# Sample & Hold: Estimate jitter delay of an ADC

I've got a question regarding the effects of sample and hold. My input signal is $$x$$, and my output signal $$x_{SH}$$. The error between the signals is $$e_{SH} = x_{SH}-x$$.

The sampling frequency is set to $$f_s = 1000kHz$$ and the jitter delay to $$t_{jitter}=4\cdot 10^{-4}s$$.

The influence of the jitter is implemented in Matlab as the following:

t = (0:N-1)/f_S;
x = sin(2*pi*f*t);
t_SH = t+Tjitter*(2*rand(size(t))-1);
x_SH = sin(2*pi*f*t_SH);
e_SH = x_SH-x;
rxx = xcorr(x,x,'biased');
ree = xcorr(e_SH,e_SH,'biased');


Now I'd like to estimate $$t_{jitter}$$ using the formula:

$$SNR = 20 \log_{10}\left[\frac{1/\sqrt{2}}{\Delta v_{rms}}\right]=20 \log_{10} \left[\frac{1}{2\pi f t_{jitter}}\right]\\$$ $$\Rightarrow t_{jitter} = \frac{1}{ 2\pi f \left(\sqrt{ \frac{P_{signal}}{P_{noise}} }\right) }=\frac{1}{ 2\pi f \left(\sqrt{ \frac{max(r_{xx})}{max(r_{ee})}} \right) }$$

Now I only achieve the same result of $$t_{jitter}$$ if I add a factor of $$\sqrt{3}$$, which I can't explain why.

$$\Rightarrow t_{jitter} = \frac{1}{ 2\pi f \left(\sqrt{ \frac{P_{signal}}{P_{noise}\cdot 3} }\right) }$$

I'd be happy for some help! Thanks in advance!

I think the jitter to SNR formula is based on a gaussian jitter.

You use "rand" which yields uniformly distributed numbers. You should use randn() instead which yields numbers distributed according to a gaussian distribution.

• Good catch Ben! And the variance of a uniform distribution [0..1] is $1/12$ If he uses randn instead it would have a variance = 1 so a difference of $1/12$ between the two in power if I am thinking correctly --- I am trying to see the $\sqrt{3}$ factor any further insight? Commented Nov 26, 2019 at 0:02
• I'm puzzled too. Maybe the bias of the uniform distribution?
– Ben
Commented Nov 26, 2019 at 0:15
• Cycle to cycle jitter vs absolute time jitter? (cycle to cycle jitter is a high pass on what would otherwise be white noise) Commented Nov 26, 2019 at 0:18
• The jitter is actually 2 * rand() - 1, which would yield a variance of 1/3 Standard deviation would be 1/sqrt(3), Maybe that's where the sqrt(3) comes from
– Ben
Commented Nov 26, 2019 at 1:37
• Awesome you figured it out- that is exactly it. The std of a uniform distribution is $d/\sqrt{12}$ and in this case d = 2 so the variance = 4/12 = 1/3. The bias doesn't do anything to the variance. But there is the 1/3 factor compared to using randn-- just like you first suggested (so using randn and without the x2 factor). The world is back in alignment. Commented Nov 26, 2019 at 1:48