Available Noise Power and Amplifiers

  • Thread starter Runei
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  • #1
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Main Question or Discussion Point

Hello there,

I'm working with the textbook by Wim van Etten, "Introduction to Random Signals and Noise". And right now I'm preparing for an exam in a stochastic processes course.

I have a question regarding some noise calculations, that I just can't wrap my head around. I hope someone will be able to shed some light!

The situation
So we have a noisy resistor of value R. By using approximation we can say that the RMS voltage across this resistor due to thermal noise, is equal to (Vr is assumed to be the rms voltage).

[itex]V_r = \sqrt{4kTRB} [/itex]

And here B is of course the bandwidth of whatever measuring system you use to determine the voltage. So far so good.

The noise spectral density of the noise voltage is

[itex]S_r= 2kTR [/itex]

Now we connect this to to an amplifier (noiseless to begin with). The amplifier has an input resistance of Ri, and output resistance of Ro, a frequency response of H(ω) and connected to a load resistor of RL. All resistors other than R are assumed to be noiseless.

Now, what I would do then is to say:

The noise spectral density at the input is then given by

[itex]S_i = S_r \frac{R_i^2}{(R_i + R)^2}[/itex]

The output spectral density (not across the load by the voltage generated before the output resistance), is then given by

[itex]S_o = S_i \cdot |H(\omega)|^2[/itex]

Looking then at the spectrum seen across the load resistor we have

[itex]S_L = S_o \frac{R_L^2}{(R_L + R_o)^2}[/itex]

The totality of this becomes the following

[itex]S_L = S_r \frac{R_i^2}{(R_i + R)^2} \frac{R_L^2}{(R_L + R_o)^2} |H(\omega)|^2 [/itex]

This is all well and good, however, what I don't understand is why he goes ahead in the book and want to create the maximum power transfer (real power).

As I see it, if the impedances are matched, the spectral density at the loads go down, as opposed to the case where the input impedance was set to infinity, and the Load resistance was also set to infinity. This would give an indication of maximum possible noise voltage we could see at the output, due to that single resistor (or noisy circuit with an equivalent temperature of T and equivalent resistance of R).

I know that if we DO match the impedances, the R dissapears, but why is this a "smart" thing to do? Don't we wanna analyze situations in which the noise signal we get out is the maximum, to see what the worst case scenario is? Or am I completely missing something?

I know it's nice that have the available noise power as

[itex]P_a = kT/2[/itex]

But this is only the case of matched impedances, which make the signal become smaller, and hence, create a smaller output noise than in the case of a larger impedance...

Am I ranting? Or am I making sense? o_O

My general problem is that I don't see why we go ahead and assume matched impedances.

Thank you for any help you might give!

Best regards,
Rune
 

Answers and Replies

  • #2
18,086
7,510
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
meBigGuy
Gold Member
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I'm not sure, but I expect real world applications are generally more concerned with noise power than noise voltage. Assuming a noiseless matching resistance is less "unreal" than assuming a noiseless infinite input impedance and load.
 

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