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A sine wave is applied to a balanced modulator. The peak output envelope power is \$ 1000 \$ times the minimum output envelope power. Then the carrier suppression in dBc will be ______?

My thoughts:
Note: dBc means Decibels(with referenced to carrier)
A.T.P\$ \to \$
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here, \$ s_{DSB-SC} (t) = A_m A_c \sin (\omega _m t) \cos (\omega_c t)\$
And it's envelope is given as: \$ A_m A_c \sin (\omega _m t) \$
so instantaneous envelope power \$ = \{ A_m A_c \sin (\omega _m t) \} ^2 \$ $$=A_m^2 A_c^2 \frac{1- \cos (2 \omega _m t)}{2} \dots \dots (1)$$ Therefore peak output envelope power \$ = A_m^2 A_c^2\$ which is attained when \$ \cos (2 \omega _m t) = -1\$ or \$ \omega _m t = \frac{\pi}{2} \$
And minimum output envelope power \$ =0 \$ which is attained when \$ \cos (2 \omega _m t) = 1\$ or \$ \omega _m t = \pi \$
Now, $$P_{out \_ p}=1000 \times P_{out \_ min} \quad \quad [Given]$$ $$\implies A_m^2 A_c^2=0$$ Thus not getting any information about carrier suppression in dBc
so any help please...
Answer given is \$ 30 dBc \$

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There are two problems here. You don't realise what is being asked. And the book has made an unforgivable mistake in the use of dB.

You are told that The peak output envelope power is 1000 times the minimum output envelope power. Yet you have asserted further down And minimum output envelope power =0 ...

In an ideal signal, the minimum is 0. In this particular signal generation scheme, using a balanced modulator, you are told that the minimum is 1/1000th of the peak. This is because in a real modulator, there is always some signal that leaks through, the perfect 0 is not obtainable.

You are asked for the carrier suppression, the ratio by which the residual output has been suppressed below the wanted output. You have been given the power ratio, 1000, so you use the 10log() formula for dB, \$10log_{10}(1000) = 30dB\$

That's your answer, 30dB, because you've been asked for the suppression ratio.

For some bizarre reason, you've been asked for the carrier suppression in dBc. This is plain ignorant, especially of a question setter who should know better. dBc is a power level, not a ratio. It's a power, with respect to the carrier. If you were asked for the residual, 'zero output' power, then the answer would be -30dBc.

If you see a measurement value in dB, it's a ratio, a number. If it's not clear what the other reference measurement is, then go searching, it will be defined somewhere. If you see a measurement in dBx, then it's a level in the same dimensions as x, dBm for radio power with a 1mW reference, dBuV for voltage with a 1uV reference, dBc for power with respect to some definition of the carrier, here it's implied (though not clearly stated) that it's the peak of the envelope of the carrier power.

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    \$\begingroup\$ @Niel_UK, If dBc is "with respect to the carrier", then is it not also a ratio? I mainly use dBc when talking about phase noise, in which case we don't even care about the power levels - just about how much lower the power is in a band at an offset. Hence, if I amplify or attenuate (with ideal, noiseless circuits) the power levels will change, but the phase noise, and hence the "dBc" numbers, will stay the same. \$\endgroup\$ – Joren Vaes Mar 15 '18 at 7:43
  • \$\begingroup\$ @JorenVaes in '-30dBc', the -30dB represents a ratio with respect to the carrier, the 'c' represents the carrier, in units of voltage, power, pressure, however it's measured, so '-30dBc' is an absolute level, having the same units as 'c' \$\endgroup\$ – Neil_UK Mar 15 '18 at 9:07

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