# Counter-example to Nyquist's Sampling Theorem?

1. Mar 23, 2005

### chingkui

The Sampling Theorem states that when you have a signal x(t) bandlimited to B Hz, then if you sample the signal at frequency f_s higher than or equal to 2B, then you can use the sample to reconstruct the original signal x(t) uniquely.
Yet, if you consider the simple sinusoidal signal x(t)=sin(2*pi*B*t), which obviously only has one frequency of B Hz, and sample with a frequency f_s=2B at time t=0,1/(2B),2/(2B),3/(2B),4/(2B),...,n/(2B),... then, all you get is a sequence x_sample(0)=x_sample(1)=...=x_sample(n)=sin(n*pi)=0, and you would naturally conclude that x(t)=0. Wouldn't it be a very simple counter-example to the sampling theorem?

2. Mar 23, 2005

### willib

How do you figure this??
then, all you get is a sequence x_sample(0)=x_sample(1)=...=x_sample(n)=sin(n*pi)=0, and you would naturally conclude that x(t)=0.

the sampling theorem states that you must sample at least twice the highest expected frequency, which you are sampling..
what you get from sampling is a digital number..

3. Mar 23, 2005

### faust9

What chingkui is getting at is correct. Nyquist need not apply. Sampling of recurring wave patterns can be done at less than the wave frequence and still give enough useful information to reconstruct the original wave.

Here's a little device which utilizes this principle: http://www.bitscope.com/

I wouldn't use the bitscope to T/S a glitch problem (I.E. a messed up pulse at a frequence 1/2 higer than the bitscope frequence) because there's no guarantee that the bitscope will pick up the glitch--especially if said glitch occured somewhat randomely. But, sampling of recurring or semirecurring wave patterns are easily and accuretly accomplished with a sample rate less than the input frequency.

4. Mar 23, 2005

### chingkui

What I mean is that if you know the signal is bandlimited to B Hz, and you sample at frequency f_s=2B, if someone send you a sine wave x(t)=sin(2*pi*B*t), and you sample at t=0,1/(2B),2/(2B),3/(2B),4/(2B),...,n/(2B),... then you will be very unlucky and getting all zeros, which you probably will think there is no signal at all and you can have no way to reconstruct the original sinusoidal signal... thus violating the Sampling Theorem which garantee you can reconstruct the original signal.

5. Mar 23, 2005

### faust9

That is incorrect. I thought you were eluding to the fact that Nyquist need not apply in all situations. Here, plot sin(x) and sin(2x) you'll see these two functions cross at multiple points--enough to reconstruct a wave.

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6. Apr 1, 2005

### rbj

no, the sampling theorem says that you must sample at a rate (f_s) strictly greater than 2B. f_s = 2B is not good enough (there are an infinite number of sinusoids at frequency B that alias to the same set of samples.

yeah, that's a bummer. f_s must be bigger than 2B.

r b-j