# Signal to Noise ratio problem

A 200mv peak to peak sinusoidal signal is applied to an ideal 12 bit A/D converter, for which Vref(v p-p full scale) is 5v. Find signal to noise ratio.

This is a problem worked out by one of my lecturer and the answer seems to be 46dB. But I am confused on his method. Is there anyone who could explain me how the result was worked out?

• What is his method? What in his method don't you understand? – user17592 Apr 20 '13 at 5:22
• he calculates the S.N.R max by S.N.R max= 6.02N + 1.76 then he finds the drop of power, and subtracts drop of power from S.N.R max – DesirePRG Apr 20 '13 at 5:37

If $N$ is the number of bits then your input signal range is divided into quantization intervals of size

$$q = \frac{V_{ref}}{2^N}$$

The maximum quantization error is $q/2$ and it is usually assumed that the quantization error is uniformly distributed between $-q/2$ and $q/2$. So the PDF of the quantization error is constant between $-q/2$ and $q/2$ with height $1/q$. With this assumption the quantization noise power is

$$P_q = \frac{1}{q}\int_{-q/2}^{q/2}x^2dx = \frac{q^2}{12} = \frac{V_{ref}^2}{12\cdot 2^{2N}}$$

The signal power is given by

$$P_x = \frac{V_{max}^2}{2} \text{ with } V_{max} = 100mV$$

The factor $1/2$ in the signal power comes from the fact that the input is sinusoidal, i.e. its power is given by half its maximum value. Note that the maximum value is the amplitude which is half the peak-to-peak value. Putting everything together we get

$$SNR = 10\log\frac{P_x}{P_q} = 10\log\frac{6V_{max}^22^{2N}}{V_{ref}^2} = \\ = 10\log\frac{6V_{max}^2}{V_{ref}^2} + 10\log 2^{2N} = 10\log\frac{6(0.1)^2}{5^2} + N\cdot 20\log 2 =\\ = -26.2 + 12\cdot 6.02 = 46\text{dB}$$

EDIT: To see how the $1.76$dB pop up we now use $V_{pp}$ instead of $V_{max}$, where $V_{pp}=2V_{max}$ is the peak-to-peak input voltage. The SNR can then be written as

$$SNR = 10\log\frac{3V_{pp}^2}{2V_{ref}^2} + 10\log 2^{2N} =\\ = 10\log\frac{V_{pp}^2}{V_{ref}^2} + 10\log\frac{3}{2} + N\cdot 20\log 2 =\\ = 10\log\frac{V_{pp}^2}{V_{ref}^2} + 1.76 + 6.02\cdot N$$

So if we use the maximum input range, i.e. $V_{pp}=V_{ref}$ we get the formula that was mentioned in the question.

• Thanx a lot @matt – DesirePRG Apr 20 '13 at 10:22

Quantization noise is the mismatch in actually presented voltage and the digital representation of the voltage. The resolution of the ADC is 12 bits and its reference voltage is 5V.

This means that the maximum resolution of the ADC is $2^{12}-1 = 4095$ steps divided.

These 4095 steps divide 5V down to: $\dfrac{5V}{4095} = {1.22\text{mV}_\text{pp}}$.

Any signal sampled will have a maximum accuracy of 1.22mV. So your noise will be:

$\dfrac{S}{N} = \dfrac{200\text{mV}_\text{pp}}{1.22\text{mV}_\text{pp}}$

$\left ( \dfrac{S}{N} \right )_\text{dB} = 20 \log\left ( \dfrac{S}{N} \right ) = 20\log\left ( \dfrac{200\text{mV}}{\frac{5\text{V}}{2^{12}-1}} \right ) = 20\log\left ( \dfrac{200\text{mV} \cdot (2^{12}-1)}{5\text{V}} \right )\approx \boxed{44.29\text{dB}}$

The above calculation accounts for the 6.02N in your comment. Not sure where the '1.76 drop of power' came from, did you miss any details in the question? Notice that 44.29 + 1.76 = 46.05

• The 1.76dB come from the relation between the amplitude of a sinusoidal signal and its power. $1.76 = 10\log(3/2)$. – Matt L. Apr 20 '13 at 9:03