At 4:56 there is a bit of simplification -> if x and y are in dB, then: log(x+y)=log(10*log(x)+10*log(y))=log(10*log(x*y)) which doesn't mean it is usefull in practice
Very nice video. One thing I noticed. A 12:44 you say: "the spectrum analyzer is telling you what the power is..." Doesn't it show the voltage of the spectrum? It it would, as you say, show the power there is no need to convert it to the power in 50 Ohms...
632mV=0.632V power=0.998mW...1mW approximately.... prof. your calculation is wrong at 12:16... you cannot square 632mV which will result in 632^2* (10^-3)^2 V which is wrong... convert to Volts first then square the voltage
No, the calculation is correct. He has squared the 1/2 and the 1/sqrt(2) terms indefinitely, which together equals 1/8. Then, he squared the 623 mV term. Finally, he multiplied them together. It is simply a shortcut to avoid having to divide by root 2
I know this is quite old, but the correct answer is 998 mW. Sure, when squaring millivolts you end up with microwatts, but 632^2 gives you 399424 or 399.424 x 10^3, which multiplies with microwatts to get milliwatts.
@@gvcallen lol... I don't even remember anything anymore about that.... I graduated a year ago and am currently working in the field of embedded systems/FPGAs 🤣
You really mess up the whole concept of dB and dBm; you have made this look more complicated. Where is the dBc concept? If you do not know something do not try to make things up worst.
Thank you for putting out such quality information! I'm really enjoying your course!
632 is the peak to peak, dividing by 2 gives you the amplitude. Then dived by 2 again due to rms^2.
At 4:56 there is a bit of simplification -> if x and y are in dB, then: log(x+y)=log(10*log(x)+10*log(y))=log(10*log(x*y)) which doesn't mean it is usefull in practice
dBC is not explained in the video
Great courses. Thank you Professor.
Very nice video.
One thing I noticed.
A 12:44 you say: "the spectrum analyzer is telling you what the power is..."
Doesn't it show the voltage of the spectrum?
It it would, as you say, show the power there is no need to convert it to the power in 50 Ohms...
At 14:-03 You say: "for dBm we simply define 1 mV....."
I feel you mean 1 mW.
Maybe you can correct this by placing an overlay text on the video.
I have the same question
632mV=0.632V power=0.998mW...1mW approximately.... prof. your calculation is wrong at 12:16... you cannot square 632mV which will result in 632^2* (10^-3)^2 V which is wrong... convert to Volts first then square the voltage
No, the calculation is correct. He has squared the 1/2 and the 1/sqrt(2) terms indefinitely, which together equals 1/8. Then, he squared the 623 mV term. Finally, he multiplied them together. It is simply a shortcut to avoid having to divide by root 2
@@gvcallenyeah it is correct
I know for calculating power, we use RMS value of current or voltage. But I am not sure why you took the peak value i.e., dividing 632 mV by 2
can you help me for my home wotk?
Lil late but what you need
a lil more late but i can help
i get 998 µW
That's what I am getting, because by squaring 10^-3 = 10^-6 = Micro W
That is the correct answer, we build that circuit in the lab and measured it.
Yes, its 998 microWatts of power.
I know this is quite old, but the correct answer is 998 mW. Sure, when squaring millivolts you end up with microwatts, but 632^2 gives you 399424 or 399.424 x 10^3, which multiplies with microwatts to get milliwatts.
@@gvcallen lol... I don't even remember anything anymore about that.... I graduated a year ago and am currently working in the field of embedded systems/FPGAs 🤣
You really mess up the whole concept of dB and dBm; you have made this look more complicated.
Where is the dBc concept?
If you do not know something do not try to make things up worst.