So now we are considering some external noise

which is incident from left hand side and altogether we want to express the system by

some equivalent noise temperature then how to calculate it So we have a 2 port network which is providing

some internal noise and it is also connected to some external source of noise which can

be given by T external So in that case then the total noise power at the output N s this

is equal to k T external by L this is for the contribution from left hand side plus

T physical into 1 by 1 minus L whatever we have seen we can represent it by some effective

noise temperature T s k Ts so N s this is equal to k T s Now consider one antenna so here you can see

here so N s it is equal to k into T s Now consider 1 antenna which is connected to a

transmitter now antenna it will have some efficiency Let us say antenna efficiency is

given as 90 percent then and if any power we feed to antenna it will be attenuated by

antenna and uhh this value it can be given by that 90 percent if we convert it to fraction

0 point 9 Or we can say L loss of the (anten) due to antenna this is 1 by Rho or 1 by 0

point 9 Then T physical is the antenna temperature let us say so this is the physical temperature

of the antenna and T external is the noise temperature of the transmitter just before

antenna So in that case we can calculate then what is noise power available from this antenna

and source which can be given by k T s Let us consider a receiver now and we are

going to calculate the overall receiver noise temperature So what we will be doing let us

first consider one amplifier a 2 port network again So the amplifier amplification factor is given

by G a it is sometimes called the available gain and T e is its effective noise temperature

Now it is connected to some external resistance which is given by R L at right hand side and

left hand side it is connected to a source of internal resistance R s And if the source

power is 0 in that case uhh the power which is incident from left hand side it is only

the noise power Also the input and output resistances they are also given for the amplifier

Now the noise voltage available uhh because of the noise power generated by R s that can

be given by e s equal to square root of 4 k T R so maximum noise power delivered to

a mesh load we are considering one mesh scenario that is P i this is equal to e s by twice

R s whole square into R s So you see here what we are considering so

R s this is the noise source I want to extract noise power from the source or we can see

in other way We are connecting a resistor here so some noise power will be delivered

to the resistor then we are considering what is the maximum noise power that is being delivered

to this resistor and that happens according to maximum power transfer theorem when this

R this is equal to R s or we can write down this is equal to R s So in that case if I

want to calculate P i this is simply I square into R then the current component went through

it this is e s by twice R s whole square multiplied by R s or we can write down this is equal

to e s square divided by 1 R s is cancelled out so 4 R s and this is equal to k T which

already we have seen Now gain of the amplifier G a we can express

in terms of the input (noid) noise power and output noise power So output noise power is

e 0 square divided by 4 into R out and the input noise power this is e s square divided

by 4 into R s So one interesting thing then Gain you see for C amplifier also if we want

what to calculate the overall gain of any C amplifier uhh it depends on R R l but simply

if I say what is the gain of a C amplifier we do not consider R l we take just the open

circuit at output voltage and then calculate what is the gain of the amplifier Now if you

change R l obviously the available power to R l it will change but when we call the G

a or gain of the amplifier then we do not consider R l but this gain it depends on the

source register R s So here also then this G a it depends on R

s but not on R l So output noise power if I want to calculate at output terminal let

us say we are denoting it by N 0 so that is equal to G a k T s so k T s this is the incident

power from left hand side which is being multiplied by gain of the amplifier G a In addition to

this the amplifier itself it will produce some noise which is given by N internal N

internal this factor we can also express in terms of T effective noise temperature which

is given T e so we can also write down that this is equal to G a k T s plus N internal

this is equal to G a k into T e or this is equal to G a k T s plus T e Let us represent it by effective temperature

G a k T 0 so T 0 this is called the operating temperature operating noise temperature So

you see here so this is the amplifier and left hand side we have a noise source connected

to this amplifier and right hand side we have a load resistor connected here then we calculate

it we expressed already a G a in terms of the input and output noise power and from

that we calculate that we calculated the output noise power from the device so output noise

power what we see that it contains uhh the contribution due to the external noise source

which is G a k T s In addition to that this amplifier itself produce some noise which

is given G a k T e so all together this effect we can represent by some operating noise temperature

T 0 so N 0 then that is equal to G a k T 0 So T 0 T e G a all functions of R s and but

they do not depend on R l And in this case also we assume that the output

R l is mesh to this and so it is the worst case scenario this N 0 this is the highest

available noise power from the amplifier on the noisy source Now let us consider several

2 port components they are connected in Cascade so that means what we are considering let

us say we have 3 amplifiers the first one gain is given by G a 1 and the effect noise

temperature effective noise temperature is given by T e1 It is followed by another amplifier for which

gain is G a 2 effective temperature is T e 2 Again we have one more G e 3 effective noise

temperature T e 3 and we want to calculate what is the overall noise contribution of

this system so if we know for 3 we can do it for N components in Cascade Then N internal

we can write down this is equal to so you see this first amplifier whatever noise it

will produce let us say that is given by N internal 1 so this N internal 1 it will be

amplified by G a 2 further it will be amplified by G a 3 but whatever noise produced by the

third component this component is not being amplified by left hand that is G a 1 or G

a 2 Then the total noise contribution N internal for the first one this is G a 3 G a 2 into

G a 1 k T e 1 So k T e 1 this is the noise produced by the

first one and it is being amplified by all these 3 amplifiers Then for the second one

G a 3 G a 2 into k into k T e 2 and for the third one this is G a 3 k T e 3 So if we represent

it this overall combination if we contribution if we represent it by some effective temperature

let us say T e in that case we can write down this is equal to k T e is the overall equivalent

noise temperature then k T e into G a 3 G a 2 G a 1 so therefore T e this is equal to

N internal divided by k G a 3 G a 2 G a 1 So what we are assuming as if one resistor

of equivalent noise temperature T e is connected to left of this system and system itself is

noise less so that is why we are multiplying by 3 gain factors G a 3 G a 2 and G a 1 So in terms of the individual noise temperature

we can also express it so you just put that value of N internal whatever we obtained in

this first equation So if we put this value here so it becomes T e 1 plus T e 2 by so

remaining term is G a 1 plus T e 3 remaining terms G a 1 G a 2 For N components if we have

N number of components simply we can extend it so it will become for Nth term T e n and

below we have (G 1) G a 1 G a 2 to till G a (N 1) so this is called the Friis law for

cascaded system So what we see that if we have many Cascade

connections the contribution of noise from the first component it is the it is dominant

over all others That is why when we design any receiver the first component is chosen

to have which will have minimum contribution of noise so that is why we use low noise amplifier

typically which will provide very low noise and after that whatever components we are

having their contribution is smaller compared to the first component So now let us consider

a receiver it is modelled by a constant gain G 0 over a band B n we are introducing now

some bandwidth given by B n Then the receiver input power is given by P s and it receives

a noise power P n from the antenna then considering matched scenario so always by default we will

be considering matched scenario so whatever power is coming from left hand side so it

is being observed right hand side half of that observed in the load So then SNR SNR at the output of the receiver

that is equal to you see this is G 0 SNR is signal to noise ratio so signal at left hand

side is P s it is being multiplied by G 0 so out at output signal available signal power

is G 0 into P s And what is the output noise power so left hand side we have receive P

n noise it is being amplified by G 0 times by the receiver chain plus we have some internal

noise contribution because of the receiver cover bandwidth B n B n multiplied by n internal

and it can be given by equal to SNR at the input divided by 1 plus B n N internal divided

by G 0 P n How we uhh come here simply you divide both denominator numerator and denominator

by G 0 P n so then the top one it becomes P s by P n which is SNR at the input So this equation is called Radar equation

for receiver it is one important quantity parameter Now considering available power

gain G a and this above definition we can also express SNR at the output so in terms

of operating temperature this is equal to G a into P s divided by G a k T 0 B n so this

is the operating temperature and simply then this is equal to P s by k T 0 B n so you remember

recall this T 0 is the operating temperature Now noise bandwidth so what if uhh we are

many times calling noise bandwidth B n and we are using it to calculate overall noise

power then what value should we take should we take for B n for any given channel Let

us say uhh we are using a channel uhh bandwidth 1 gigahertz so should we take simply B n equal

to 1 gigahertz So whatever definition the significance of

B n till now we have used it is like one simple window function which has some value over

the given bandwidth and out of this band it has 0 value but it is not the practical scenario

So let us consider any practical case so in this expression when we use some value of

B n so that means let us say we are considering the ideal (scenario) scenario this is the

1 gigahertz channel bandwidth it is given B n and then model it by a constant gain G

a when we uhh talk about the gain of the amplifier or gain of the receiver G 0 whatever and out

of this band this G a is exactly 0 But for any practical system consider any uhh band

pass filter or anything so for that it is not constant but it becomes a function of

frequency and we have contribution due to you see other frequency components so it will

it will vary So gain G a it is a function of frequency

and bandwidth B n then what value we should take for this type of practical scenario so

what we do here there is a way of calculation For a real system obviously it will be defined

by its transfer function H which is a function of j omega and the corresponding temperature

which previously we consider constant over the frequency but practically it should be

a function of frequency Then actual output noise power we should consider all the frequency

bands starting from 0 to infinity and we integrate it over 0 to infinity H j Omega square k T

0 d f Now if T 0 is constant over the band and let us say mid band gain is given by G

G a equal to H j Omega 0 square in that case so we are assuming T 0 is constant otherwise

it is becoming complicated for simplification this assumption So B n this is equal to then total noise power

P n divided by k t 00 G a or 0 to infinity this integration value divided by H j Omega

0 whole square So from this expression we can calculate B n value but even then you

see here we consider T 0 is constant over this given bandwidth Next noise factor so

total noise contribution it can be calculated in terms of noise temperature also sometimes

another parameter is used to quantify the noise contribution that is called noise figure

we can calculate it for any given 2 port devices so usually if it is a lossy device with some

insertion loss let us say L in dB and then noise figure in dB it is simply becomes becomes

L If you buy any active component like an amplifier or some active mixer usually the

noise figure value will be written or given by the manufacturer it can be also measured

in uhh labs by using in lab experiment by using spectrum analyser and standard noise

source values or VNA Vector Network Analyser So in general then what is first let us define

noise factor of a uhh 2 port network Noise factor F this is the ratio of signal to noise

ratio at the input terminal divided by signal to noise ratio at the output terminal so SNR

at the input divided by SNR at the output Now signal to noise ratio obviously is better

at left hand side compared to right hand side at the output why Because at output side we

have additional noise generated by the component itself so that is why F this is always greater

than 1 If we express it in dB we call it noise figure and that is always a positive quantity

so for a given system then already we did it let me express mathematically So then the noise factor F this is equal to

SNR at the input terminal divided by SNR at the output terminal so at the input terminal

let us say the power already whatever we assumed in previous example P s is the incident power

and let us say N i is the incident uhh noise from left hand side we are considering per

unit hertz So this is the SNR at the input side now we can calculate SNR at the output

side so then the available power is the application factor G a multiplied by P s divided by total

noise total noise is some noise internally generated inside the 2 port device plus the

contribution from left hand side which is G a into N i So this is equal to then G a

k T 0 plus N internal divided by G a k T 0 Now it is a function of temperature you see

what value we should take for T 0 so if the noise from my left hand side it changes obviously

this value will change So then uhh if we buy let us say one amplifier

from market some noise figure or noise factor is given with that then what it denotes Because

it becomes a function of input noise So usually a standard temperature is considered to calculate

noise factor and hence figure that is 290 Kelvin So if I do not consider simply actual

scenario practical scenario by noise factor we mean the noise contributed by the component

it has been calculated at a given T 0 equal to 290 Kelvin So from this then you see if

I put N internal that is given by G a k T e T e is the equivalent noise temperature

so then it becomes simply 1 plus T e by T 0 So T e this is the effective noise temperature

of the 2 port device itself T 0 uhh this is for standard calculation we will consider

290 Kelvin Then we can express T e also in terms of F

so T e equal to F minus 1 into T 0 this is another important expression So if F is given

we can calculate what is the effective noise temperature T e if T e is given we can calculate

what is the noise factor F so they are related by this important relationship This noise

factor depends on source impedance and operating frequency obviously if we have a lossy network

as I discussed it will provide you some insertion loss or power will attenuated by that 2 port

network If the loss is given by L then simply its noise factor is equal to L And now we

are considering a different scenario let us say this source temperature is not T 0 it

is given by T s whatever we considered in the previous case Okay so let us take a break

then uhh we will consider a practical scenario with a given temperature T s thank you