If you have an analog TV channel whose visual carrier level is 0 dBmV and a 6 MHz wide 64-QAM (quadrature amplitude modulation) signal whose digital channel power is 0 dBmV, which one has more power? Let’s look at this subject more closely and see what the numbers say. Grab a cup of coffee and your scientific calculator. Yep, there’s math coming up. Grrr, math… Calculating power (P) in direct current (DC) circuits is straightforward. It’s the product of voltage (E) and current (I), or P = EI. Another way to express power is voltage squared divided by resistance, or P = E2/R. When we move to alternating current (AC), things aren’t quite so simple. In an AC circuit, the instantaneous values of voltage and current are varying continuously in time. How can useful quantities for these varying values be defined? To get AC voltage and current quantities that are equivalent to DC, we use root mean square (RMS) values. For instance, 10 millivolts (mV) RMS AC voltage causes the same average power dissipation in a resistor as does a 10 mV DC voltage. Likewise, 0.133 milliamperes (mA) RMS of AC has the same heating effect as 0.133 mA of DC. What about power in an AC circuit? Because the instantaneous values of AC voltage and current vary in time, we generally measure average power. Otherwise, each time the instantaneous values of voltage and current are multiplied to calculate power, the answer would be different. So average power it is. If voltage and current are in phase in an AC circuit, average power is the product of RMS voltage and RMS current, or Pavg = ErmsIrms. But the formula has to be tweaked when voltage and current aren’t in phase: Pavg = ErmsIrmscos[theta symbol], where [theta symbol] is the phase angle difference. For this example, let’s assume voltage and current are in phase. What power does a 0 dBmV signal represent? We know that 0 dBmV in a 75 ohm impedance cable network is 1 mV (0.001 volt) RMS because that’s the definition of 0 dBmV. The power can be calculated using the formula Pavg = Erms2/R:  = 0.0012/75
= 0.000001/75
= 0.000000013333 watt (W) or 13.33 nanowatts (nW) If the analog TV channel’s visual carrier is 13.33 nW and the 64-QAM signal’s digital channel power is 13.33 nW, the two power levels look the same! But are they? When we measure an analog TV channel’s visual carrier level, what is being measured? Technically, it’s the RMS amplitude of the instantaneous sync peaks. This happens to be equal to peak envelope power (PEP), which is the average power (watts) during one cycle at the crest of the modulation envelope. With a modulated visual carrier, the crest of the modulation envelope occurs during sync peaks. Mathematically, PEP = (PEV x 0.707)2/R. This formula is a variation of Pavg = Erms2/R. A sine wave’s RMS voltage is 0.707 of the peak voltage. For example, if the RMS voltage is 0.001 V, the peak voltage is 0.001414 V. This is a peak to average ratio of 0.0014141/0.001 = 1.414, or, in decibels, 20log(1.414) = 3 dB. The PEP of a modulated visual carrier whose peak envelope voltage (PEV) is 0.001414 volt during sync peaks is: = (0.001414 x 0.707)2/75
= (0.001)2/75
= 0.000001/75

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