Is BPSK signal DC free?

Hi there

I got mentally stuck when I try to interpret the PSD of BPSK signal.

Let's define the baseband BPSK signal by
s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k = -1, 1
of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.

It's stated in each textbook that the PSD of s(t) takes its global maximum
at DC while I can also show that s(t) has zero DC power. Well, you may tell
me that PSD indicates power/Hz and not power itself, but it's still hard to
me  to imagine that a signal has absolutely null power at DC while
possessing the maximum power density at the vicinity of DC. Is there an
intuitive explanation? 

Regards
Marco

taicon <taicon@sunrise.ch> wrote:

>I got mentally stuck when I try to interpret the PSD of BPSK signal.

>s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k = -1, 1
>of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
>
>It's stated in each textbook that the PSD of s(t) takes its global maximum
>at DC while I can also show that s(t) has zero DC power. Well, you may tell
>me that PSD indicates power/Hz and not power itself, but it's still hard to
>me  to imagine that a signal has absolutely null power at DC while
>possessing the maximum power density at the vicinity of DC. Is there an
>intuitive explanation? 

Sure.  Consider a white noise signal.  It has constant PSD
everywhere, but has no power exactly at DC.  It in fact has
no power exactly anywhere. Same with the BPSK signal.

I would say, offhand, any non-time-limited signal that 
is not simply the sum of a finite number of sinusoids will have
zero power at any given exact frequency.  It still may have
a non-zero PSD in the vicinity of those frequencies.   Someone 
correct me if this isn't right.

Steve

On Fri, 12 Sep 2008 18:32:53 -0500, taicon wrote:

> Hi there
> 
> I got mentally stuck when I try to interpret the PSD of BPSK signal.
> 
> Let's define the baseband BPSK signal by
> s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k = -1,
> 1 of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
> 
> It's stated in each textbook that the PSD of s(t) takes its global
> maximum at DC while I can also show that s(t) has zero DC power. Well,
> you may tell me that PSD indicates power/Hz and not power itself, but
> it's still hard to me  to imagine that a signal has absolutely null
> power at DC while possessing the maximum power density at the vicinity
> of DC. Is there an intuitive explanation?
> 
Actually you hit the nail on the head when you said "(in) the vicinity of 
DC".

Consider an iron bar two inches thick.  It's heavy as -- well -- an iron 
bar.  Now cut off a inch thick disk from that bar -- not so heavy.  Now 
cut off a 1mm thick disk from that bar -- it's starting to get downright 
light.  Now (assuming you can find a tool to do it), cut a slice of .01 
inch thick shim stock -- drop it and it doesn't fall, it wafts.

Do you see where I'm going?  Iron is dense, but an infinitesimal amount 
of it still only weighs an infinitesimal amount.  Take that amount down 
to a 2 inch diameter by no-inches-thick disk that has no iron atoms in it 
at all, and it's weight is zero, just as the DC content of your BPSK 
signal is zero.

-- 
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

>On Fri, 12 Sep 2008 18:32:53 -0500, taicon wrote:
>
>> Hi there
>> 
>> I got mentally stuck when I try to interpret the PSD of BPSK signal.
>> 
>> Let's define the baseband BPSK signal by
>> s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k =
-1,
>> 1 of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
>> 
>> It's stated in each textbook that the PSD of s(t) takes its global
>> maximum at DC while I can also show that s(t) has zero DC power. Well,
>> you may tell me that PSD indicates power/Hz and not power itself, but
>> it's still hard to me  to imagine that a signal has absolutely null
>> power at DC while possessing the maximum power density at the vicinity
>> of DC. Is there an intuitive explanation?
>> 
>Actually you hit the nail on the head when you said "(in) the vicinity of

>DC".
>
>Consider an iron bar two inches thick.  It's heavy as -- well -- an iron

>bar.  Now cut off a inch thick disk from that bar -- not so heavy.  Now 
>cut off a 1mm thick disk from that bar -- it's starting to get downright

>light.  Now (assuming you can find a tool to do it), cut a slice of .01 
>inch thick shim stock -- drop it and it doesn't fall, it wafts.
>
>Do you see where I'm going?  Iron is dense, but an infinitesimal amount 
>of it still only weighs an infinitesimal amount.  Take that amount down 
>to a 2 inch diameter by no-inches-thick disk that has no iron atoms in it

>at all, and it's weight is zero, just as the DC content of your BPSK 
>signal is zero.
>
>-- 
>Tim Wescott
>Control systems and communications consulting
>http://www.wescottdesign.com
>
>Need to learn how to apply control theory in your embedded system?
>"Applied Control Theory for Embedded Systems" by Tim Wescott
>Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
>

Thank both authors above. I can see what you mean. But a similar strange
thing happens in s(t) at frequency 1/T, where PSD vanisches at the first
null of sinc^2(.). Integrating in the vicinity of null in a continuous
function yields null, meaning there's no power at 1/T, while my instinkt 
says (nail in head) there must be a frequency component there with nonzero
power because p(t) repeats in period T. How to explain that?

Marco

"taicon" <taicon@sunrise.ch> writes:

> Hi there
>
> I got mentally stuck when I try to interpret the PSD of BPSK signal.
>
> Let's define the baseband BPSK signal by
> s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k = -1, 1
> of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
>
> It's stated in each textbook that the PSD of s(t) takes its global maximum
> at DC while I can also show that s(t) has zero DC power. Well, you may tell
> me that PSD indicates power/Hz and not power itself, but it's still hard to
> me  to imagine that a signal has absolutely null power at DC while
> possessing the maximum power density at the vicinity of DC. Is there an
> intuitive explanation? 

Hi Marco,

First of all, great question! Second of all, what I say here adds to
what has already been posted.

You've already hit several of the key ideas in your post. The thing to
understand thoroughly here is the difference between energy signals and
power signals. I believe Brown [brown] has a great coverage of this.

Quite simply, the PSD is the "energy spectrum." Think of the units:
watts / Hz = joules / second / (1 / second) = joules. Any signal that is
sinusoidal (including DC) is a finite-power, infinite-energy signal, so
it must have a spike (a Dirac delta function) in the energy spectrum.

OK, so why is there even energy at DC? Because of the pulse shape (g(t)
in Proakis [proakiscomm]). The information sequence is like a signal
that keys on and off the pulse shape, which means it modulates it in
time and convolves it in frequency (roughly - I'm not accounting for 
the information there but just thinking of it like an infinite impulse 
train). 

Hope this makes some sense.

--Randy

@book{brown,
  title = "Introduction to Random Signal Analysis and Kalman Filtering",
  author = "{Robert~Grover~Brown}",
  publisher = "John Wiley and Sons",
  year = "1983"}
@BOOK{proakiscomm,
  title = "{Digital Communications}",
  author = "John~G.~Proakis",
  publisher = "McGraw-Hill",
  edition = "fourth",
  year = "2001"}

-- 
%  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
%% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
%%% 919-577-9882                %  Verdi's always creepin' from her room." 
%%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO   
http://www.digitalsignallabs.com

>"taicon" <taicon@sunrise.ch> writes:
>
>> Hi there
>>
>> I got mentally stuck when I try to interpret the PSD of BPSK signal.
>>
>> Let's define the baseband BPSK signal by
>> s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k =
-1, 1
>> of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
>>
>> It's stated in each textbook that the PSD of s(t) takes its global
maximum
>> at DC while I can also show that s(t) has zero DC power. Well, you may
tell
>> me that PSD indicates power/Hz and not power itself, but it's still
hard to
>> me  to imagine that a signal has absolutely null power at DC while
>> possessing the maximum power density at the vicinity of DC. Is there
an
>> intuitive explanation? 
>
>Hi Marco,
>
>First of all, great question! Second of all, what I say here adds to
>what has already been posted.
>
>You've already hit several of the key ideas in your post. The thing to
>understand thoroughly here is the difference between energy signals and
>power signals. I believe Brown [brown] has a great coverage of this.
>
>Quite simply, the PSD is the "energy spectrum." Think of the units:
>watts / Hz = joules / second / (1 / second) = joules. Any signal that is
>sinusoidal (including DC) is a finite-power, infinite-energy signal, so
>it must have a spike (a Dirac delta function) in the energy spectrum.
>
>OK, so why is there even energy at DC? Because of the pulse shape (g(t)
>in Proakis [proakiscomm]). The information sequence is like a signal
>that keys on and off the pulse shape, which means it modulates it in
>time and convolves it in frequency (roughly - I'm not accounting for 
>the information there but just thinking of it like an infinite impulse 
>train). 
>
>Hope this makes some sense.
>
>--Randy
>
>@book{brown,
>  title = "Introduction to Random Signal Analysis and Kalman Filtering",
>  author = "{Robert~Grover~Brown}",
>  publisher = "John Wiley and Sons",
>  year = "1983"}
>@BOOK{proakiscomm,
>  title = "{Digital Communications}",
>  author = "John~G.~Proakis",
>  publisher = "McGraw-Hill",
>  edition = "fourth",
>  year = "2001"}
>
>-- 
>%  Randy Yates                  % "She's sweet on Wagner-I think she'd
die for Beethoven.
>%% Fuquay-Varina, NC            %  She love the way Puccini lays down a
tune, and
>%%% 919-577-9882                %  Verdi's always creepin' from her
room." 
>%%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO  

>http://www.digitalsignallabs.com
>

Randy

Thank you for the tip about units. Yes, I agree the unit of PSD is Joules,
but does it mean the PSD value at a frequency equals the energy of the
corresponding spectral component? I don't think so. If you look at the BPSK
signal s(t), it even does not have a specific DC value since the limit 

\bar{s}(t) = \lim_{\tau->\infty}\int_{t=0}^{\tau}s(t)dt

simply does not exist. Actually it varies within a range depending on the
integration period. Therefore, the DC energy of s(t), |\bar{s}(t)|^2,  is
not a constant but limited within a range, while the PSD at frequency zero
is a constant. Why? Correct me if I am wrong.

Marco

"taicon" <taicon@sunrise.ch> writes:

>>"taicon" <taicon@sunrise.ch> writes:
>>
>>> Hi there
>>>
>>> I got mentally stuck when I try to interpret the PSD of BPSK signal.
>>>
>>> Let's define the baseband BPSK signal by
>>> s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k =
> -1, 1
>>> of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
>>>
>>> It's stated in each textbook that the PSD of s(t) takes its global
> maximum
>>> at DC while I can also show that s(t) has zero DC power. Well, you may
> tell
>>> me that PSD indicates power/Hz and not power itself, but it's still
> hard to
>>> me  to imagine that a signal has absolutely null power at DC while
>>> possessing the maximum power density at the vicinity of DC. Is there
> an
>>> intuitive explanation? 
>>
>>Hi Marco,
>>
>>First of all, great question! Second of all, what I say here adds to
>>what has already been posted.
>>
>>You've already hit several of the key ideas in your post. The thing to
>>understand thoroughly here is the difference between energy signals and
>>power signals. I believe Brown [brown] has a great coverage of this.
>>
>>Quite simply, the PSD is the "energy spectrum." Think of the units:
>>watts / Hz = joules / second / (1 / second) = joules. Any signal that is
>>sinusoidal (including DC) is a finite-power, infinite-energy signal, so
>>it must have a spike (a Dirac delta function) in the energy spectrum.
>>
>>OK, so why is there even energy at DC? Because of the pulse shape (g(t)
>>in Proakis [proakiscomm]). The information sequence is like a signal
>>that keys on and off the pulse shape, which means it modulates it in
>>time and convolves it in frequency (roughly - I'm not accounting for 
>>the information there but just thinking of it like an infinite impulse 
>>train). 
>>
>>Hope this makes some sense.
>>
>>--Randy
>>
>>@book{brown,
>>  title = "Introduction to Random Signal Analysis and Kalman Filtering",
>>  author = "{Robert~Grover~Brown}",
>>  publisher = "John Wiley and Sons",
>>  year = "1983"}
>>@BOOK{proakiscomm,
>>  title = "{Digital Communications}",
>>  author = "John~G.~Proakis",
>>  publisher = "McGraw-Hill",
>>  edition = "fourth",
>>  year = "2001"}
>>
>>-- 
>>%  Randy Yates                  % "She's sweet on Wagner-I think she'd
> die for Beethoven.
>>%% Fuquay-Varina, NC            %  She love the way Puccini lays down a
> tune, and
>>%%% 919-577-9882                %  Verdi's always creepin' from her
> room." 
>>%%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO  
>
>>http://www.digitalsignallabs.com
>>
>
> Randy
>
> Thank you for the tip about units. Yes, I agree the unit of PSD is Joules,
> but does it mean the PSD value at a frequency equals the energy of the
> corresponding spectral component? 

Well, yes, I think it does. 

> I don't think so. If you look at the BPSK signal s(t), it even does
> not have a specific DC value since the limit
>
> \bar{s}(t) = \lim_{\tau->\infty}\int_{t=0}^{\tau}s(t)dt
>
> simply does not exist. 

Ahh, but that's true for any infinite-energy signal. In fact you can't
really take the Fourier transform of any infinite-energy signal - it
violates the Dirichlet conditions [signalsandsystems]. We break the
rules a bit by using Dirac delta functions for sinusoids, but the
situation is more complicated when you're talking about a random signal
such as this. Random signals are infinite energy, finite power signals.

Wiener-Khinchine tells us that the energy spectrum of a weakly
stationary random signal is given by the Fourier transform of its
autocorrelation, and I believe them.

--Randy

@BOOK{signalsandsystems,
  title = "{Signals and Systems}",
  author = "{Alan~V.~Oppenheim, Alan~S.~Willsky, with Ian~T.~Young}",
  publisher = "Prentice Hall",
  year = "1983"}

-- 
%  Randy Yates                  % "And all that I can do
%% Fuquay-Varina, NC            %  is say I'm sorry, 
%%% 919-577-9882                %  that's the way it goes..."
%%%% <yates@ieee.org>           % Getting To The Point', *Balance of Power*, ELO
http://www.digitalsignallabs.com

Randy Yates wrote:
(after someone wrote)

>>\bar{s}(t) = \lim_{\tau->\infty}\int_{t=0}^{\tau}s(t)dt

> Ahh, but that's true for any infinite-energy signal. In fact you can't
> really take the Fourier transform of any infinite-energy signal - it
> violates the Dirichlet conditions [signalsandsystems]. We break the
> rules a bit by using Dirac delta functions for sinusoids, but the
> situation is more complicated when you're talking about a random signal
> such as this. Random signals are infinite energy, finite power signals.

A delta function in time would be infinite power, but finite
energy, with energy = integral of power dt.

So yes, a delta in frequency would be infinite energy,
and finite power, with power = integral of energy dw
But that isn't delta's fault, an infinite sinusoid
has infinite energy.

-- glen

glen herrmannsfeldt <gah@ugcs.caltech.edu> writes:

> Randy Yates wrote:
> (after someone wrote)
>
>>>\bar{s}(t) = \lim_{\tau->\infty}\int_{t=0}^{\tau}s(t)dt
>
>> Ahh, but that's true for any infinite-energy signal. In fact you can't
>> really take the Fourier transform of any infinite-energy signal - it
>> violates the Dirichlet conditions [signalsandsystems]. We break the
>> rules a bit by using Dirac delta functions for sinusoids, but the
>> situation is more complicated when you're talking about a random signal
>> such as this. Random signals are infinite energy, finite power signals.
>
> A delta function in time would be infinite power, but finite
> energy, 

I disagree with this.

> with energy = integral of power dt.

I agree with that. 

I think the problem here glen is that you've neglected
to take the square of the signal, assuming that the
signal represents something typical like voltage.

What's the integral of \delta(t)^2 dt? I think that
is also infinite.

How can infinite power not imply infinite energy when
energy = integral of power dt?
-- 
%  Randy Yates                  % "And all that I can do
%% Fuquay-Varina, NC            %  is say I'm sorry, 
%%% 919-577-9882                %  that's the way it goes..."
%%%% <yates@ieee.org>           % Getting To The Point', *Balance of Power*, ELO
http://www.digitalsignallabs.com

>Hi there
>
>I got mentally stuck when I try to interpret the PSD of BPSK signal.
>
>Let's define the baseband BPSK signal by
>s(t)=\sum_{k=-\infty}^{\infty}b_k p(t-kT) with random variable b_k = -1,
1
>of equal probability and p(t) = 1 if 0<t<=T and 0 otherwise.
>
>It's stated in each textbook that the PSD of s(t) takes its global
maximum
>at DC while I can also show that s(t) has zero DC power. Well, you may
tell
>me that PSD indicates power/Hz and not power itself, but it's still hard
to
>me  to imagine that a signal has absolutely null power at DC while
>possessing the maximum power density at the vicinity of DC. Is there an
>intuitive explanation? 
>
>Regards
>Marco
>
>
>
>

Hi Marco,

Please keep in mind that you should compare the auto-correlation function
of s(t) with it's corresponding PSD (instead of comparing the PSD with s(t)
itself)

If you'll do that you'll see that the auto-correlation function DOES have
a DC component (it is not averaged to zero) and this is why it's
corresponding PSD has a positive value at zero.

Best regards,
Yaron Dorman,
Israel.