f
c equal to 60 gigahertz it comes 68 dB Average nice power per bit N this is equal to minus
174 10 log 10 of R b So this 174 how it comes you see thermal noise at the input it depends
on the noise temperature T and the noise bandwidth B kTB Now if we put the value of k and T and calculate
thermal noise power per hertz then you put k equal to Boltzmanns constant that is 1 point
381 into 10 to the power minus 23 watt per hertz per Kelvin And if nothing is specified
we will consider T equal to 290 Kelvin it is called the room temperature support the
values here then it will come 4 point 005 into 10 to the power minus 21 watt per hertz
and if I convert it to decibel scale dBm minus 174 dBm per hertz So B bar we are calling
normalised bandwidth of 1 hertz Now if we have a rate R b then we need to add the contribution
because of this bandwidth and it is 10 log 10 of R b R b (GB) this is the system payload
bit rate So then average noise power per bit in dBm P N this is equal to N plus R x noise
figure in reference to the antenna terminal Tolerable path loss P L the previous formula
in addition to that we are considering M shadowing that shadowing effect so for human body it
can be 15 to 20 dB so depending on scenario we have to put a suitable value for M shadowing
and here S is the minimum signal to noise ratio for E b by small n 0 for the additive
white Gaussian noise noise channel And I L this is the implementation lost in dB including
filter dispersion phase noise if there is any frequency error so everything is taken
into account in IL then we can calculate total loss from here Then maximum operating range d it can be given
by 10 to the power PL divided by 10 into n where n is the path loss exponent So for ideal
case in free path in free space n equal to 2 but for indoor model n depends and it varies
from room to room we have to use some experimentally fit value for small n Here are some examples
suggested for IEEE 802 dot 15 dot 3c standard For line of sight scenario consider 68dB path
loss exponent n equal to 2 and M shadowing just 1 dB this is called shadowing link margin
But for non line of sight scenario path loss at 1 meter same value 68 dB consider path
loss exponent small n equal to 2 point 5 and shadowing link margin equal to 5 dB so these
are some suggested value for this IEEE standard Okay so we will take a break then we will
calculate link budget for a given system thank you

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