How Does Parseval's Theorem Apply to Noise Amplitude Calculations?

[MartynaJ]

Homework Statement

You have a signal with a maximum frequency of 40 MHz. The average noise amplitude is 2 mV in the time domain. The noise is uniformly distributed from 0 to 40 MHz in the frequency domain. Now you apply a filter to remove all the frequency components above 10 MHz. What is the average amplitude of the noise in the filtered signal in the time domain?

Relevant Equations

$\int_\infty ^\infty {\left| f(x) \right|}^2 dx=\int_\infty ^\infty {\left| F(u) \right|}^2 du$

Before:
$\int_\infty ^\infty {\left| F_i(u) \right|}^2 du=\int_0 ^{40} a^2 du=40 a^2$

Therefore: $a^2= \frac{1}{40}\int_\infty ^\infty {\left| F_i(u) \right|}^2 du$

After:
$\int_\infty ^\infty {\left| F_f(u) \right|}^2 du=\int_0 ^{10} a^2 du=10 a^2$

Therefore: $a^2= \frac{1}{10}\int_\infty ^\infty {\left| F_f(u) \right|}^2 du$

Equating the two equations:
$\frac{1}{40}\int_\infty ^\infty {\left| F_i(u) \right|}^2 du=\frac{1}{10}\int_\infty ^\infty {\left| F_f(u) \right|}^2 du$

Therefore:
$\int_\infty ^\infty {\left| F_i(u) \right|}^2 du=4\int_\infty ^\infty {\left| F_f(u) \right|}^2 du$

using Parseval’s theorem which states that: $\int_\infty ^\infty {\left| f(x) \right|}^2 dx=\int_\infty ^\infty {\left| F(u) \right|}^2 du$, we get:

$\int_\infty ^\infty {\left| f_i(x) \right|}^2 dx=4\int_\infty ^\infty {\left| f_f(x) \right|}^2 dx$

The question states that the average noise amplitude is 2 mV in the time domain. Does this mean that:
$\int_\infty ^\infty {\left| f_i(x) \right|}^2 dx=2$ ?

 

[Master1022]

Sorry I know this is an old thread, but are you still looking for an answer to this?

The question states that the average noise amplitude is 2 mV in the time domain. Does this mean that:
$\int_\infty ^\infty {\left| f_i(x) \right|}^2 dx=2$ ?

I think the 2 mV refers to $|f(x)|$, so the power would be proportional to $|f(x)|^2$ just like $P = \frac{V^2}{R}$. I haven't seen the term 'noise amplitude' before to be honest, but it seems like they are asking for the rms value. We can say that the power is proportional to 4 of some arbitrary units (we don't need to worry about them given that the power spectral density is constant).

Relevant Equations:: $\int_\infty ^\infty {\left| f(x) \right|}^2 dx=\int_\infty ^\infty {\left| F(u) \right|}^2 du$

Before:
$\int_\infty ^\infty {\left| F_i(u) \right|}^2 du=\int_0 ^{40} a^2 du=40 a^2$

I think you can get to the end result sooner. We are told that the noise is uniformly distributed from 0 to 40 MHz (area of a rectangle, as you have found). Therefore, when only considering the noise from 0 to 10 MHz, you can immediately tell that the noise power is now 1/4 of what it previously was (there isn't necessarily a need to write out the integrals with $a^2$).

Once we have the power is 1/4 of the previous value, we can find the noise amplitudes. You were basically at the solution. We want to find the value that solves: $$x^2 \times 4 = 4$$

Hope that is of some help.