hey guys, we are exploring noise in cascaded circuits in telecomms at the moment and this business of a matched source is righteously confusing me!

Can I begin by stating that the general formula for noise in a cascaded circuit is easy to understand and I have no problem with it.

Perhaps I will explain my problem with an example question that we were given:

A TV receiver is connected to an antenna of effective noise temperature Te = 290 K via a coaxial cable of insertion loss L = 6 DB.
The coaxial cable is assumed to be matched to the receiver and the antenna. The receiver has a noise figure F2 = 6DB and a frequency bandwidth B = 5.5 MHz. The received sigbnal power by the antenna Si = 3.6 nW.

a) prove that the insertion loss (L) and the noise figure (F1) of the matched cable have equal values.

there is more to this question but part a is what i have a problem with

In answering this question the input is said to be a matched source and therefore have an input noise power equal to ktsB where k is Boltzmann's constant and B is the bandwidth of the signal. The added noise power is referred to as kTeB where Te is the effective noise temperature.

The noisy circuit is modelled by a noiseless circuit with the added noise being referred to as kTeB and multiplied by the noisy circuit's gain.

therefore the noise figure can be said to equal 1 + Te/Ts

The solution to part a) reads as follows:

The addd noise power Na is replaced by an equivalent noise power at the input kTeB added to Ni, and the noisy cable becomes a noiseless cable. The cable is also considered as a matched source to the receiver, thus the output noise power of the noiseless cable is : No = Ap1KB(Ts + Te) = kBTs

Therefore Ap1(Ts+Te) = Ts
therefore 1 + Te/Ts = 1/Ap1
where Ap1 is the gain of the circuit

This line is at the heart of my confusion:
Therefore Ap1(Ts+Te) = Ts
Because the cable is a matched source it cant introduce more noise than the standard noise?? How does the line being matched guarantee this?


Thanks for the help!

 

hey guys, we are exploring noise in cascaded circuits in telecomms at the moment and this business of a matched source is righteously confusing me!

Can I begin by stating that the general formula for noise in a cascaded circuit is easy to understand and I have no problem with it.

Perhaps I will explain my problem with an example question that we were given:



In answering this question the input is said to be a matched source and therefore have an input noise power equal to ktsB where k is Boltzmann's constant and B is the bandwidth of the signal. The added noise power is referred to as kTeB where Te is the effective noise temperature.

The noisy circuit is modelled by a noiseless circuit with the added noise being referred to as kTeB and multiplied by the noisy circuit's gain.

therefore the noise figure can be said to equal 1 + Te/Ts

The solution to part a) reads as follows:


This line is at the heart of my confusion:
Therefore Ap1(Ts+Te) = Ts
Because the cable is a matched source it cant introduce more noise than the standard noise?? How does the line being matched guarantee this?


Thanks for the help!

I haven't tried to unravel your derivation, but consider that the matched source gives kTsB noise power. When you connect your cable to the source, if the cable is also at a temperature Ts then when you look into the end of the cable you see a matched source at a temperature Ts, so the noise power must also be kTsB. When you make your model give you this result you get F=L.

The noise from a 50Ω resistor at a particular temperature is fixed. If I give you a box with two terminals that measure 50Ω and contains only passive components, the noise you measure should be only a function of the box temperature. You can't tell if inside there is a simple 50Ω resistor across the terminals or if there is a long length of 50Ω coax terminated in 50Ω.

 

Hi and thank you for the response. I have been thinking about what you were saying and yes, I agree that if the channel or circuit that the signal is traveling through has the same noise temperature as the input device then it makes sense that the output noise will be the same but my question is how do we know that the channel or circuit has the same noise temperature?

And in the line mentioned in my first post, the formula here assumes that the the effective noise temperature of the circuit and the input noise to the circuit is additive and somehow equals the standard noise density!?

Therefore Ap1(Ts+Te) = Ts

How do we know that the noise temperature of the circuit is standard just because it is matched?

 

Hi and thank you for the response. I have been thinking about what you were saying and yes, I agree that if the channel or circuit that the signal is traveling through has the same noise temperature as the input device then it makes sense that the output noise will be the same but my question is how do we know that the channel or circuit has the same noise temperature?

And in the line mentioned in my first post, the formula here assumes that the the effective noise temperature of the circuit and the input noise to the circuit is additive and somehow equals the standard noise density!?

Therefore Ap1(Ts+Te) = Ts

How do we know that the noise temperature of the circuit is standard just because it is matched?

You seem to be confusing the temperature of a device with noise temperature which is an idealised model parameter.

We don't know if the cable is at the same temperature as the source, all I've said is that if it is then the loss equals the noise figure. As you change the temperature of the cable the noise temperature and noise figure change. A very long length of 50Ω cable looks like a 50Ω resistor, if it was held near absolute zero it would be almost noiseless.

Off the top of my head, I think that you can show that the effective noise temperature of a passive attenuator / cable at temperature Tc with power gain G (<1) is
Te = (L-1)Tc where L = 1/G

so if Tc = Ts then F = L

 

Ok i have spent many hours trying to fully understand this idea of the noise figure. Can i just clarify something to confirm my understanding of this. In the question i asked above, we assumed that the cable was a matched source to its load and as such supplied the load with noise power or kTsB such that the output noise power of the coaxial cable: Ap1(Ts+Te) = Ts. I am assuming this only applies to a passive circuit as if Ap1 was a gain then the output noise power of the network would have to be more than Ts right? Such that the output would be Ap1(Ts+Te)

I had posted this recently but I asked my lecturer about this as well and he confirmed that it doesnt matter if the circuit is passive or not and that the same formula will apply. I have to stress again the the formulae and derivation for the noise figure I understand perfectly as they make perfect sence and I understand what is meant by the effective noise temeperature of a network as well.

What I DONT understand is the use of these concepts in the extract from a book "telecommunications engineering" by Chapman. In the extract mentioned and supplied in this post, the insertion loss of a matched passive network is proved to be also the network's noise figure. Again the only line causing me premature hair greying is that we assume that the output noise from the network is KTs regardless of the noise at the input and how noisy the network is itself. All descriptions of the noise figure and of effective noise temperature I have no problem with whatsoever and I understand the need to standardise the noise input of a network so that the noise figure will have a meaningfull value but I believe that when dealing with the noise in one particular network in the cascade, then surely the output noise will be Ap*Ni + Na. I fail to understand how we can just let the output of this network equal KTsB just because it acts like a matched source for the next network???

Any explanation of this would be greatly appreaciated!!!

 

Ok, just in case anyone was confuzzled by this, I got an explanation.
The formula: Ap(KTsB + KTeB) = KTsB only applies to PASSIVE circuits with no amplification. (only attenuation) I wasted so many hours on this purely because I was given the wrong information!!!!

 

...when you look into the end of the cable you see a matched source at a temperature Ts...

Why is that? Isn't a practical coax also resistive? What assumptions have you made then?