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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: 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) | + | <math> a_k exp(j k w_0 t) <\math> |
from the complex Fourier series into the formula for RMS | from the complex Fourier series into the formula for RMS | ||
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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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A: To combine the RMS voltages to get the net RMS of the first nine spectral components, use the following: | A: To combine the RMS voltages to get the net RMS of the first nine spectral components, use the following: | ||
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<math>V_{rms,net} = \sqrt{ v_0^2 + v_1^2 + ... + v_9^2}</math> | <math>V_{rms,net} = \sqrt{ v_0^2 + v_1^2 + ... + v_9^2}</math> | ||
Revision as of 06:52, 4 September 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) <\math> from the complex Fourier series into the formula for RMS <math> 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:
$ 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} W $ and taking the $ 10\log $ of each side, we get:
$ 10\log\left(\frac{P}{10^{-3}W}\right)=10\log\left(\left(\frac{V}{\sqrt(R_{ref}\times 10^{-3}W)}\right)^2\right)=20\log\left(\frac{V}{\sqrt(R_{ref}\times 10^{-3} W})\right) $
Any $ R_{ref} $ can do, however, it is easiest if we pick $ R_{ref}=1000 \Omega $ in which case we get:
$ \text{Power reading in dBm}=\text{Voltage of signal in dBV} $.
