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We got this question in our test :-

$$S(t) = 5\cos(10πt - π/2) + 6\sin(15πt)\cos(15πt)$$

My Solution :-

$$S(t) = 5\cos(10πt - π/2) + 3\sin(30πt)\qquad \textrm{[Using $\sin(2A) = 2\sin(A)\cos(A)$]}$$

=> Power$ = 5x5/2 + 3x3/2 = 25/2 + 9/2 = 17$

But in the solution, the correct answer was given to be 21.5 watt.

What mistake did I do?

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  • $\begingroup$ I haven't checked your calculations, but, you cannot convert digital signals to power in Watts at all (without defining a DAC sensitivity, a load, and possibly a sampling rate). I know this is probably not your fault – there's a million courses out there that started out with statements like "assuming a digital 1 corresponds to 1V from the DAC and the load is 1Ω" or something like that, and then forgot to state that – including a lot of of different funky constant factors like 2 W/1², 4W/1², $\sqrt2$W/1², $\frac1{f_s}$ W/1² that I've seen people assume between their digital values and analog $\endgroup$ – Marcus Müller Mar 9 at 8:34
  • $\begingroup$ power. I've literally earned 1000s our € (not an exaggeration) explaining that no, digital power is not per se linked to analog power by a constant you know until you've calibrated your whole signal chain. So, look for the definition of how a digital count translates to an analog amplitude (and is that amplitude a voltage? a current? a light intensity? and if one of the two former, what's the load impedance?), and look at what they're doing. $\endgroup$ – Marcus Müller Mar 9 at 8:37
  • $\begingroup$ This doesn't mean converting between digital and analog power is wrong by any means! It's just that the unit of digital power is 1, not W. The fact that your solution is given in W just tells me that there's probably a conversion factor in between, which simply might not be 1W/1 (1 W per 1 digital count). $\endgroup$ – Marcus Müller Mar 9 at 8:41
  • $\begingroup$ Thanks for this extra insight. It's great to learn this additional information which in not taught in the class. Maybe the teacher themselves tend to make such mistakes. Additionally, I sent my solution to the teacher for recheck, and viola.... My solution was actually correct. Also, thanks for taking out time to write huge content. It was really helpful, and will definitely keep that in mind. $\endgroup$ – Hentai Ousama Mar 9 at 19:11
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Apart from possible dimensional problems as discussed in the comments, your result is correct and the given solution is wrong. In general with such exercises, don't forget to check if the cross-term really cancels out (in this case it does).

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  • $\begingroup$ Thanks for the reply. I sent my solution to the teacher for recheck, and viola.... My solution was actually correct. $\endgroup$ – Hentai Ousama Mar 9 at 19:13

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