Revision as of 08:09, 15 January 2014

Discussion page for Lab 1

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

$ x_k (t) = a_k e^{j k w_0 t} $

from the complex Fourier series into the formula for RMS

$ x_{k,RMS} = \sqrt{ \frac{1}{T} \int_T | x_k (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 + ... } $,

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_{avg}=\frac{V_{rms}^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} $.

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