Jerry Avins <jya@ieee.org> wrote in message news:<3ffee316$0$6735$61fed72c@news.rcn.com>...
> ganesh wrote:
>
> ...
> >
> > Very true, but you'll have to determine the phase for every
> > bit/symbol transmitted. Now, if u r designing ur own synthesizer, and
> > if u r willing to have this additional complexity
> > in ur logic due to that (for some other benefit ), its fine.
>
> I don't understand. Say you generate your carrier by indexing through a
> table of 2^n sines with a phase accumulator of n bits. Then the
> frequency is determined by the output rate (a conatant) and the
> increment to the phase accumulator. The "complexity" you speak of is
> simply retaining the value of the phase accumulator from one output to
> the next. Since that's needed even for a steady frequency, it's no added
> complexity at all.
hmmm, well, I was thinking of the simplest simplest possible system,
where
u dont have 2^n tables, but only one table of sines and no phase
accumulator.
This is possible if u use f1 = k1/T and f2 = k2/T, with k1 and k2
integers.
Whether this system actually satisfies ur other requirements, is a
different issue. Anyway, with the current levels
of integration in a chip, u may argue that 2^n sine tables r no big
deal..
>
> > Second, is the orthogonality. Non orthogonality is undesirable.
> > Now orthogonality and constant phase is satisfied v. nicely
> > if f1 = k1/T and f2 = k2/T (where k1 and k2 are integers).
> >
> > Now, if k1 and k2 are not exactly integers, but vary a little bit (say delta),
> > i.e say k2 = m.k1 + delta, the signals r not v. orthogonal. For the same
> > delta, larger the values of k1 and k2 the lesser the non-orthogonality.
>
> But k1 and k2 must be integers in any digital system when one looks
> deeply enough. It's not necessary that f1 = k1/T and f2 = k2/T, but only
> that f1 = k1/nT and f2 = k2/nT, where n can be very large.
U mean make n large to reduce the probability of error?? Agreed.
>
> > Jerry, I dont think I ever said there are two oscillators running. As u say,
> > I have a v. high freq clock. I use some n samples per bit/symbol transmitted.
> > Then send the samples to a DAC. And I'm sure u'll get a v. smooth waveform
> > (i.e at no point the derivative becomes infinity) if k1 and k2 are integers,
> > and for a given phase.
>
> So we see it pretty much the same way, then?
yeah...
ganesh
Reply by Jerry Avins●January 9, 20042004-01-09
Jerry Avins wrote:
...
> But k1 and k2 must be integers in any digital system when one looks
> deeply enough. It's not necessary that f1 = k1/T and f2 = k2/T, but only
> that f1 = k1/nT and f2 = k2/nT, where n can be very large.
KLONG! It should be
"that f1 = k1/(T/n) and f2 = k2/(T/n), where n can be very large."
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Jerry Avins●January 9, 20042004-01-09
ganesh wrote:
...
>
> Very true, but you'll have to determine the phase for every
> bit/symbol transmitted. Now, if u r designing ur own synthesizer, and
> if u r willing to have this additional complexity
> in ur logic due to that (for some other benefit ), its fine.
I don't understand. Say you generate your carrier by indexing through a
table of 2^n sines with a phase accumulator of n bits. Then the
frequency is determined by the output rate (a conatant) and the
increment to the phase accumulator. The "complexity" you speak of is
simply retaining the value of the phase accumulator from one output to
the next. Since that's needed even for a steady frequency, it's no added
complexity at all.
> Second, is the orthogonality. Non orthogonality is undesirable.
> Now orthogonality and constant phase is satisfied v. nicely
> if f1 = k1/T and f2 = k2/T (where k1 and k2 are integers).
>
> Now, if k1 and k2 are not exactly integers, but vary a little bit (say delta),
> i.e say k2 = m.k1 + delta, the signals r not v. orthogonal. For the same
> delta, larger the values of k1 and k2 the lesser the non-orthogonality.
But k1 and k2 must be integers in any digital system when one looks
deeply enough. It's not necessary that f1 = k1/T and f2 = k2/T, but only
that f1 = k1/nT and f2 = k2/nT, where n can be very large.
>
> Jerry, I dont think I ever said there are two oscillators running. As u say,
> I have a v. high freq clock. I use some n samples per bit/symbol transmitted.
> Then send the samples to a DAC. And I'm sure u'll get a v. smooth waveform
> (i.e at no point the derivative becomes infinity) if k1 and k2 are integers,
> and for a given phase.
So we see it pretty much the same way, then?
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by ganesh●January 9, 20042004-01-09
"Bevan Weiss" <kaizen__@NOSPAMhotmail.com> wrote in message news:<xtdLb.5671$9k7.118211@news.xtra.co.nz>...
> > > I can't see why k should be an integer. I accept that it allows for
> much
> > > easier maths in most situations, but don't see why it should be 2 as
> opposed
> > > to 2.5
> >
> > If k is not an integer, the resulting waveform is not smooth. To see
> why,
> > take the bit pattern 010101, with f1 for 0 and f2 for 1. If k is not an
> > integer the waveform is not smooth, there are abrupt changes. This results
> > in lot more power residing in the higher frequency ranges (if u take a
> > fourier transform of the waveform). If this signal were to pass thru
> > any comm channel, the distortion wud be much higher.
>
> I thought that k was only to do with the frequency deviation, and that
> whether or not the change was smooth was based more on the method used for
> modulation and whether it was phase coherant or otherwise.
> ie using a Direct Digital Synthesiser (DDS) with phase accumulator as the
> frequency modulator would produce a phase coherant FSK signal, no abrupt
> amplitude changes would be present, however an abrupt frequency change would
> obviously be present at the point the frequency word of the DDS is loaded.
>
> In the above example it doesn't matter what the new frequency word is loaded
> to, there will still be no abrupt amplitude change, the amplitude will
> simply change at the newly loaded frequency.
Very true, but you'll have to determine the phase for every
bit/symbol transmitted. Now, if u r designing ur own synthesizer, and
if u r willing to have this additional complexity
in ur logic due to that (for some other benefit ), its fine.
Second, is the orthogonality. Non orthogonality is undesirable.
Now orthogonality and constant phase is satisfied v. nicely
if f1 = k1/T and f2 = k2/T (where k1 and k2 are integers).
Now, if k1 and k2 are not exactly integers, but vary a little bit (say delta),
i.e say k2 = m.k1 + delta, the signals r not v. orthogonal. For the same
delta, larger the values of k1 and k2 the lesser the non-orthogonality.
Jerry, I dont think I ever said there are two oscillators running. As u say,
I have a v. high freq clock. I use some n samples per bit/symbol transmitted.
Then send the samples to a DAC. And I'm sure u'll get a v. smooth waveform
(i.e at no point the derivative becomes infinity) if k1 and k2 are integers,
and for a given phase.
ganesh
Reply by Jerry Avins●January 8, 20042004-01-08
ganesh wrote:
> "Bevan Weiss" <kaizen__@NOSPAMhotmail.com> wrote in message news:<AHmKb.2936$9k7.68990@news.xtra.co.nz>...
>
>
>>I can't see why k should be an integer. I accept that it allows for much
>>easier maths in most situations, but don't see why it should be 2 as opposed
>>to 2.5
>
>
> If k is not an integer, the resulting waveform is not smooth. To see why,
> take the bit pattern 010101, with f1 for 0 and f2 for 1. If k is not an
> integer the waveform is not smooth, there are abrupt changes. This results
> in lot more power residing in the higher frequency ranges (if u take a
> fourier transform of the waveform). If this signal were to pass thru
> any comm channel, the distortion wud be much higher.
>
> ganesh
Surely there aren't two oscillators running continuously, with the one
to be transmitted selected by the bit to be transmitted. A simple way to
describe a better way is to ask you to imagine that both frequencies are
created by dividing a common clock, and that the bit (1 or 0) selects
the divide ratio. Then whether k is an integer or not, the waveform is
continuous and its derivative is not.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Bevan Weiss●January 8, 20042004-01-08
> > I can't see why k should be an integer. I accept that it allows for
much
> > easier maths in most situations, but don't see why it should be 2 as
opposed
> > to 2.5
>
> If k is not an integer, the resulting waveform is not smooth. To see
why,
> take the bit pattern 010101, with f1 for 0 and f2 for 1. If k is not an
> integer the waveform is not smooth, there are abrupt changes. This results
> in lot more power residing in the higher frequency ranges (if u take a
> fourier transform of the waveform). If this signal were to pass thru
> any comm channel, the distortion wud be much higher.
I thought that k was only to do with the frequency deviation, and that
whether or not the change was smooth was based more on the method used for
modulation and whether it was phase coherant or otherwise.
ie using a Direct Digital Synthesiser (DDS) with phase accumulator as the
frequency modulator would produce a phase coherant FSK signal, no abrupt
amplitude changes would be present, however an abrupt frequency change would
obviously be present at the point the frequency word of the DDS is loaded.
In the above example it doesn't matter what the new frequency word is loaded
to, there will still be no abrupt amplitude change, the amplitude will
simply change at the newly loaded frequency.
Reply by ganesh●January 8, 20042004-01-08
"Bevan Weiss" <kaizen__@NOSPAMhotmail.com> wrote in message news:<AHmKb.2936$9k7.68990@news.xtra.co.nz>...
> I can't see why k should be an integer. I accept that it allows for much
> easier maths in most situations, but don't see why it should be 2 as opposed
> to 2.5
If k is not an integer, the resulting waveform is not smooth. To see why,
take the bit pattern 010101, with f1 for 0 and f2 for 1. If k is not an
integer the waveform is not smooth, there are abrupt changes. This results
in lot more power residing in the higher frequency ranges (if u take a
fourier transform of the waveform). If this signal were to pass thru
any comm channel, the distortion wud be much higher.
ganesh
Reply by Parthasarathy●January 6, 20042004-01-06
>
> With T = 20 microseconds and f1 and f2 being 2 kHz or less, you
> are rather far from the ideal (f1 - f2) = k/T with k being an integer.
> So it is not clear why you are expecting FSK to work as predicted
> by the theory of orthogonal signaling. And, in answer to a question
> in your original post:
Sorry, my 'T' is around 20 millisecond.
> Ideally the performance of the FSK system does not depend on the
> actual value of k as long as it is an integer. However, if (f1 - f2)T is
> a large number, then it does not matter as much if (f1 - f2)T is not
> exactly an integer. The difference in performance between a system
> with (f1 - f2)T = 10 and a system with (f1 - f2)T = 10.1 is quite small;
> the difference in performance between a system with (f1 - f2)T = 1
> and a system with (f1 - f2)T = 1.1 (or (f1-f2)T = 0.04 as in your case)
> is more substantial.
Thanks Dilip, for the comments.
what about the comments on the question 'choice of demodulators'? Viz..
3. In literature, we come across two types of noncoherent
FSKdemodulators
(a) Quadrature receiver with product integrators.
(b)Receiver with bandpass filters and envelop detectors.
Is there any obligations in choosing any one of the two demodulators?
{Note my frequencies are < 2kHz}
please help
Partha
Reply by Bevan Weiss●January 5, 20042004-01-05
"Dilip V. Sarwate" <sarwate@YouEyeYouSee.edu> wrote in message
news:btcdh8$rnd$1@news.ks.uiuc.edu...
>
> "Parthasarathy" <parth175@yahoo.co.in> wrote in message
> news:7f126353.0401042104.30cd9a39@posting.google.com...
> > In one of the projects, where we are using FSK, is not working as
> > predicted by us. So I wanted to find out, whether, there was still
> > something elusive as far as the intricacies of FSK demodulation is
> > concerned.
>
> With T = 20 microseconds and f1 and f2 being 2 kHz or less, you
> are rather far from the ideal (f1 - f2) = k/T with k being an integer.
> So it is not clear why you are expecting FSK to work as predicted
> by the theory of orthogonal signaling. And, in answer to a question
> in your original post:
>
> Ideally the performance of the FSK system does not depend on the
> actual value of k as long as it is an integer. However, if (f1 - f2)T is
> a large number, then it does not matter as much if (f1 - f2)T is not
> exactly an integer. The difference in performance between a system
> with (f1 - f2)T = 10 and a system with (f1 - f2)T = 10.1 is quite small;
> the difference in performance between a system with (f1 - f2)T = 1
> and a system with (f1 - f2)T = 1.1 (or (f1-f2)T = 0.04 as in your case)
> is more substantial.
I can't see why k should be an integer. I accept that it allows for much
easier maths in most situations, but don't see why it should be 2 as opposed
to 2.5 surely if you have the available clean bandwidth then as large a k
as possible would be preferred? to allow for less dependance on frequency
stability, ie a small frequency jitter in either the transmitter or the
receiver would equal a very small change in the received signal, ie very
much far below the change from the comparatively huge change in frequency
with a larger k.
Can anyone enlighten me? I'm not fully conversant with communication systems
yet :s
Reply by ●January 5, 20042004-01-05
"Parthasarathy" <parth175@yahoo.co.in> wrote in message
news:7f126353.0401042104.30cd9a39@posting.google.com...
> In one of the projects, where we are using FSK, is not working as
> predicted by us. So I wanted to find out, whether, there was still
> something elusive as far as the intricacies of FSK demodulation is
> concerned.
With T = 20 microseconds and f1 and f2 being 2 kHz or less, you
are rather far from the ideal (f1 - f2) = k/T with k being an integer.
So it is not clear why you are expecting FSK to work as predicted
by the theory of orthogonal signaling. And, in answer to a question
in your original post:
Ideally the performance of the FSK system does not depend on the
actual value of k as long as it is an integer. However, if (f1 - f2)T is
a large number, then it does not matter as much if (f1 - f2)T is not
exactly an integer. The difference in performance between a system
with (f1 - f2)T = 10 and a system with (f1 - f2)T = 10.1 is quite small;
the difference in performance between a system with (f1 - f2)T = 1
and a system with (f1 - f2)T = 1.1 (or (f1-f2)T = 0.04 as in your case)
is more substantial.