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<math> x_{RMS} = \sqrt{ \frac{1}{T} \int_T | x (t)|^2 dt } </math> | <math> x_{RMS} = \sqrt{ \frac{1}{T} \int_T | x (t)|^2 dt } </math> | ||
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where v_i is the RMS of the i-th spectral component. By Parseval's theorem, Vrms,net should converge to the RMS voltage of the overall signal as the number of spectral components included in the sum approaches infinity. | where v_i is the RMS of the i-th spectral component. By Parseval's theorem, Vrms,net should converge to the RMS voltage of the overall signal as the number of spectral components included in the sum approaches infinity. | ||
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| + | ----- | ||
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| + | Q: What is reference resistance? How to get the dBV of a signal from a dBm reading? | ||
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| + | A: A reference resistance is a resistance used in an instrument (e.g., Wavetek RMS Voltmeter) to measure the power of a signal. The power, P in watts, and voltage, V in volts, of the signal are related according to | ||
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| + | <math>P=\frac{V^2}{R_{ref}}</math>, | ||
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| + | where <math>R_{ref}</math> is the reference resistance in ohms. Now, from the above relationship we can get the dBV of a signal from a dBm reading. Dividing each side by <math>10^{-3}</math> and taking the <math>10\log </math> of each side, we get: | ||
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| + | <math>10\log\left(\frac{P}{10^{-3}}\right)=10\log\left(\left(\frac{V}{\sqrt(R_{ref}\times 10^{-3})}\right)^2\right)=20\log\left(\frac{V}{\sqrt(R_{ref}\times 10^{-3}}\right)</math> | ||
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| + | Any <math>R_{ref}</math> can do, however, it is shortest if we pick <math>R_{ref}=1000 \Omega</math> in which case we get: | ||
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| + | <math> \text{Power reading in dBm}=\text{Voltage of signal in dBV} </math>. | ||
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---- | ---- | ||
[[ECE440F13LoveLab1|Back to Lab 1 ECE440 Fall 2013]] | [[ECE440F13LoveLab1|Back to Lab 1 ECE440 Fall 2013]] | ||
Revision as of 08:45, 29 August 2013
Discussion page for Lab 1 ECE440 Fall 2013
This discussion board is used to clarify issues on the lab and pre-lab.
Q: How do I get the RMS voltage of spectral components on the 5th question?
A: One can show that the RMS voltage of each spectral component is equal to the corresponding coefficient of the complex Fourier series. To show this, plug in the spectral component
a_k exp(j k w_0 t)
from the complex Fourier series into the formula for RMS
x_RMS = sqrt(1/T int_T( | x (t)|2 dt ) ),
where int_T denotes the integral over one period, i.e.
$ x_{RMS} = \sqrt{ \frac{1}{T} \int_T | x (t)|^2 dt } $
Q: Do the RMS voltage of the spectral components need to sum up to the total RMS voltage?
A: To combine the RMS voltages to get the net RMS of the first nine spectral components, use the following:
Vrms,net = sqrt( v_0^2 + v_1^2 + ... + v_9^2),
$ V_{rms,net} = \sqrt{ v_0^2 + v_1^2 + ... + v_9^2} $
where v_i is the RMS of the i-th spectral component. By Parseval's theorem, Vrms,net should converge to the RMS voltage of the overall signal as the number of spectral components included in the sum approaches infinity.
Q: What is reference resistance? How to get the dBV of a signal from a dBm reading?
A: A reference resistance is a resistance used in an instrument (e.g., Wavetek RMS Voltmeter) to measure the power of a signal. The power, P in watts, and voltage, V in volts, of the signal are related according to
$ P=\frac{V^2}{R_{ref}} $,
where $ R_{ref} $ is the reference resistance in ohms. Now, from the above relationship we can get the dBV of a signal from a dBm reading. Dividing each side by $ 10^{-3} $ and taking the $ 10\log $ of each side, we get:
$ 10\log\left(\frac{P}{10^{-3}}\right)=10\log\left(\left(\frac{V}{\sqrt(R_{ref}\times 10^{-3})}\right)^2\right)=20\log\left(\frac{V}{\sqrt(R_{ref}\times 10^{-3}}\right) $
Any $ R_{ref} $ can do, however, it is shortest if we pick $ R_{ref}=1000 \Omega $ in which case we get:
$ \text{Power reading in dBm}=\text{Voltage of signal in dBV} $.
