bandpass sampling

[anu]

hello friends
i need help regarding bandpass sampling i can see that already lot of
discussion had taken in this topic and but i am getting confused after
reading those discussion
i am having a signal of BW =5MHZ and my Fc=60 Mhz i want to sample this
signal with Fsamp=16MHZ ,(according to sampling theorem my sampling
frequency should be Fsamp>2BW.) and after sampling my signal will lie
in the first nyquist zone (0 to FS/2).
My problem is i am not able to visualize this signal in frequency
domain how the spectrum will look like.

[Bhaskar Thiagarajan]

"anu" <[email protected]> wrote in message
news:[email protected] oups.com...
> hello friends
> i need help regarding bandpass sampling i can see that already lot of
> discussion had taken in this topic and but i am getting confused after
> reading those discussion
> i am having a signal of BW =5MHZ and my Fc=60 Mhz i want to sample this
> signal with Fsamp=16MHZ ,(according to sampling theorem my sampling
> frequency should be Fsamp>2BW.) and after sampling my signal will lie
> in the first nyquist zone (0 to FS/2).
> My problem is i am not able to visualize this signal in frequency
> domain how the spectrum will look like.

What is Fc here?

[Phil]

Anu,

I assume that Fc is the carrier frequency. If you look up aliasing,
you should be able to find what you are looking for. Be aware that
your signal will not be properly centered about DC once you sample it
at 16 MHz.

If you are using an actual ADC (analog to digital converter) you will
have to make sure that its analog bandwidth is greater than 60 MHz. If
not, your sampled signal will be attenuated.

You can probably ignore this part, but if you are interested... your
sampled signal will experience the effect of jitter around the 60 MHz
range so keep this in mind when you look at the signal quality.

I hope that this helps.

[Jerry Avins]

anu wrote:
> hello friends

> ... BW =5MHZ ...

Is that the bandwidth of interest, or the band outside which there is no
signal?

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

[anu]

MY band of interest is 5 MHz and Fc=60MHz.My problem is i am not able
to visulaize the signal after sampling.As after sampling my information
should lie in first Nyquist zone ie between 0 to Fs/2.

[krishna_sun82]

>MY band of interest is 5 MHz and Fc=60MHz.My problem is i am not able
>to visulaize the signal after sampling.As after sampling my information
>should lie in first Nyquist zone ie between 0 to Fs/2.
>
>

For a time being forget the bandpass sampling theorem and consider th
original theorem. If you have a sinusoid with a frequency fb and you ar
sampling it at a frequency fs, such that fs >= 2 * fb. Now you represen
the sampled signal in frequency domain, at the frequency fb, between 0 an
fs/2. Also it will be mirrored in between -fs/2 and 0. But the replica o
the sample can be found as fs + fc and fs - fc, also in 2fs + fc and 2fs
fc. So it is the property of the sampling that in frequency domain a
infinite replica of the sampled signal will be present in an intervel o
fs in between.

Now, let us come back to the baseband sampling theorem. If you say tha
your signal is in the band 2B, then your highest frequency is fc + B an
the lowest is fc - B. This is nothing but a signal in between -B and +
with zero as the center frequency, which is modulated with the carrie
frequency fc, resulting in a frequency shift. We can consider this as
set of sinusoids with highest frequency as B. This is true, since an
signal can be represented as a set of sinusoids as per Fourier. So yo
need to sample it with fs = 2 * B. Drawing the same analogy as in th
original sampling theorem, now you can say that the frequency spectrum o
the sampled signal will be found in between -fs/2 and fs/2. Similarl
there will be infinite such replica.

Can you visualize it now?

- Krishna

[anu]

thanks for ur help.what i was thinking is that my entire band of signal
ie 2B will lie Between 0 to fs/2 and same is mirrored in between 0 to
-fs/2.one more help if after sampling my output of sampler is going to
a mixer than what should be the beating frequency of my
oscillator.thanks again

[Jerry Avins]

anu wrote:
> MY band of interest is 5 MHz and Fc=60MHz.My problem is i am not able
> to visulaize the signal after sampling.As after sampling my information
> should lie in first Nyquist zone ie between 0 to Fs/2.

The best overall discussion of sub-band sampling is early in Lyons'
"Understanding Digital Signal Processing.". You will need to filter out
extraneous signals not in your band of interest before sampling.

One way to look at sub-band sampling is as controlled aliasing. When
sampling at baseband, frequencies higher than Fs/2 appear shifted or
reflected and shifted into baseband. When the frequencies above Fs/2 are
the only ones present in the sampler input, the alias frequencies,
rather than interfering, can accurately represent the sampled signal.

There are restrictions on the range of sampling frequencies that can
accurately capture the information in any particular band. Sampling at
more than twice the bandwidth of interest is a given -- Nyquist will not
be denied -- but it is also necessary to avoid any "folding". Lyons lays
it all out very clearly.

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

[Thomas Magma]

In the equation for negative images f(K)=IF + K * Fs where IF is the
intermittent frequency being sampled and K is an integer, it appears that
your third negative image is falling at 3/4 Fs. If you want it to fall at
1/4 Fs you have to sample your 60MHz at 18.46MHz.

Thomas

"anu" <[email protected]> wrote in message
news:[email protected] oups.com...
> MY band of interest is 5 MHz and Fc=60MHz.My problem is i am not able
> to visulaize the signal after sampling.As after sampling my information
> should lie in first Nyquist zone ie between 0 to Fs/2.
>

[Mike Yarwood]

"anu" <[email protected]> wrote in message
news:[email protected] oups.com...
> MY band of interest is 5 MHz and Fc=60MHz.
But you have already said Fc=60MHz ; what Bhaskar was asking is what do you
mean by Fc? Is it the center of your band of interest for example?
> My problem is i am not able
> to visulaize the signal after sampling.As after sampling my information
> should lie in first Nyquist zone ie between 0 to Fs/2.
>
So why not get hold of a free soft-sys like scilib then generate two or more
uneven amplitude sinusoids at Fc+/-2.5MHz and see what they look like when
you only sample every 1/16 microseconds? You can plot them, you can add
noise, jitter what have you and look at the output, look at DFT's of the
output, try different frequencies, all sorts of nice aids for visualisation.

Best of Luck - Mike

[Thomas Magma]

"intermediate frequency" I meant.

Dam you spell checker, why can't you correct my stupidity as well.

"Thomas Magma" <[email protected]> wrote in message
news:v781g.35488$7a.32313@pd7tw1no...
> In the equation for negative images f(K)=IF + K * Fs where IF is the
> intermittent frequency being sampled and K is an integer, it appears that
> your third negative image is falling at 3/4 Fs. If you want it to fall at
> 1/4 Fs you have to sample your 60MHz at 18.46MHz.
>
> Thomas
>
> "anu" <[email protected]> wrote in message
> news:[email protected] oups.com...
>> MY band of interest is 5 MHz and Fc=60MHz.My problem is i am not able
>> to visulaize the signal after sampling.As after sampling my information
>> should lie in first Nyquist zone ie between 0 to Fs/2.
>>
>
>