- #1
MartynaJ
- 19
- 1
- 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## ?
##\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## ?
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