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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!

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1 Answer 1

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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.

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    $\begingroup$ 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? $\endgroup$ Commented Nov 26, 2019 at 0:02
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    $\begingroup$ I'm puzzled too. Maybe the bias of the uniform distribution? $\endgroup$
    – Ben
    Commented Nov 26, 2019 at 0:15
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    $\begingroup$ Cycle to cycle jitter vs absolute time jitter? (cycle to cycle jitter is a high pass on what would otherwise be white noise) $\endgroup$ Commented Nov 26, 2019 at 0:18
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    $\begingroup$ 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 $\endgroup$
    – Ben
    Commented Nov 26, 2019 at 1:37
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    $\begingroup$ 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. $\endgroup$ Commented Nov 26, 2019 at 1:48

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