Re: [IRCA] MW-IBOC carrier power

["CHARLES HUTTON" <charlesh3@xxxxxxx>]

You're so right - the figure I gave is the power in a "primary sideband" to 
use iBiquity's nomenclature. Consider me guilty of grabbing the wrong value 
out of my spreadsheet.

So here is the correct value: total power per sideband is 1592 Watts.

16QAM is used in order to get more bits through the system.

As to the "loudness factor" of IBOC, it depends..... The sidebands up to 10 
kHz are transmitted as conjugates to maintain a constant envelope to normal 
AM detectors. Whether you receive both sidebands and whether you receive 
them perfectly depends on your receiver, propagation, and where you have 
your receiver tuned relative to the IBOC station. Then there's the nature of 
the QAM beast: some transitions are pure phase transitions so the envelope 
detector sees no difference. It all adds up to a 7 or 8 dB difference (with 
conditions I do not know) in the "audibility" of the IBOC carriers and 
analog programming (measured empirically) if I remember some info from Barry 
McLarnon correctly.


Chuck


>From: "Dan Strassberg" <dan.strassberg@xxxxxxx>
>Reply-To: Mailing list for the International Radio Club of 
>America<irca@xxxxxxxxxxxxxxxx>
>To: <irca@xxxxxxxxxxxxxxxx>
>CC: Scott Fybush <scott@xxxxxxxxxx>
>Subject: [IRCA] MW-IBOC carrier power
>Date: Mon, 19 Sep 2005 08:59:13 -0400
>
>12 dB down from 50 kW is NOT 625W! It is 1/16 of 50,000W or 3333W. That's
>assuming we're talking dB power and not dB voltage. Although dBs are a way
>of expressing the logarithms of dimensionless ratios, they were originally
>defined for power ratios. dB = 10 log (base 10) P2/P1. Since the power
>dissipated in a resistor is poportional to the square of the voltage across
>the resistor (P = V^2/R), it was a very small step to dBV = 10 log (base
>10) V2^2/V1^2 = 20 log (base 10) V2/V1. But don't mix dB and dBV within a
>calculation or formula because the numeric results will be wrong!
>
>I suspect that I came up with 15 dB, where the correct answer was 
>apparently
>12 db, because, in all likelihood, the 12-dB number applies to the ratio of
>the main-carrier power to the combined power in both digital sidebands. The
>power in both sidebands is twice the power in one of the sidebands. A power
>ratio of 2:1 = 3 dB.
>
>Most likely, we're also talking about power at zero modulation, for both 
>the
>main (analog) carrier and the digital sidebands. Although the FCC allows
>125% modulation of the AM carrier on positive peaks, I believe that most
>stations average modulation index is considerably lower--probably a bit 
>less
>than 40%. In the ridiculous extreme of modulating an AM carrier with an
>audio-frequency square wave and hitting 125% on the positive peaks, the
>power on the positive peaks would be almost exactly five times the nominal
>carrier power (2.25^2 = 5.0625) and the average power would be about 2.5
>times nominal carrier power. Since, when broadcasting normal program
>material, modern highly efficient 50-kW AM transmitters usually have a 
>power
>input of about 85 kW (that is, 1.7 times the nominal carrier power), it is
>pretty clear that the normal average modulation level is substantially 
>lower
>than 0.5*(125%) or 62.5%.
>
>I think that the reason the MW-IBOC digital sidebands sound so much louder
>than their relatively low power (~1600W per sideband for a 50 kW AM)
>suggests, is that the average index of modulation is much higher than that
>of the main (analog) carrier. iBiquity's HD Radio systems are designed
>around a system called OFDM (orithogonal frequency-division multiplexing).
>OFDM uses multiple carriers. The HD Radio digital sidebands thus contain
>multiple subcarriers, each of which conveys information via a complex
>digital-modulation scheme--at least BPSK (bipolar phase-shift keying), but
>more likely QPSK (quadrature phase-shift keying) or 16 QAM (16-level
>quadrature amplitude modulation).
>
>If I'm wrong about any of this, please don't hesitate to correct me.
>
>--
>Dan Strassberg, dan.strassberg@xxxxxxx
>eFax 707-215-6367
>
>
>
>
>
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