# Homework Help: Signal to Noise Ratio

1. May 12, 2014

### roam

1. The problem statement, all variables and given/known data

I need some help with the following problem (from an old exam):

A 16-bit single-ramp ADC has a sampling interval of T=1 μs and an input voltage range of 0 to 10V.

Since for this ADC the input must be positive, its design includes a fixed DC input of 5 V upon which the varying signal is superimposed (the 5V DC is not part of the signal). If the varying signal is a sinusoidal voltage of amplitude 1.8V, what is the SNR in dB?

2. Relevant equations

Signal to noise ratio in dB is given by:

$|(SNR)|_{dB} = 7.8+6b+20 \log(\frac{A}{R})$

R is the range, A is the amplitude, b is the bits.

3. The attempt at a solution

Using the above equation I got

$|(SNR)|_{dB} = 7.8+6\times 16 +20 \log(\frac{1.8}{10}) =88.90 \ dB$

But my answer was marked as incorrect! I'm very confused here.

So, do I need to add the 5V fixed DC to the 1.8V amplitude? But the question clearly states "the 5V DC is not part of the signal".

If I add 5V I will get:

$|(SNR)|_{dB} = 7.8+6\times 16 +20 \log(\frac{6.8}{10}) =100.45 \ dB$

But which method is correct?

Any help is greatly appreciated.

2. May 12, 2014

### rude man

I don't know where you got that formula, but I suppose since your answer was marked wrong, so must be the formula.

Take a direct approach:
what is the rms quantization noise for a 16 bit a/d with a 10V input span?
what is the rms voltage of a 1.8V amplitude sine voltage?
then divide the second by the first and change to dB.

(My answer came out a bit bigger than your first and a whole lot smaller than your second. The 5V offset does not enter the computations.)

Hint: how many bits are cobered with a sine wave of 3.6V amplitude?

3. May 12, 2014

### roam

Where does the 3.6V come from? Do you mean a 1.8V sine wave? I think $(1.8/10)\times 16 =2.88 \approx 3 \ bits$ are covered.

The formula comes from my notes. I've attached the derivation to this post.

So using your method for a 16 bit and 10V I get

$SNR = 6 \times 2^{2(16)} \times \left( \frac{10}{10} \right)^2 = 2.57 \times 10^{10}$

For the 1.8V I get

$SNR = 6 \times 2^{2(3)} \times \left( \frac{1.8}{10} \right)^2 = 12.44$

Dividing them gives 4.84 x 10-10

In decibel this is $10 \log(4.84 \times 10^{-10}) = -93.15 \ dB$

What was the value you had? And is it possible to somehow use my own formula?

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4. May 12, 2014

### roam

Frankly, I'm not sure what you meant by 3.6V. If you have subtracted 5 from 1.8 to get 3.2V, then using my formula I got:

$7.8+6\times 16+20 \log_{10} \left( \frac{3.2}{10} \right) = 93.9 \ dB$

Is this the value you've got?

I believe in the formula "b" is the number of bits of the ADC, not how many bits are covered by a signal of a particular amplitude. That's why I used 16.

I think the formula is correct, it worked on other problems. Is it possible to somehow modify it for this problem?

5. May 12, 2014

### rude man

Sorry, I had meant to erase the hint of my 1st post. The 3.6V is the range of input voltages but I should never have mentioned it. The rest of my post was what I wanted.

Our methods should be the same, and they very nearly are. I think your original value of 88.90 dB is correct to within roundoff errors plus the fact that 10log6 = 7.78, not 7.8. My answer is 89.22 dB which is close to yours. So, bottom line, I claim you did it right the first time given a somewhat imprecise formula.

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• ###### a-d snr.pdf
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6. May 12, 2014

### roam

Thank you so much for your post. So, should we not take the 5V offset into account somehow? Personally I also think my answer was correct (maybe slightly rounded off).

7. May 12, 2014

### rude man

The 5Vdc offset does not enter the picture at all.

Could you post your prof's explanation of why our methods are wrong?