Signalto Noise Ratio SN Versus Carrierto Noise Ratio CN in CATV Systems

We have been using S/N and C/N many times in previous chapters. In CATV systems S/N has a slightly different definition as follows (Ref. 2):

This relationship is expressed by the "signal-to-noise ratio," which is the difference between the signal level measured in dBmV, and the noise level, also measured in dBmV, both levels being measured at the same point in the system.

S/N can be related to C/N on CATV systems as

This is based on Carson (Ref. 4), where the premise is "noise just perceptible" by a population of TV viewers, with an NTSC 4.2-MHz TV signal. Adding noise weighting3 improvement (6.8 dB), we find

It should be noted that S/N is measured where the signal level is peak-to-peak4 and the noise level is rms. For C/N measurement, both the carrier and the noise levels are rms.

3 Weighting (IEEE). The artificial adjustment of measurements in order to account for factors that in normal use of the device would otherwise be different from the conditions during measurement. In the case of TV, the lower baseband frequencies (i.e., from 20 Hz to 15 kHz) are much more sensitive to noise than the higher frequencies (i.e., >15 kHz).

4Peak-to-peak voltage refers, in this case, to the measurement of voltage over its maximum excursion, which is the voltage of the "sync tips." See Figures 16.3 and 16.4.

These values are based on a VSB-AM (vestigial sideband, amplitude modulation) with an 87.5% modulation index.

The values for S/N should be compared to those derived by the Television Allocations Study Organization (TASO) and published in their report to the US FCC in 1959. Their ratings, corrected for a 4-MHz bandwidth, instead of the 6-MHz bandwidth that was used previously, are shown in Section 16.4.2.

Once a tolerable noise level is determined, the levels required in a CATV system can be specified. If the desired S/N has been set at 43 dB at a subscriber TV set, the minimum signal level required at the first amplifier would be —9 dBmV + 43 dB or -16 dBmV, considering thermal noise only. Actual levels would be quite a bit higher because of the noise generated by subsequent amplifiers in cascade.

It has been found that the optimum gain of a CATV amplifier is about 22 dB. When the gain is increased, IM/Xm products become excessive. For gains below this value, thermal noise increases, and system length is shortened or the number of amplifiers must be increased—neither of which is desirable.

There is another rule-of-thumb of which we should be cognizant. Every time the gain of an amplifier is increased 1 dB, IM products and "beats" increase their levels by 2 dB. And the converse is true: every time gain is decreased 1 dB, IM products and beat levels are decreased by 2 dB.

With most CATV systems, coaxial cable trunk amplifiers are identical. This, of course, eases noise calculations. We can calculate the noise level at the output of one trunk amplifier. This is

where NF is the noise figure of the amplifier in decibels.

In the case of two amplifiers in cascade (tandem), the noise level (voltage) is

If we have M identical amplifiers in cascade, the noise level (voltage) at the output of the last amplifier is

This assumes that all system noise is generated by the amplifiers, and none is generated by the intervening sections of coaxial cable.

Example 1. A CATV system has 30 amplifiers in tandem; each amplifier has a noise figure of 7 dB. Assume that the input of the first amplifier is terminated in 75 Q resistive. What is the thermal noise level (voltage) at the last amplifier output? Use Eq. (17.8):

Nv = —59 dBmV + 7 dB + 10 log 30 = —59 dBmV + 7 dB + 14.77 dB = —37.23 dBmV.

For carrier-to-noise ratio (C/N) calculations, we can use the following procedures. To calculate the C/N at the output of one amplifier,

Example 2. If the input level of a CATV amplifier is +5 dBmV and its noise figure is 7 dB, what is the C/N at the amplifier output?

With N cascaded amplifiers, we can calculate the C/N at the output of the last amplifier, assuming all the amplifiers are identical, by the following equation:

Example 3. Determine the C/N at the output of the last amplifier with a cascade (in tandem) of 20 amplifiers, where the C/N of a single amplifier is 62 dB. Use Eq. (17.10):

C/Nl = 62 dB - 10 log 20 = 62 dB - 13.0 dB = 49 dB.

+1 0


  • Judith
    What is c/n ratio in catv?
    6 years ago

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