>Hi Tanriover,
>
>Then I would suggest shifting the incoming carriers down to a lower
>rate by multiplying with a frequency such as 105Khz so as to get
>carriers at 3Khz and 6Khz This will give a notional carrier of 4.5Khz.
>If you then downsample to get a sampling rate of 18Khz then a 90deg
>phase shift is 1 sample so you can just use the quadrature demodulation
>- multiply the 3-6Khz signal by a 1 sample delayed  copy of itself and
>lowpass filter with a cutoff of 1.5Khz before slicing. The original
>sampling rate could be as low as 288KHz (16 * 18Khz) - it does not
>matter that the sum of 111+105 will alias as this image is quite far
>from the 3-6Khz range and can easily be filtered out.
>
>The guru's on comp.dsp could probably suggest a more elegant method -
>such as subsampling or similar...
>
>Regards - Robert
>
>

Hi Robert,

Thanks for your suggestion. Due to the system requirements, my carriers
have to fall within the 95 kHz and 125 kHz range. Also, for limitations on
the frequency generation on the microprocessor, 108 and 111 kHz carriers
were found to be suitable.

Kind regards,
Cagri (Tanriover)

		
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Hi Tanriover,

Then I would suggest shifting the incoming carriers down to a lower
rate by multiplying with a frequency such as 105Khz so as to get
carriers at 3Khz and 6Khz This will give a notional carrier of 4.5Khz.
If you then downsample to get a sampling rate of 18Khz then a 90deg
phase shift is 1 sample so you can just use the quadrature demodulation
- multiply the 3-6Khz signal by a 1 sample delayed  copy of itself and
lowpass filter with a cutoff of 1.5Khz before slicing. The original
sampling rate could be as low as 288KHz (16 * 18Khz) - it does not
matter that the sum of 111+105 will alias as this image is quite far
from the 3-6Khz range and can easily be filtered out.

The guru's on comp.dsp could probably suggest a more elegant method -
such as subsampling or similar...

Regards - Robert

>What is the data rate you are trying to demodulate?
>
>Robert
>
>

The data rate is not very high Robert:- about 1 kbits/sec.

Cagri
		
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"Tanriover" <c.tanriover@ieee.org> wrote in message 
news:JIqdnZIIyJ85-WHfRVn-tw@giganews.com...
>>>
>>> This message was sent using the Comp.DSP web interface on
>>> www.DSPRelated.com
>>
>>You just need to satisfy the Nyquist sampling frequency, and this is
> already
>>the case in your setup.
>>
>>I think both signals are correctly sampled and it is quite normal that
> you
>>think to see amplitude canges in sampled f1.Because f2 = fs/4, you always
>
>>sample f2 at the positive top, zero, negative top, zero, ... (assuming no
>
>>phase offset). Hence you probably see 1, 0 , +1, 0, ... or so for f2.
>>
>>For f1 however, you do not sample the sin at the same 'places'all the
> time.
>>If for example sample x would be at the top of the cos (--> +1) then
> sample
>>x+4 will be slightly left from the top of the cos, hence you get only
> 0.98
>>or so. Sample x+8 will be shifted again more to the left from the top,
> hence
>>maybe you only get 0.94 or so..
>>Hence visually you see some kind of oscillating envelope. This is just
> fine.
>>
>>I think that the oscillation you see is actually 111-108 = 3Hz
>>
>>
>>
>
> I see what you mean. I agree with your comments.
>
> I have checked the oscillation frequency and it is around 12 kHz, which I
> think is correct as the sampling frequency is 4 times the carrier
> frequencies. Hence 4(111k-108k)=12k is what one would expect to see.
>
> The matched filtering strategy does not work as effectively as expected
> when the carriers are too closely spaced and the number of sample points
> is low.
>
> Cagri

It may be helpful to consider just what you're trying to accomplish and just 
what the parameters of your "system" are.  I'm thinking you need to consider 
the bandwidth of the signals of interest, their separation and the frequency 
resolution of the arrangement.  (It appears that the bandwidth of the two 
carriers is narrow compared to their separation of 3kHz).

To resolve the two carriers, you need a frequency resolution of at least 
3kHz and a time record or observation time of at least 1/3kHz = 0.000333 
seconds.  At 444kHz sample rate this is around (i.e. at least) 1500 samples. 
Since you have only 444/108 = 4.1111 samples per cycle this means you need 
around (i.e. at least) 1500/4.1111 = 364 periods of the 108kHz carrier to 
resolve it relative to the 111kHz carrier.

Well, that's interesting and important I guess but perhaps not really to the 
point....

If your application requires that both carriers "look good" that implies (to 
me) sampling at 10x the carrier frequency or greater.  You will still have 
some "modulation" due to sampling if a carrier being sampled is no at an 
integer multiple of the sampling frequency.

The observed "oscillation at 12kHz" must actually be different as follows:
The 108 kHz signal is sampled at 444kHz.  This is 4.11111 samples per cycle. 
It repeats exactly every 37 cycles of the sample frequency and every 9 
cycles of the carrier frequency.  This occurs at 37*(1/444,000) seconds = 
0.000083333 seconds which is a period of 12kHz.

This observation is independent of the other carrier.  It might be viewed as 
the beat frequency between the 444 and 108 - it is the highest frequency 
that is a submultiple of both 444 and 108.

Now, let's consider the other carrier at 111,000Hz.  The sampled version of 
this one repeats every 4 samples.  We might ask the question, how many 
samples are needed so that both waveforms repeat exactly?  We saw that 37 
samples were adequate for 108kHz.  We observe that there must be a multiple 
of 4 samples for 111kHz to repeat so we need 4x37 or 148 samples.

148 samples at 444000 is 0.0003333 seconds long, representing a frequency 
resolution of 3kHz.  A resulting FFT has the two carriers immediately 
adjacent with equal amplitude and there is no leakage.  The time waveform 
(assuming both are cosines) goes to zero at midpoint where the two sinusoids 
destructively interfere.

Fred 

What is the data rate you are trying to demodulate?

Robert

>>
>> This message was sent using the Comp.DSP web interface on
>> www.DSPRelated.com
>
>You just need to satisfy the Nyquist sampling frequency, and this is
already 
>the case in your setup.
>
>I think both signals are correctly sampled and it is quite normal that
you 
>think to see amplitude canges in sampled f1.Because f2 = fs/4, you always

>sample f2 at the positive top, zero, negative top, zero, ... (assuming no

>phase offset). Hence you probably see 1, 0 , +1, 0, ... or so for f2.
>
>For f1 however, you do not sample the sin at the same 'places'all the
time. 
>If for example sample x would be at the top of the cos (--> +1) then
sample 
>x+4 will be slightly left from the top of the cos, hence you get only
0.98 
>or so. Sample x+8 will be shifted again more to the left from the top,
hence 
>maybe you only get 0.94 or so..
>Hence visually you see some kind of oscillating envelope. This is just
fine.
>
>I think that the oscillation you see is actually 111-108 = 3Hz
>
>
>

I see what you mean. I agree with your comments.

I have checked the oscillation frequency and it is around 12 kHz, which I
think is correct as the sampling frequency is 4 times the carrier
frequencies. Hence 4(111k-108k)=12k is what one would expect to see.

The matched filtering strategy does not work as effectively as expected
when the carriers are too closely spaced and the number of sample points
is low.

Cagri
		
This message was sent using the Comp.DSP web interface on
www.DSPRelated.com

>
> I am using 444 kHz sampling frequency (fs) to sample two different carrier
> frequencies: f1=108 kHz and f2=111 kHz.
>
> When I look at the sampled f2, the waveform seems to be pretty accurately
> represented. However, for f1 this is not the case. Even thought the
> frequency of f1 is represented accurately, the amplitudes seem to
> oscillate.
>
> My interpretation is that because fs is not an integer multiple of f1 and
> fs is not high enough, the sampled f1 does not look as good as f2.
>
> If the above is true then I need to choose my sampling frequency as the
> least common multiple (LCM) of f1 and f2 while satisfying Nyquist sampling
> theorem.
>
> Is my thinking correct or there is something that I am missing? Is it
> really necessary to choose the sampling frequency as an integer multiple
> of the waveform to be sampled? If this is the case, then sampling multiple
> frequency waveforms must be a nightmare!
>
>
> Many thanks,
> Cagri
>
>
>
>
>
>
> This message was sent using the Comp.DSP web interface on
> www.DSPRelated.com

You just need to satisfy the Nyquist sampling frequency, and this is already 
the case in your setup.

I think both signals are correctly sampled and it is quite normal that you 
think to see amplitude canges in sampled f1.Because f2 = fs/4, you always 
sample f2 at the positive top, zero, negative top, zero, ... (assuming no 
phase offset). Hence you probably see 1, 0 , +1, 0, ... or so for f2.

For f1 however, you do not sample the sin at the same 'places'all the time. 
If for example sample x would be at the top of the cos (--> +1) then sample 
x+4 will be slightly left from the top of the cos, hence you get only 0.98 
or so. Sample x+8 will be shifted again more to the left from the top, hence 
maybe you only get 0.94 or so..
Hence visually you see some kind of oscillating envelope. This is just fine.

I think that the oscillation you see is actually 111-108 = 3Hz

Tanriover wrote:
> Hi,
> 
> I am using 444 kHz sampling frequency (fs) to sample two different carrier
> frequencies: f1=108 kHz and f2=111 kHz.
> 
> When I look at the sampled f2, the waveform seems to be pretty accurately
> represented. However, for f1 this is not the case. Even thought the
> frequency of f1 is represented accurately, the amplitudes seem to
> oscillate.
> 
> My interpretation is that because fs is not an integer multiple of f1 and
> fs is not high enough, the sampled f1 does not look as good as f2.
> 
> If the above is true then I need to choose my sampling frequency as the
> least common multiple (LCM) of f1 and f2 while satisfying Nyquist sampling
> theorem.
> 
> Is my thinking correct or there is something that I am missing? Is it
> really necessary to choose the sampling frequency as an integer multiple
> of the waveform to be sampled? If this is the case, then sampling multiple
> frequency waveforms must be a nightmare!
> 
> 
> Many thanks,
> Cagri
> 
> 
> 
> 
> 
> 		
> This message was sent using the Comp.DSP web interface on
> www.DSPRelated.com

Why not use a pair of filters, one tuned to 111, the other to 108, then 
subtract their outputs. Their Qs will need to be fairly low, reflecting 
the small number of samples in a bit, but it ought to work fine. However 
you go about it, you want to detect the signals as frequencies, not 
operate on the samples with logic.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Hi,

I am using 444 kHz sampling frequency (fs) to sample two different carrier
frequencies: f1=108 kHz and f2=111 kHz.

When I look at the sampled f2, the waveform seems to be pretty accurately
represented. However, for f1 this is not the case. Even thought the
frequency of f1 is represented accurately, the amplitudes seem to
oscillate.

My interpretation is that because fs is not an integer multiple of f1 and
fs is not high enough, the sampled f1 does not look as good as f2.

If the above is true then I need to choose my sampling frequency as the
least common multiple (LCM) of f1 and f2 while satisfying Nyquist sampling
theorem.

Is my thinking correct or there is something that I am missing? Is it
really necessary to choose the sampling frequency as an integer multiple
of the waveform to be sampled? If this is the case, then sampling multiple
frequency waveforms must be a nightmare!


Many thanks,
Cagri





		
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