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Thursday, April 16, 2009. Saturday, March 7, 2009. 8226;For example: suppose the time samples are as shown in the diagram: Values: 0 7 10 7 0 -7 -10 -7. Squares: 0,49,100,49,0,49,100,49 Sum of squares = 396 Average of squares = 396/8 = almost 50 Square root 7 7 = 0.7 x 10 (peak voltage). With more intervals the r.m.s. average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 x peak value. ExamplesIf V = 24 volts, R = 6 ohms then I = 24/6 (V/R) = 4 amps and P = (24)2/6 (V2/R) = 576/6 = 96 watts.

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Thursday, April 16, 2009. Saturday, March 7, 2009. 8226;For example: suppose the time samples are as shown in the diagram: Values: 0 7 10 7 0 -7 -10 -7. Squares: 0,49,100,49,0,49,100,49 Sum of squares = 396 Average of squares = 396/8 = almost 50 Square root 7 7 = 0.7 x 10 (peak voltage). With more intervals the r.m.s. average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 x peak value. ExamplesIf V = 24 volts, R = 6 ohms then I = 24/6 (V/R) = 4 amps and P = (24)2/6 (V2/R) = 576/6 = 96 watts.
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HFC | fundamentalsofhfc.blogspot.com Reviews

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Thursday, April 16, 2009. Saturday, March 7, 2009. 8226;For example: suppose the time samples are as shown in the diagram: Values: 0 7 10 7 0 -7 -10 -7. Squares: 0,49,100,49,0,49,100,49 Sum of squares = 396 Average of squares = 396/8 = almost 50 Square root 7 7 = 0.7 x 10 (peak voltage). With more intervals the r.m.s. average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 x peak value. ExamplesIf V = 24 volts, R = 6 ohms then I = 24/6 (V/R) = 4 amps and P = (24)2/6 (V2/R) = 576/6 = 96 watts.

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HFC: March 2009

http://fundamentalsofhfc.blogspot.com/2009_03_01_archive.html

Saturday, March 7, 2009. 8226;For example: suppose the time samples are as shown in the diagram: Values: 0 7 10 7 0 -7 -10 -7. Squares: 0,49,100,49,0,49,100,49 Sum of squares = 396 Average of squares = 396/8 = almost 50 Square root 7 7 = 0.7 x 10 (peak voltage). With more intervals the r.m.s. average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 x peak value. ExamplesIf V = 24 volts, R = 6 ohms then I = 24/6 (V/R) = 4 amps and P = (24)2/6 (V2/R) = 576/6 = 96 watts. Why Log, Antilog? DBmV = ...

2

HFC: Fundamentals of HFC

http://fundamentalsofhfc.blogspot.com/2009/03/fundamentals-of-hfc.html

Saturday, March 7, 2009. 8226;For example: suppose the time samples are as shown in the diagram: Values: 0 7 10 7 0 -7 -10 -7. Squares: 0,49,100,49,0,49,100,49 Sum of squares = 396 Average of squares = 396/8 = almost 50 Square root 7 7 = 0.7 x 10 (peak voltage). With more intervals the r.m.s. average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 x peak value. ExamplesIf V = 24 volts, R = 6 ohms then I = 24/6 (V/R) = 4 amps and P = (24)2/6 (V2/R) = 576/6 = 96 watts. Why Log, Antilog? DBmV = ...

3

HFC

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HFC: April 2009

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HFC

Thursday, April 16, 2009. Saturday, March 7, 2009. 8226;For example: suppose the time samples are as shown in the diagram: Values: 0 7 10 7 0 -7 -10 -7. Squares: 0,49,100,49,0,49,100,49 Sum of squares = 396 Average of squares = 396/8 = almost 50 Square root 7 7 = 0.7 x 10 (peak voltage). With more intervals the r.m.s. average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 x peak value. ExamplesIf V = 24 volts, R = 6 ohms then I = 24/6 (V/R) = 4 amps and P = (24)2/6 (V2/R) = 576/6 = 96 watts.

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