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A2 Physics | Post your doubts here

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the signal is a sine wave, so it has is like one wave.. which has a min and a max, which will cause a region of high frequency when its a maximum and a region of low frequency when its a min.
Hence at 12 us frquency is min i.e a trough on the signal wave, at 4 us , frequency is min i.e trough on the signal wave. From trough to trough, its one signal wave.
 
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From the application booklet,
"There is a continuous distribution of wavelengths with a
sharp cut-off at short wavelength and also a series of high-intensity spikes that are characteristic of the
target material."
I dont get the difference between the "sharp cut-offs" and the "high intensity peaks" since according to the text, they are different things! :S
 
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From the application booklet,
"There is a continuous distribution of wavelengths with a
sharp cut-off at short wavelength and also a series of high-intensity spikes that are characteristic of the
target material."
I dont get the difference between the "sharp cut-offs" and the "high intensity peaks" since according to the text, they are different things! :S
Assalamoalaikum wr wb
Check the notes from XPFMember...that has that graph and the answers to the different features of the graph...may be that'll help..
 
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From the application booklet,
"There is a continuous distribution of wavelengths with a
sharp cut-off at short wavelength and also a series of high-intensity spikes that are characteristic of the
target material."
I dont get the difference between the "sharp cut-offs" and the "high intensity peaks" since according to the text, they are different things! :S
sharp cut-off at short wavelength is due towhen e give all their energy to photon and this is min wavelenght
whereas "high intensity peaks are when e are excited n then deexcited hope u understood the d/f
 
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Assalamoalaikum wr wb
Check the notes from XPFMember...that has that graph and the answers to the different features of the graph...may be that'll help..
link plz?
sharp cut-off at short wavelength is due towhen e give all their energy to photon and this is min wavelenght
whereas "high intensity peaks are when e are excited n then deexcited hope u understood the d/f
I know that, but I mean I cant differentiate between them on the graph!
 
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From the application booklet,
"There is a continuous distribution of wavelengths with a
sharp cut-off at short wavelength and also a series of high-intensity spikes that are characteristic of the
target material."
I dont get the difference between the "sharp cut-offs" and the "high intensity peaks" since according to the text, they are different things! :S
watch this
 
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Assalamoalaikum wr wb!

need help with frequency modulation :( plzz...

and also i dont get the sidebands part in the amplitude modulation...!! :(
 
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plz explain wo9 p41 q 6 b i
and q7 b i n ii
Assalamoalaikum wr wb once again..
wel this is what i understand...
usse right hand grip rule to get the field direction ...it would be anticlockwise...(circular)

so the points gonna be to the left of that current...
 
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Assalamoalaikum wr wb once again..
wel this is what i understand...
usse right hand grip rule to get the field direction ...it would be anticlockwise...(circular)

so the points gonna be to the left of that current...
walkum aslam.
thanx nd wht abt that alternating voltage. how to draw on that graph????
jazak Allah
 
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walkum aslam.
thanx nd wht abt that alternating voltage. how to draw on that graph????
jazak Allah
wa eyyakum..
half wave rectification...
with height reduced i think...[dont rremember why..if u get to know..lemme know tooo..]
and in one it'll be only the positive part and no negative part...'
in the second it'll be only the negative part not the positve...u can see this when yo look in the diagram and see how they're gonna work..as the input is ac..it'll be one time up one is positive..and then negative..and goes on..

btw need help with frequency modulation :( plzz...

and also i dont get the sidebands part in the amplitude modulation...!! :(
 
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wa eyyakum..
half wave rectification...
with height reduced i think...[dont rremember why..if u get to know..lemme know tooo..]
and in one it'll be only the positive part and no negative part...'
in the second it'll be only the negative part not the positve...u can see this when yo look in the diagram and see how they're gonna work..as the input is ac..it'll be one time up one is positive..and then negative..and goes on..

btw need help with frequency modulation :( plzz...

and also i dont get the sidebands part in the amplitude modulation...!! :(
i still didnt get it..:( can you put a sketch of it over here...

and wht do you want to knw abt frquency modulation and sidebands?? p.s i dont knw much abt them too
 
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