Data Comms question i can't figure - Please help me!!

Peace

Hi there,

I hope the maths boards it the right one for this question:

I have a Data Comms assingment due in next week and i'm stuck on a question i obviously can't figure out.

So the question is as follows:
"Consider an audio signal with spectral compnents in the range 300 - 3000hz. Assum that a sampling rate of 7000 samples per second will be used to generate a PCM signal.
(A) For SNR = 30dB, what is the number of uniform quantizing levels needed?
(B) What is the data rate?
Note: SNR for quantizing noise can be expressed as SNR = 20log2^n + 1.76dB"

Ok, so my thoughts on this are thus:

(A) the SNR = 30dB is the same as (20log2^n + 1.76dB). So if i figure out for "n" then thats the number of levels that are being used?? Am i right on this one. I'm kinda flying blind sine the book i bought isn't great (Data Comm's + Networking by Behourozan).

For part (B) the data rate is the (number of bits) n * 7000 (sampling rate)....so i assume when i solve for "n" then i just multiply n*7000hz and get my kbs?

I know there are a lot of chancers asking people to do their assingments up here but what i'm really looking for is guidance. If you just post up an answer its pretty much useless as i'll still get 0 in my assingment!

So am i right in what i've said above? If not can somone let me know how it should be worked out please

Thanks in advance!

Capt'n Midnight

Claude Shannon

Peace

Originally posted by Capt'n Midnight
Claude Shannon

Its the Nyquist Theorem thats metioned in the book. We covered Shannon earlier in the course and i'm reasonably sure its not involved. However if you post up more information then i might get a better idea of what you are refering to.

Capt'n Midnight

it's a telephone line using a modem

(A) 30dB SNR = 1000:1

so work out how many bits you need first.

The highest Freq is 3000Hz but you are told 7000Hz is the sampling freq.
Is this enough - what is the minimum freq needed ?

n is the number of bits. - put it back into the equation to see what pops out and then you'll have to decide if it is enough - if not then you may need to add another bit to to ensure that none of the 1000 different analogue levels could be translated to a different bit. (eg: how does the noise level on quantisiation compare to 1/1000 of the signal )


(B) That looks ok. (except you'd multiply by 7k to get Kbs)

Peace

Originally posted by Capt'n Midnight
it's a telephone line using a modem

(A) 30dB SNR = 1000:1

so work out how many bits you need first.

The highest Freq is 3000Hz but you are told 7000Hz is the sampling freq.
Is this enough - what is the minimum freq needed ?

Cap'n Midnight, thanks for your help so far....i'm a bit lost at sea on this one

As far as my understanding goes, the Nyquist Theorem states that the sampling freqency must be at least twice the highest possible frequency of the signal you are sampling. Therefore the minimum frequency that would be required this time would be 6000 samples per second. - however they seemed to have taken 7000 but thats probably because they want the answer to work out at 56k!!

I kinda guessed the bit that it was a modem working on a phone line, looks like a 56k modem. So looking at what my final answer should be i'm thinking the number of bits will be 8. But herein lies the problem, i need to prove how i got to this answer (assuming i'm right )...pity i can't just say, "yeah - it looked like it was going to work out so i went with it!!"

Now i'm having another thought. Would the number of levels be 3???? since its 2^n (2^3 = 8) - so the number of sampling levels would be 3. And that means that there can be 8bits per sample (umm i'm kinda playing it fast and loose here) -> pleasetell me i cracked it!!

That would also leave the data rate at 56kbs (8*7000b).

If i'm wrong, yeah i said if!, can someone explain it to me like a 2yr old?? I won't feel patronised i get it a lot

Peace

I had a chat with my lecturer and its simply a case of plugging in higher valuse for n until the SNR of 30db is about the same as 20log2^n + 1.76

Which means that a value of 5 for n gives the correct number of levels in the signal.

The data rate is then (2^5 ) * 7000 , or so i believe.