basic bpsk bandwidth question

Dear All,
 Assume that we have 1kbps data, if we BPSK modulate it on a 2 khz
carrier  with 2 period per bit, we have a 1 khz wide signal centered
at 2 khz so the 3 dB bandwidth of the signal is 1 khz.
 If we increase the carrier to 4 khz with 2 period per bit, we have a
data speed of 2 kbps and a 1khz wide signal centered at 4 khz.The 3 db
bandwidth is still 1 khz for 2 kbps.
 Can you point out my error?

On 3/7/2012 9:13 AM, recoder wrote:
> Dear All,
>   Assume that we have 1kbps data, if we BPSK modulate it on a 2 khz
> carrier  with 2 period per bit, we have a 1 khz wide signal centered
> at 2 khz so the 3 dB bandwidth of the signal is 1 khz.

Phase-shift keying is not AM. You need to understand "modulation index".

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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On Mar 7, 9:13=A0am, recoder <kurtulmeh...@gmail.com> wrote:
> Dear All,
> =A0Assume that we have 1kbps data, if we BPSK modulate it on a 2 khz
> carrier =A0with 2 period per bit, we have a 1 khz wide signal centered
> at 2 khz so the 3 dB bandwidth of the signal is 1 khz.
.....
> =A0Can you point out my error?


Err, *how* do you get 1 kbps binary data into a 500 Hz bandwidth?
BPSK modulation doubles the bandwidth of the baseband BPSK
pulse and centers it at the carrier frequency.

On Wed, 7 Mar 2012 06:13:22 -0800 (PST), recoder
<kurtulmehtap@gmail.com> wrote:

>Dear All,
> Assume that we have 1kbps data, if we BPSK modulate it on a 2 khz
>carrier  with 2 period per bit, we have a 1 khz wide signal centered
>at 2 khz so the 3 dB bandwidth of the signal is 1 khz.

Sounds right to me.

> If we increase the carrier to 4 khz with 2 period per bit, we have a
>data speed of 2 kbps and a 1khz wide signal centered at 4 khz.The 3 db
>bandwidth is still 1 khz for 2 kbps.
> Can you point out my error?

The period of the carrier at 4kHz is half that of the carrier at 2kHz,
so the symbol rate has doubled.   This doubles the 3dB bandwidth of
the signal.

   
Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

On Wed, 07 Mar 2012 06:13:22 -0800, recoder wrote:

> Dear All,
>  Assume that we have 1kbps data, if we BPSK modulate it on a 2 khz
> carrier  with 2 period per bit, we have a 1 khz wide signal centered at
> 2 khz so the 3 dB bandwidth of the signal is 1 khz.
>  If we increase the carrier to 4 khz with 2 period per bit, we have a
> data speed of 2 kbps and a 1khz wide signal centered at 4 khz.The 3 db
> bandwidth is still 1 khz for 2 kbps.
>  Can you point out my error?

Which one?

If you have 1kbps data, BPSK modulated on a 2kHz carrier, (with, by the 
definitions so far has 2 carrier periods per bit), then you have a signal 
that's far wider than 1kHz.  It's not really centered at 2kHz any more, 
because the sidelobes on the BPSK subtract between 1kHz and 2kHz, but add 
between 2kHz and 3kHz.  I'm too damn lazy to compute where the 3dB down 
points are, but they're not at 1.5kHz and 2.5kHz, at least not 
simultaneously.

Moreover, when you do your modulation at such a low proportion of the 
carrier frequency, the relative phase of the switching point matters -- 
switch phase at the zero crossings of the carrier and you'll have less 
high-frequency content in your signal than if you switch phase at the 
maxima.  Again, I'm too damn lazy to do the math, but those sharp corners 
when you switch at the maxima have to count for something.

If you scale the whole problem up in frequency by a factor of two, then 
you have 2kHz BPSK modulation, so of course your 3dB down points 
(wherever they were to start with) will be twice as far apart in 
frequency as they were before.

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

On Thu, 08 Mar 2012 01:35:46 -0600, Tim Wescott
<tim@seemywebsite.please> wrote:

>On Wed, 07 Mar 2012 06:13:22 -0800, recoder wrote:
>
>> Dear All,
>>  Assume that we have 1kbps data, if we BPSK modulate it on a 2 khz
>> carrier  with 2 period per bit, we have a 1 khz wide signal centered at
>> 2 khz so the 3 dB bandwidth of the signal is 1 khz.
>>  If we increase the carrier to 4 khz with 2 period per bit, we have a
>> data speed of 2 kbps and a 1khz wide signal centered at 4 khz.The 3 db
>> bandwidth is still 1 khz for 2 kbps.
>>  Can you point out my error?
>
>Which one?
>
>If you have 1kbps data, BPSK modulated on a 2kHz carrier, (with, by the 
>definitions so far has 2 carrier periods per bit), then you have a signal 
>that's far wider than 1kHz.  It's not really centered at 2kHz any more, 
>because the sidelobes on the BPSK subtract between 1kHz and 2kHz, but add 
>between 2kHz and 3kHz.  I'm too damn lazy to compute where the 3dB down 
>points are, but they're not at 1.5kHz and 2.5kHz, at least not 
>simultaneously.

I'm not quite sure what you're thinking, but what he described is
consistent with a 1kHz symbol frequency on a 2kHz center frequency.
For BPSK, QPSK, QAM, or most single-carrier modulations the 3dB
bandwidth is essentially the symbol rate (depending only a little bit
on how the filtering is done).

>Moreover, when you do your modulation at such a low proportion of the 
>carrier frequency, the relative phase of the switching point matters -- 
>switch phase at the zero crossings of the carrier and you'll have less 
>high-frequency content in your signal than if you switch phase at the 
>maxima.  Again, I'm too damn lazy to do the math, but those sharp corners 
>when you switch at the maxima have to count for something.

If one generates the BPSK signal at baseband and then mixes it up to
fc, one has the same BPSK baseband signal (with 3dB bandwidth
essentially equal to the symbol rate), translated to fc.    It doesn't
really matter how high or low fc is, it just translates the signal
spectrum to that frequency.   2kHz is not at all impractical for a
center frequency for a BPSK signal with a 1kHz symbol rate.

>If you scale the whole problem up in frequency by a factor of two, then 
>you have 2kHz BPSK modulation, so of course your 3dB down points 
>(wherever they were to start with) will be twice as far apart in 
>frequency as they were before.

When he kept the constraint that the symbol period was two cycles of
the carrier frequency it doubled the symbol rate when the carrier
frequency was doubled.   Doubling the symbol rate doubles the 3dB
bandwidth of the signal.


Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

On Mar 8, 12:13=A0pm, eric.jacob...@ieee.org (Eric Jacobsen) wrote:

>
> If one generates the BPSK signal at baseband and then mixes it up to
> fc, one has the same BPSK baseband signal (with 3dB bandwidth
> essentially equal to the symbol rate), translated to fc. =A0 =A0It doesn'=
t
> really matter how high or low fc is, it just translates the signal
> spectrum to that frequency. =A0 2kHz is not at all impractical for a
> center frequency for a BPSK signal with a 1kHz symbol rate.


Eric:

If the 3dB bandwidth of a signal at baseband is fs Hz,
then, for those who believe in negative frequencies, the
baseband spectrum extends from -fs Hz to +fs Hz
(at least in the 3dB sense). When modulated on to a
carrier at fc Hz, the 3 dB signal bandwidth extends
from fc - fs to fc + fs (and from -fc-fs to -fc+fs) so that
the modulated signal has a 3dB bandwidth of 2fs, not fs.
So, if a passband 1 kbps BPSK signal with 3dB bandwidth
of 1kHz is desired, the baseband BPSK signal needs to
have a 3dB bandwidth of 500 Hz in the ordinary sense
of amplitude modulation x(t)cos(2 pi fc t) which translates
the baseband spectrum to be centered at the carrier
frequency fc.

Or you can say that the baseband 1 kbps BPSK signal
really has 3 dB bandwidth 1 kHz, but we are going to
use single sideband modulation to create a passband
signal of 3dB bandwidth 1 kHz centered at frequency fc,
which is quite different from the amplitude modulation
described above.  In particular, the carrier frequency
corresponding to this SSB modulation would be fc-500
and not fc, right?

So, could you describe in a little more detail exactly
what the 1 kbps baseband BPSK signal x(t) with 500 Hz
3dB bandwidth looks like so that x(t)cos(2 pi fc t) gives
a BPSK signal with 3dB bandwidth 1 kHz centered at
fc Hz, or the modulation process for creating the SSB
signal?

Dilip Sarwate

On Thu, 8 Mar 2012 09:55:41 -0800 (PST), dvsarwate
<dvsarwate@yahoo.com> wrote:

>On Mar 8, 12:13=A0pm, eric.jacob...@ieee.org (Eric Jacobsen) wrote:
>
>>
>> If one generates the BPSK signal at baseband and then mixes it up to
>> fc, one has the same BPSK baseband signal (with 3dB bandwidth
>> essentially equal to the symbol rate), translated to fc. =A0 =A0It doesn'=
>t
>> really matter how high or low fc is, it just translates the signal
>> spectrum to that frequency. =A0 2kHz is not at all impractical for a
>> center frequency for a BPSK signal with a 1kHz symbol rate.
>
>
>Eric:
>
>If the 3dB bandwidth of a signal at baseband is fs Hz,
>then, for those who believe in negative frequencies, the
>baseband spectrum extends from -fs Hz to +fs Hz
>(at least in the 3dB sense). When modulated on to a
>carrier at fc Hz, the 3 dB signal bandwidth extends
>from fc - fs to fc + fs (and from -fc-fs to -fc+fs) so that
>the modulated signal has a 3dB bandwidth of 2fs, not fs.
>So, if a passband 1 kbps BPSK signal with 3dB bandwidth
>of 1kHz is desired, the baseband BPSK signal needs to
>have a 3dB bandwidth of 500 Hz in the ordinary sense
>of amplitude modulation x(t)cos(2 pi fc t) which translates
>the baseband spectrum to be centered at the carrier
>frequency fc.
>
>Or you can say that the baseband 1 kbps BPSK signal
>really has 3 dB bandwidth 1 kHz, but we are going to
>use single sideband modulation to create a passband
>signal of 3dB bandwidth 1 kHz centered at frequency fc,
>which is quite different from the amplitude modulation
>described above.  In particular, the carrier frequency
>corresponding to this SSB modulation would be fc-500
>and not fc, right?
>
>So, could you describe in a little more detail exactly
>what the 1 kbps baseband BPSK signal x(t) with 500 Hz
>3dB bandwidth looks like so that x(t)cos(2 pi fc t) gives
>a BPSK signal with 3dB bandwidth 1 kHz centered at
>fc Hz, or the modulation process for creating the SSB
>signal?
>
>Dilip Sarwate
>
>

I'll give it a shot. The short answer is that it's not SSB.

The 3dB "bandwidth" of a BPSK signal with a symbol rate (and therefore
bit rate) of 1kHz to a comm guy like me is 1 kHz, even when it's at
baseband (centered at DC).   BPSK and QPSK signals differ only by the
presence or absence of the "imaginary" or "Q" or "orthogonal" second
channel, which does not affect the bandwidth, only the symmetry, or
lack thereof, of the occupied signal spectrum.

So most comm guys, I would argue, don't consider the BW of a BPSK
signal to be one-sided at baseband, even though it's symmetric about
DC, because that's the ONLY time it would arguably have the narrower
bandwidth.   It would also mean that BSPK would be treated (or thought
of) differently than it's close cousins QPSK and QAM, and there's no
real reason to treat it differently, even though it's essentially
real-valued.

Also, like QPSK, when the BPSK signal spectrum is mixed to an IF
frequency, 2 kHz in the first example, it is usually a complex mix so
that the two-sided bandwidth is preserved.   Again, this is the same
as would be done with a QPSK or QAM signal (or essentially any comm
signal with a quadrature component).   The BPSK signal isn't treated
differently even though it does not have the associated quadrature
component.

Since the "real" part of the output of the complex upconversion is
taken as the ultimate transmit signal that actually goes out the
antenna, for BPSK this boils down to a simple DSB upconversion as
could be done in an AM system.   The output is similarly a DSB,
symmetric, two-sided signal spectrum.   So a transmitter that only
ever transmits BPSK doesn't need a complex upconverter, since it boils
down to the same as DSB AM upconversion.  The only real difference in
this case is that at very low center frequencies (e.g., less than
~500Hz+rolloff margin for the 1kHz symbol rate), the DSB system would
have a negative-frequency image that would inferfere with the desired
signal, while a complex upconverter would not.

My understanding is that the OP's system has a symbol (bit) rate of
1kHz, so the baseband signal extends from -500 to +500 Hz.   Mixing
that to 2kHz, either with a complex mix or a real DSB mix, results in
a signal extending from 1.5kHz to 2.5kHz, still with 1kHz 3dB BW.

In the OP's second case the center frequency is 4kHz, but the symbol
rate is held to two periods of the carrier frequency, or Tsym =
2/4kHz, so that the symbol rate is now 2kHz.  The bit rate doubles and
the signal bandwidth doubles with the doubling of the symbol rate.
If the symbol rate were held constant in time, rather than constant in
carrier period intervals, the bandwidth and bit rate would be held to
1kHz.

I suspect it was a test or homework problem because it requires one to
think about the relationship between the carrier frequency and symbol
state stated in the problem.  There usually isn't such a relationship,
so it seems like a test or homework problem to me.  It's a reasonably
clever one, too, IMHO.   It seems to have tripped up the OP at least a
little bit.




Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com

On Mar 8, 7:27=A0pm, eric.jacob...@ieee.org (Eric Jacobsen) wrote:

Eric:

Thanks for taking the time to write such a detailed answer.
Unfortunately, I still have some questions and hope that
you will indulge me further because there are still some
gaps in my understanding of the issues.

Let us ignore the OP's question and the 2 kHz carrier versus
the 4 kHz carrier and any associated problems that might be
caused by the low carrier frequency.  You say

> The 3dB "bandwidth" of a BPSK signal with a symbol rate (and therefore
> bit rate) of 1kHz to a comm guy like me is 1 kHz, even when it's at
> baseband (centered at DC).

I think there is a trivial typo and that you meant to say
that a BPSK signal with symbol rate and bit rate of
1 kbps (not 1 kHz) at baseband (that is, a center
frequency of 0 Hz or DC) has a 1 kHz bandwidth.  To
me, your statement also says that the spectrum of the
baseband BPSK signal extends from -1 kHz to +1 kHz.
But to you and presumably most other comm guys, it
means that the the spectrum extends only from -500 Hz
to +500 Hz since you say later that

>for BPSK this boils down to a simple DSB upconversion as
could be done in an AM system.   The output is similarly a DSB,
symmetric, two-sided signal spectrum.

If the passband BPSK signal at center frequency fc Hz
(and let's not drag in the OP's 2 kHz or 4kHz here) also
has a bandwidth of 1 kHz, then its baseband signal must
be extending from -500 Hz to +500 Hz only in order that
the passband signal x(t)cos(2 pi fc t) will extend from
fc-500 Hz to fc+500 Hz for a passband bandwidth of 1 kHz.
(There is a mirror image at negative frequency -fc Hz that
is irrelevant to this discussion).  This follows from the Fourier
transform of x(t)cos(2 pi fc t), plain vanilla DSB-SC modulation,
is (1/2)[X(f - fc) + X(f + fc)] where X(f-fc) is the translation
of the baseband spectrum X(f) to be centered at fc.

> My understanding is that the OP's system has a symbol (bit) rate of
> 1kHz, so the baseband signal extends from -500 to +500 Hz.

So, could you please tell me with symbols and equations
rather than in words and generalities, a formula that
can be used to figure out what x(t) is during a
one-millisecond interval (say from t =3D -0.0005 to t =3D +0.0005)
during which time interval, the BPSK signal transmits
one bit?  Since this is (antipodal) BPSK, perhaps
the formula will include something like (-1)^(b)p(t)  where b
is the bit (0 or 1) so that (-1)^b evaluates to +1 or -1
giving antipodality and p(t) will be the baseband pulse.
It would be very helpful if you could tell me the time
domain shape of p(t), and the Fourier transform P(f)
would be helpful too.

The reason for asking for all this, and hoping that
you will be kind enough to reply with the desired
formula is that there seems to be a disconnect
between the meaning of bandwidth of baseband
digital signals and low-frequency analog audio
signals.  An audio signal has spectrum extending out
to 20 kHz (and to -20 KHz too) and is said to have
a bandwidth of 20 kHz (which is why CDs use
sampling at 44.1 kHz, somewhat larger than twice
the highest frequency) but a 1 kbps baseband BPSK
signal has a bandwidth of 1 kHz but its spectrum
extends out to only 500 Hz (and to - 500 Hz).  Thus,
bandwidth seems to mean different things depending
on whether we are talking of analog signals or digitally
modulated signals.

Dilip Sarwate

On Thu, 8 Mar 2012 19:37:16 -0800 (PST), dvsarwate
<dvsarwate@yahoo.com> wrote:

>On Mar 8, 7:27=A0pm, eric.jacob...@ieee.org (Eric Jacobsen) wrote:
>
>Eric:
>
>Thanks for taking the time to write such a detailed answer.
>Unfortunately, I still have some questions and hope that
>you will indulge me further because there are still some
>gaps in my understanding of the issues.
>
>Let us ignore the OP's question and the 2 kHz carrier versus
>the 4 kHz carrier and any associated problems that might be
>caused by the low carrier frequency.  You say
>
>> The 3dB "bandwidth" of a BPSK signal with a symbol rate (and therefore
>> bit rate) of 1kHz to a comm guy like me is 1 kHz, even when it's at
>> baseband (centered at DC).
>
>I think there is a trivial typo and that you meant to say
>that a BPSK signal with symbol rate and bit rate of
>1 kbps (not 1 kHz) at baseband (that is, a center
>frequency of 0 Hz or DC) has a 1 kHz bandwidth.

Generally the symbol rate is expressed in Hz (or baud, for the more
picky), so in this case with 1 bit/symbol when talking about the
symbol rate and bit rate in the same breath one could use either bps
or Hz. IMHO, of course.

>To
>me, your statement also says that the spectrum of the
>baseband BPSK signal extends from -1 kHz to +1 kHz.
>But to you and presumably most other comm guys, it
>means that the the spectrum extends only from -500 Hz
>to +500 Hz since you say later that
>
>>for BPSK this boils down to a simple DSB upconversion as
>could be done in an AM system.   The output is similarly a DSB,
>symmetric, two-sided signal spectrum.
>
>If the passband BPSK signal at center frequency fc Hz
>(and let's not drag in the OP's 2 kHz or 4kHz here) also
>has a bandwidth of 1 kHz, then its baseband signal must
>be extending from -500 Hz to +500 Hz only in order that
>the passband signal x(t)cos(2 pi fc t) will extend from
>fc-500 Hz to fc+500 Hz for a passband bandwidth of 1 kHz.
>(There is a mirror image at negative frequency -fc Hz that
>is irrelevant to this discussion).  This follows from the Fourier
>transform of x(t)cos(2 pi fc t), plain vanilla DSB-SC modulation,
>is (1/2)[X(f - fc) + X(f + fc)] where X(f-fc) is the translation
>of the baseband spectrum X(f) to be centered at fc.
>
>> My understanding is that the OP's system has a symbol (bit) rate of
>> 1kHz, so the baseband signal extends from -500 to +500 Hz.
>
>So, could you please tell me with symbols and equations
>rather than in words and generalities, a formula that
>can be used to figure out what x(t) is during a
>one-millisecond interval (say from t =3D -0.0005 to t =3D +0.0005)
>during which time interval, the BPSK signal transmits
>one bit?  Since this is (antipodal) BPSK, perhaps
>the formula will include something like (-1)^(b)p(t)  where b
>is the bit (0 or 1) so that (-1)^b evaluates to +1 or -1
>giving antipodality and p(t) will be the baseband pulse.
>It would be very helpful if you could tell me the time
>domain shape of p(t), and the Fourier transform P(f)
>would be helpful too.

I think there's a quicker answer that I can give below, plus it's late
in the day and at the moment I'm too lazy to dig up the formal
equations.   I can say, though, that the expressions for BPSK and QPSK
are the same with the exception that QPSK has two copies of the BPSK
representation, summed with one of them with a coefficient of 'j'.
QAM is just QPSK with additional, independent, amplitude terms on both
the real and imaginary (or, if you're a comm guy, the in-phase and
quadrature) components.

The time domain shape of the pulses varies widely, from rectangular
pulses to root-raised cosine to all manner of things in between and
elsewise.  That's always a system engineering decision to choose
between one of the infinite number of possible pulse shapes, assuming
it's not specified for standard compliance or something.

>The reason for asking for all this, and hoping that
>you will be kind enough to reply with the desired
>formula is that there seems to be a disconnect
>between the meaning of bandwidth of baseband
>digital signals and low-frequency analog audio
>signals.  An audio signal has spectrum extending out
>to 20 kHz (and to -20 KHz too) and is said to have
>a bandwidth of 20 kHz (which is why CDs use
>sampling at 44.1 kHz, somewhat larger than twice
>the highest frequency) but a 1 kbps baseband BPSK
>signal has a bandwidth of 1 kHz but its spectrum
>extends out to only 500 Hz (and to - 500 Hz).  Thus,
>bandwidth seems to mean different things depending
>on whether we are talking of analog signals or digitally
>modulated signals.
>
>Dilip Sarwate

Yes, the terms are used differently here, for the case of a signal "at
baseband", to a digital comm guy, or an audio signal.   There's a good
reason for this, I think.   The audio signal sitting between 0 and y
Hz lives there natively, occurs there naturally, is consumed there,
and if generally processed there (within that frequency range).   So
the primary area of interest, in all phases of generation, processing,
and consumption, occur within that band and with general disregard for
the frequency components mirrored on the negative side of DC.  Audio
signals are often thought of and processed as real-valued.

Comm signals, on the other hand, may be generated at baseband, but
live much of their lives elsewhere at IF or RF and even other exotic
frequencies, where the bandwidth of the signal exists independently
far away from DC.   In order to be consistent, the "bandwidth" of the
signal is generally taken to be determined by the 3dB points on the
rolloff on either side of the center frequency.  When the signal is
generated at baseband, or mixed back down there in a receiver,
measuring "bandwidth" from DC to the positive-frequency 3dB rolloff
point becomes inconsistent with the "bandwidth" that the signal had at
IF or RF.

Also, since comm signals are routinely and, these days, almost always
processed at baseband with both real and imaginary components (even
when it's BPSK), the baseband signal is NOT assumed to be symmetric
about DC.   It will be for a pristine BPSK signal generated in a
modulator, but in a demodulator it often won't be due to channel
distortion, interference, and all manner of evil and abuse that may
have befallen it on it's way through the dark forests of the channel.
Any higher-order modulation will have independent information in each
of the in-phase and quadrature channels, so even with the possibility
that BPSK might be symmetric the general thinking and processing for
comm is done with the assumption that the baseband spectrum cannot be
assumed to be symmetric about DC except under special circumstances.

So, yeah, a comm guy will measure "bandwidth" from -fx to +fx at
baseband, and an audio guy will have no reason to do that and plenty
of reason not to. 

It is, however, just a matter of translating zero Herz to where one is
most comfortable.   For many, including comm people, having zero Hz
(or, for those so inclined,  zero rad/sec)  in the middle of the
spectrum of interest is often very natural and desirable.  I think for
anybody who routinely deals with analytic signals this may be the
case.   For others it might be more natural to keep the frequency
origin comfortably far to the left, so that no negative frequencies
have to be imagined or dealt with, all signals are real-valued, and
the direct connection to related physical phenomena is kept as close
as possible.

Or something like that.   ;)


Eric Jacobsen
Anchor Hill Communications
www.anchorhill.com