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.

 

[rezeew]

To determine the value of L that will result in 70% of the noise power being dissipated in R, we can use the fact that the total noise power is equal to the integral of the PSD over the frequency range. In this case, the total noise power is equal to 1, since the PSD is equal to 1 for all frequencies below 140 kHz.

So, we can set up the following equation:

0.7 = ∫ Pn(f) * |H(f)|^2 df

Where |H(f)|^2 is the squared magnitude of the transfer function, and df is the frequency range over which we are integrating.

Substituting in the values for R and L from the given filter (R = 18 kΩ and L = 120 mH), we get:

0.7 = ∫ 1 * |H(f)|^2 df
= ∫ 1 * (1/(1+(f/23.9)^2)) df

Integrating this equation from 0 to infinity (since the PSD is only non-zero for frequencies below 140 kHz), we get:

0.7 = 1 - arctan(140/23.9)
= 0.512

Solving for L, we get:

L = 26.9 mH

Therefore, for L = 26.9 mH, 70% of the noise power will be dissipated in R.