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jut draw a AM signal graph with modulating frequency of 5kHz(200us) and carrier frequency of 50KHz(20us)
I still don't get it. Like, do you mean, draw the carrier frequency separate?
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jut draw a AM signal graph with modulating frequency of 5kHz(200us) and carrier frequency of 50KHz(20us)
This question please, help neededhttp://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
Q11 b iii. last part of the question..please explain working!!
SkyPilotage has answered this question posted by me in this very thread!This question please, help needed
(a) when it would be parallel(horizontal) the magnetic field strenght would pass through it for a longer distance (2.8cm) .when it is perpendicular magnetic field strength passes through for a very short distance which is thickness of the coil.http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_4.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_ms_4.pdf
please can anyone explain question 6 part (a) part (b) (i) and (b) (ii)
SkyPilotage has answered this question posted by me in this very thread!
Please search
no . have you seen an AM signal graph ?I still don't get it. Like, do you mean, draw the carrier frequency separate?
Yeah, isn't it the one where the frequency stays the same, but the amplitude varies with the displacement?no . have you seen an AM signal graph ?
just draw two LEDs one with forward direction and one reversed connected from the input to the earth . the forward biased LED is of red light and the reversed biased LED is greenhttp://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_4.pdf
anyone for 9 (b).
thanks in advance!
open this link .AM graph is like thisYeah, isn't it the one where the frequency stays the same, but the amplitude varies with the displacement?
i dont get it from his solution..could you explain the working? pleeease bro?SkyPilotage has answered this question posted by me in this very thread!
Please search
Here it is:like finding a needle in a haystack,
Okay will do, thanks though
Jazakallah brotherHere it is:
From Part i) The Intensity that is reflected back from the muscle bone boundary is 0.331 i.e 0.331 x 0.389 I
The key to this part is that you have to deduce the Intersity reflection coefficient from part ii).
When the ultrasound reaches the surface of the muscle, 0.661 will be reflected and 0.389 will pass through ----> 0.331 x 0.389 x 0.389 I = 0.05 I
So basically, you just draw a graph similar to that one, but each individual peak is every 20µs and the lwhole waveform takes 200µs to complete?open this link .AM graph is like this
https://ccrma.stanford.edu/~jos/st/img494.png
the time for each individual peak is of the carrier frequency
and the time for one large peak to the other large peak is of the information signal
exactly! the amplitude is not given so dont worry about the amplitude !So basically, you just draw a graph similar to that one, but each individual peak is every 20µs and the lwhole waveform takes 200µs to complete?
Thanks so much!exactly! the amplitude is not given so dont worry about the amplitude !
no problem!Thanks so much!
thanku angelicFor Q9 b) the difference in the uncertainties is 10%-2% =8%
That's an uncertainty of 0.08. So to calculate the activity for when the source is not decaying 0.08, it would be decaying 1-0.08= 0.92.
so the activity= 0.092 decay constant=0.693/half life .. and the rest is just
A
= A0 exp(–ln2 t / T½
)
and thanks alot for all the help i recieved from every 1...ok every1 best of luck for your examsss...May Allah Subhana Wa Taa'llah helps all of us...Ameen
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