Signal to Noise Ratio: Solving 16-bit ADC Problem

[roam]

Homework Statement

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?

Homework 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.

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.

 

[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?

 

[roam]

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?

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?

 

[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?

 

[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.

 

[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).

 

[rude man]

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).

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

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