Find L such that 70% of the noise power is dissipated in R

[Gweniiix]

I've been given the following homework exercise and I really need some help with part C.

Question1.

In figure 1. (attachment), an LR-lowpass filter is given with L = 120mH and R = 18 kΩ.
This filter is characterized by the transfer function: H(f) = V_uit(F)/V_in(f)
Determine the following paramters for this filter:

A) The 3dB bandwidth.
I found the answer to this question as follows: R/2∏L = 23.9 kHz

B) The equivalent noise bandwidth.
The answer to this question is: R/4L = 37.5 kHzNow let the input signal of the lowpass filter be a bandlimited noise signal with power spectral density (PSD):

P_n(f) = 1 for f < 140 kHz
0 for f > 140 kHz

C) Determine a new value for L such that 70% of the total available noise power is dissipated in R (100% of the total available noise power is dissipated in R for L =0).

The answer to this question should be: The value of L = 26.9 mH, but I can't figure out why. Can someone please help me?

 

[rude man]

1. What is the relation between the input psd (power spectral density) and the output, with x the input and y the output, given your psd of x and G(s) = 1/(Ts+1)?

2. What is the integral that gives you E(y²), the mean-square value of the output (the expectation of the output y)?

3. Compute the integral for the original L and then take 70% of that and recompute the value of L. Hint: E(y²) will be a function only of T = L/R and the cutoff frequency.

(Alternatiely, use the table of mean-square expectation integrals formulated by R.S. Phillips. But in this case the integral is easily solved).

It should be apparent that E(y²) is the power across R. For one thing, it's tyhe only dissipative element in your network. For another, R forms the output of the transfer function G(s).

 

[Gweniiix]

I'm sorry, I'm just a beginner in this course and I don't completely understand what you're saying. I tried some things, but that doesn't seem to work.
I also don't really know how to compute the integral for the original L (point 3). I tried things, but what I've got now can't be right (can't type it her.. my phone won't allow me).
Can you maybe show me how to do it? I already handed in my homework exercise, but my teacher hasnt discussed this question (yet?) and I really want to know how it is done! :)

 

[rude man]

The input to your network is characterized by a power spectral density function rather than a voltage. The input is flat noise up to the cutoff frequency, then it's zero. Power ~ v² and the units of your input is volts-square per Hz. This is called a power spectral density function because, in 1 Hz bandwidth, the power (1 ohm assumed) is V^2. (This is not quite precise in general but it is if the noise spectrum is flat as it is here.) So let's call the input psd S_i(jw) and the output psd is S_o(jw).

To get the output the best way for you is to take G(s) = 1/(Ts+1), let s = jw, then
S_i(jw)*|G(jw)|² = S_o(jw). Then
the expression for power is (1/2π) times the integral from -w_c to +w_c of the output power spectral density times dw. You are just integrating the output psd in other words, with the 1/2π needed for mathematical reasons.

You already had to evaluate this integral to compute the noise-equivalent bandwidth of your network BTW, or you should have. The integral is the same except for the limits of integration.

I really cannot go any further. This is a problem that should not have been assigned without sufficient background in the exotic world of random signals, IMO.