Simple Sampling question

Hi, 
Sorry if this question is simple. I am trying to get this confusion out of
my head. I have a hard time visualizing and understanding this simpling
question.

Let's say that we have only one tone signal (eg. 100MHz). So, when you look
in the frequecny domain you will see a tone at 100MHz. 
Now, when we do sample it let's say at 400MHz sampling frequency, why do we
see tones at (400-100=300MHz) and (400+100=500MHz)?
I understand the math behind it, and I also know that sampling deosn't
change the signal frequency, then how come we see tones at those
frequencies ? 

--Rudy	 

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Posted through www.DSPRelated.com

On Fri, 17 Jan 2014 16:34:12 -0600
"rudykeram" <51467@dsprelated> wrote:

> Hi, 
> Sorry if this question is simple. I am trying to get this confusion out of
> my head. I have a hard time visualizing and understanding this simpling
> question.
> 
> Let's say that we have only one tone signal (eg. 100MHz). So, when you look
> in the frequecny domain you will see a tone at 100MHz. 
> Now, when we do sample it let's say at 400MHz sampling frequency, why do we
> see tones at (400-100=300MHz) and (400+100=500MHz)?
> I understand the math behind it, and I also know that sampling deosn't
> change the signal frequency, then how come we see tones at those
> frequencies ? 
> 
> --Rudy	 
> 
> _____________________________		
> Posted through www.DSPRelated.com

You don't.  There is no tone you see at 300 or 500 MHz.  What you see
is a tone at 1/2*pi in the digital frequency domain.  That tone line is
also where a 300 or 500 MHz signal would wind up if that's what you had
sampled instead.  Therefore, data in the digital frequency domain is
meaningless without additional context, such as knowing that the signal
being sampled had been bandlimited to the range DC-200 MHz.

-- 
Rob Gaddi, Highland Technology -- www.highlandtechnology.com
Email address domain is currently out of order.  See above to fix.

rudykeram <51467@dsprelated> wrote:

(snip)

> Sorry if this question is simple. I am trying to get this 
> confusion out of my head. I have a hard time visualizing and 
> understanding this simpling question.
> 
> Let's say that we have only one tone signal (eg. 100MHz). 
> So, when you look in the frequecny domain you will see a tone 
> at 100MHz. 

> Now, when we do sample it let's say at 400MHz sampling frequency, 
> why do we see tones at (400-100=300MHz) and (400+100=500MHz)?
> I understand the math behind it, and I also know that sampling deosn't
> change the signal frequency, then how come we see tones at those
> frequencies ? 

You have to explain the system a little more. If you are sampling
baseband at 400MHz, the input and output should be band limited
at 200MHz, so there shouldn't be any 300MHz or 500MHz. 

But sampling is a non-linear process, so if you think about it
in non-linear analog terms, those are the cross (intermodulation)
frequencies. 

-- glen

Den fredag den 17. januar 2014 23.34.12 UTC+1 skrev rudykeram:
> Hi, 
> 
> Sorry if this question is simple. I am trying to get this confusion out of
> 
> my head. I have a hard time visualizing and understanding this simpling
> 
> question.
> 
> 
> 
> Let's say that we have only one tone signal (eg. 100MHz). So, when you look
> 
> in the frequecny domain you will see a tone at 100MHz. 
> 
> Now, when we do sample it let's say at 400MHz sampling frequency, why do we
> 
> see tones at (400-100=300MHz) and (400+100=500MHz)?
> 
> I understand the math behind it, and I also know that sampling deosn't
> 
> change the signal frequency, then how come we see tones at those
> 
> frequencies ? 
> 

simply put you don't see those frequencies, but unless your input is bandlimited you can't tell if what you think is 100MHz is really 100MHz, the sequence of samples would be the same if it was 300MHz or 500MHz etc..

-Lasse

>rudykeram <51467@dsprelated> wrote:
>
>(snip)
>
>> Sorry if this question is simple. I am trying to get this 
>> confusion out of my head. I have a hard time visualizing and 
>> understanding this simpling question.
>> 
>> Let's say that we have only one tone signal (eg. 100MHz). 
>> So, when you look in the frequecny domain you will see a tone 
>> at 100MHz. 
>
>> Now, when we do sample it let's say at 400MHz sampling frequency, 
>> why do we see tones at (400-100=300MHz) and (400+100=500MHz)?
>> I understand the math behind it, and I also know that sampling deosn't
>> change the signal frequency, then how come we see tones at those
>> frequencies ? 
>
>You have to explain the system a little more. If you are sampling
>baseband at 400MHz, the input and output should be band limited
>at 200MHz, so there shouldn't be any 300MHz or 500MHz. 
>
>But sampling is a non-linear process, so if you think about it
>in non-linear analog terms, those are the cross (intermodulation)
>frequencies. 
>
>-- glen
>

I geuss I am still little confused. (look at the following image please).

http://fourier.eng.hmc.edu/e101/lectures/sampling_theorem_f.gif

To make the above simpler, I said that we are dealing with only one tone
(100MHz). So, when we sample it at 400MHz, we would see the same tone
around the sampling frequency. So, we would see 300MHz and 500MHz. But I
cannot see why this is the case?

I guess on a bigger picture, what I am asking is I am not sure why when
sample a signale, we copy the same baseband bandwidth arund the sampling
frequency?
--Rudy	 

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Posted through www.DSPRelated.com

On 2014-01-17 23:34, rudykeram wrote:
> Hi, 
> Sorry if this question is simple. I am trying to get this confusion out of
> my head. I have a hard time visualizing and understanding this simpling
> question.
> 
> Let's say that we have only one tone signal (eg. 100MHz). So, when you look
> in the frequecny domain you will see a tone at 100MHz. 
> Now, when we do sample it let's say at 400MHz sampling frequency, why do we
> see tones at (400-100=300MHz) and (400+100=500MHz)?

You'll "see" also at -100MHz a tone, and one at each
200MHz with 100MHz offset, positive and negative axis.

> I understand the math behind it, and I also know that sampling deosn't

No offence, but I doubt that.

> change the signal frequency, then how come we see tones at those
> frequencies ? 

Mathematically speaking, a *sampled* signal has
a _periodic_ spectrum (not sure if it needs to
be real too).

In your example, the period is 200MHz (half the
sampling frequency) and since you've only a tone
at 100MHz, you'll "see" at 100, 300, 500, 700, etc.
the very same tone. Same for -100, -300, etc.

Now, you might ask: why sampling results in periodic
spectrum?
This is the way sampling is defined in terms of
convolution. For example see:

http://www.cg.tuwien.ac.at/~theussl/DA/node25.html

bye,

-- 

piergiorgio

>On 2014-01-17 23:34, rudykeram wrote:
>> Hi, 
>> Sorry if this question is simple. I am trying to get this confusion out
of
>> my head. I have a hard time visualizing and understanding this simpling
>> question.
>> 
>> Let's say that we have only one tone signal (eg. 100MHz). So, when you
look
>> in the frequecny domain you will see a tone at 100MHz. 
>> Now, when we do sample it let's say at 400MHz sampling frequency, why do
we
>> see tones at (400-100=300MHz) and (400+100=500MHz)?
>
>You'll "see" also at -100MHz a tone, and one at each
>200MHz with 100MHz offset, positive and negative axis.
>
>> I understand the math behind it, and I also know that sampling deosn't
>
>No offence, but I doubt that.
>
>> change the signal frequency, then how come we see tones at those
>> frequencies ? 
>
>Mathematically speaking, a *sampled* signal has
>a _periodic_ spectrum (not sure if it needs to
>be real too).
>
>In your example, the period is 200MHz (half the
>sampling frequency) and since you've only a tone
>at 100MHz, you'll "see" at 100, 300, 500, 700, etc.
>the very same tone. Same for -100, -300, etc.
>
>Now, you might ask: why sampling results in periodic
>spectrum?
>This is the way sampling is defined in terms of
>convolution. For example see:
>
>http://www.cg.tuwien.ac.at/~theussl/DA/node25.html
>
>bye,
>
>-- 
>
>piergiorgio
>

Okay. I think I see it now. So, if I have a 100MHz signal, and a 300 MHz
signal or a 500MHz signal, and if I were to smaple them at 400MHz, I will
get the exact same sample set. So, I wouldn't know which one was my
original signal. But, if I abide to this rule that the sampling rate has to
be twice the original signal bandwidth, then I know that my original singal
was 100MHz.
But when we look at the frequency domain, we include all these tones,
because with respect to smapling frequency (400MHz in this case), they are
all the same.

Thanks for all the answers.
--Rudy	 

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Posted through www.DSPRelated.com

On Fri, 17 Jan 2014 17:44:19 -0600, "rudykeram" <51467@dsprelated>
wrote:

   [Snipped by Lyons]
>
>Okay. I think I see it now. So, if I have a 100MHz signal, and a 300 MHz
>signal or a 500MHz signal, and if I were to smaple them at 400MHz, I will
>get the exact same sample set. So, I wouldn't know which one was my
>original signal. But, if I abide to this rule that the sampling rate has to
>be twice the original signal bandwidth, then I know that my original singal
>was 100MHz.
>But when we look at the frequency domain, we include all these tones,
>because with respect to smapling frequency (400MHz in this case), they are
>all the same.
>
>Thanks for all the answers.
>--Rudy	 

Hi Rudy,
   You're struggling to understand the peculiar 
spectral characteristics of sampled signals.
(That is, a kind of spectral behavior that 
DOES NOT exist in the continuous-signal world.)

Don't worry Rudy, every guy here has been through 
that SAME struggle at one time or another.  And it 
looks like your understanding is becoming stronger 
and stronger.

I just thought I'd mention to you that if you sample 
a continuous 100 MHz cosine wave and a continuous 
300 MHz cosine wave, you'll obtain identical 'sampled' 
x(n) sequences.

However, if you sample a continuous 100 MHz sine wave 
and a continuous 300 MHz sine wave, you'll obtain two 
different 'sampled' sequences.

[-Rick-]

On Fri, 17 Jan 2014 22:48:42 +0000, glen herrmannsfeldt wrote:

> rudykeram <51467@dsprelated> wrote:
> 
> (snip)
> 
>> Sorry if this question is simple. I am trying to get this confusion out
>> of my head. I have a hard time visualizing and understanding this
>> simpling question.
>> 
>> Let's say that we have only one tone signal (eg. 100MHz).
>> So, when you look in the frequecny domain you will see a tone at
>> 100MHz.
> 
>> Now, when we do sample it let's say at 400MHz sampling frequency,
>> why do we see tones at (400-100=300MHz) and (400+100=500MHz)?
>> I understand the math behind it, and I also know that sampling deosn't
>> change the signal frequency, then how come we see tones at those
>> frequencies ?
> 
> You have to explain the system a little more. If you are sampling
> baseband at 400MHz, the input and output should be band limited at
> 200MHz, so there shouldn't be any 300MHz or 500MHz.
> 
> But sampling is a non-linear process, so if you think about it in
> non-linear analog terms, those are the cross (intermodulation)
> frequencies.

Ideal sampling is a perfectly linear process, and does not generate any 
intermodulation frequencies.  If you doubt me, check the mathematical 
representation of sampling to see if it obeys superposition -- if you do 
your math right, you will find that it does.

Intermodulation distortion is, indeed, a nonlinear phenomenon, but it is 
completely different from aliasing.  It may well happen with a real 
sampler, but it is because the sampler has nonlinearities, not because 
the sampler is a sampler.

What sampling _is_, is time-varying.  That accounts for all of its 
behavior, without incorrectly applying irrelevant math.

-- 
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

On Fri, 17 Jan 2014 17:17:03 -0600, rudykeram wrote:

>>rudykeram <51467@dsprelated> wrote:
>>
>>(snip)
>>
>>> Sorry if this question is simple. I am trying to get this confusion
>>> out of my head. I have a hard time visualizing and understanding this
>>> simpling question.
>>> 
>>> Let's say that we have only one tone signal (eg. 100MHz).
>>> So, when you look in the frequecny domain you will see a tone at
>>> 100MHz.
>>
>>> Now, when we do sample it let's say at 400MHz sampling frequency,
>>> why do we see tones at (400-100=300MHz) and (400+100=500MHz)?
>>> I understand the math behind it, and I also know that sampling deosn't
>>> change the signal frequency, then how come we see tones at those
>>> frequencies ?
>>
>>You have to explain the system a little more. If you are sampling
>>baseband at 400MHz, the input and output should be band limited at
>>200MHz, so there shouldn't be any 300MHz or 500MHz.
>>
>>But sampling is a non-linear process, so if you think about it in
>>non-linear analog terms, those are the cross (intermodulation)
>>frequencies.
>>
>>-- glen
>>
>>
> I geuss I am still little confused. (look at the following image
> please).
> 
> http://fourier.eng.hmc.edu/e101/lectures/sampling_theorem_f.gif
> 
> To make the above simpler, I said that we are dealing with only one tone
> (100MHz). So, when we sample it at 400MHz, we would see the same tone
> around the sampling frequency. So, we would see 300MHz and 500MHz. But I
> cannot see why this is the case?
> 
> I guess on a bigger picture, what I am asking is I am not sure why when
> sample a signale, we copy the same baseband bandwidth arund the sampling
> frequency?
> --Rudy

I'm not sure where that graphic came from, but it's not right.

Try reading this, and see if it helps.  It's more intended to debunk some 
common misconceptions about what the Nyquist/Shannon sampling theorem 
implies, but it may answer your questions.

http://www.wescottdesign.com/articles/Sampling/sampling.pdf

Note that when you hear things about sampling, there are at least two 
different, fundamentally different, mathematical models used to represent 
it:

The one that prevailed when I was an undergrad was that sampling "is" 
multiplication by a series of dirac impulses, and that a sampled-time 
signal is just a continuous-time signal with all these impulses running 
through it.  This method has the advantages that it gives one unified 
treatment of sampled- and continuous-time systems, but it requires you to 
be a good mathematical contortionist when it comes time to go from one 
domain to the other.

The other view is that a sampled-time domain is simply a completely 
different critter than the continuous-time domain, so that some 
statements that are valid about sampled-time are invalid about continuous-
time, and visa versa.  It treats sampling and reconstruction as separate 
processes from everything else going on in the system.

Each of these models have their own advantages and their own adherents.  
I find it best to use the one that's most convenient at any given time.  
You probably don't want to try to learn all the subtleties at once, but 
be aware that some people are stuck in one or the other; it's up to you 
to be flexible in your listening.

-- 
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com