# Signal-to-noise ratio



## z06dustin (Oct 6, 2009)

Sect 43, Problem 10:



> A digital meter uses the number system base b and has n digits. The noise is one half the place value of the least significiant digit. (a) Determine the signal-to-noise ratio for the case where the reading is half the full-scale value. (B) Obtain the signal to noise ratios for the following cases
> n b S/N(dB)
> 
> 3 10 20log(S/N)
> ...


Does anyone know how to do this, and is willing to explain it to me? I'm lost.


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## Dark Knight (Oct 6, 2009)

Can you verify the section and the problem number please? I have a different one at 43-10


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## z06dustin (Oct 6, 2009)

Yes, 43 (Measurement &amp; Instrumentation) problem #10. It's my understanding they changed for the '09 rev of the test, could you have an older one? I have the answer too (see below) but I was hoping someone could explain it to me..... the algorithm makes no sense to me. I can extrapolate, plug and chug, but I"d really rather understand what's going on such that if the problem on the test is different I'll be able to adapt.



> ( a )Full-scale is 999base10 for the first case. The noise is 0.5. At half-scale the signal is 500base10, so the signal-to--noise ratio is 500/0.5= 10^3, or S/N|db = 20log(10^3) = 60dB( b ) For the octal, full scase in base 10 is (7)(8)^3+(7)(8)^2+(7)(8)^1+7 = 4095
> 
> Half-scale is 2048
> 
> ...


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## benbo (Oct 6, 2009)

z06dustin said:


> Yes, 43 (Measurement &amp; Instrumentation) problem #10. It's my understanding they changed for the '09 rev of the test, could you have an older one? I have the answer too (see below) but I was hoping someone could explain it to me..... the algorithm makes no sense to me. I can extrapolate, plug and chug, but I"d really rather understand what's going on such that if the problem on the test is different I'll be able to adapt.


Stop me if I'm explaining something easy. You might be reading too much into this.

It looks like they are just taking the number of bits in a certain base, converting into base 10 with a half bit noise, and then plugging into the formula.

Like for the 4 bit, base 8 they are just converting 7777 to base 10, dividing by 2 to get half scale, dividing by the noise (.5) and then converting to dB.

In otherwords, they are doing a D/A conversion to get the Analog signal, dividing by the noise, and converting to dB.

Am I missing something?


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## z06dustin (Oct 6, 2009)

benbo said:


> Stop me if I'm explaining something easy. You might be reading too much into this.
> It looks like they are just taking the number of bits in a certain base, converting into base 10 with a half bit noise, and then plugging into the formula.
> 
> Like for the 4 bit, base 8 they are just converting 7777 to base 10, dividing by 2 to get half scale, dividing by the noise (.5) and then converting to dB.
> ...


You probably are, and I'm probably just a moron. Where did you get the 7777 from? Can you post the formulae you're speaking of?


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## benbo (Oct 6, 2009)

z06dustin said:


> You probably are, and I'm probably just a moron. Where did you get the 7777 from? Can you post the formulae you're speaking of?


From the question.

The second example down. They want 4 bits, in base 8. So the largest thing that can show up is a seven. Four of them

Just like if I want 6 bits in base 2, it would be 111111

It's a little confusing because you expect it to be harder. I thought it was something else until I saw the answer. Maybe it is something else and I'm oversimplifying.

By the way, I'm calling it bits, but whatever you call a digit in base 8.


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## benbo (Oct 6, 2009)

Or just think about your calculator. If it has 7 places, and is base 10 the full scale reading would be 9999999


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## z06dustin (Oct 6, 2009)

ahck ok i got it. thanks benbo.


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## benbo (Oct 6, 2009)

z06dustin said:


> ahck ok i got it. thanks benbo.


It's horribly worded. I think the problems are much clearer on the exam.


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## Flyer_PE (Oct 6, 2009)

benbo said:


> It's horribly worded. I think the problems are much clearer on the exam.


Concur. Camara's problems go into a lot more detail than you will likely see in the actual exam.


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## Dark Knight (Oct 6, 2009)

^^^^^Concur

I do not know how mch time you have left but I found that the Kaplan's Practice Test was the best way to prepare, next to the NCEES Test. Kaplan's problems are not PE style but I can guarantee you that will make you review fundamentals that you will need during the test.

Just my :2cents:


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## benbo (Oct 6, 2009)

Dark Knight said:


> ^^^^^Concur
> I do not know how mch time you have left but I found that the Kaplan's Practice Test was the best way to prepare, next to the NCEES Test. Kaplan's problems are not PE style but I can guarantee you that will make you review fundamentals that you will need during the test.
> 
> Just my :2cents:


I also liked Kaplan but to some it is frustrating because it has a lot of errata, and it is pretty complicated compared to the actual exam.

By the way - if you are taking the power emphasis (like most people do) I doubt you will ever see a problem on this type of stuff on the exam. But now that you understand what the heck they were asking for you probably will wish that you do see one like this.


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## z06dustin (Oct 7, 2009)

> I do not know how mch time you have left but


as of this morning i have 15 days, 23 hours, 18min and 1 second left.


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