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