Reply by March 5, 20182018-03-05
```>W dniu 03.03.2018 o 23:43, Richard (Rick) Lyons pisze:

> > Hi Roman. I forgot to ask you: Do you agree that the
> > addition of the two sequences,

> >      Seq# 1 = x(0), x(2), x(4), x(6)
> >
> >      Seq# 2 = x(1), x(3), x(5), x(7)
> >
> > produces the four-sample sequence:
> >
> >       x(0)+x(1), x(2)+x(3), x(4)+x(5), x(6)+x(7)

>Hi Richard,
>
>yes, of course.

This is of course how two sequences are added.  The question
however is how two signals are added.

Rick, you were the first to inject the word sequence into this
discussion, which had until then been discussing signals, when
you wrote:

The paper you cited is interesting. On page
1810 of the paper the author wrote: "Since
multiplexing of discrete-time signals is equivalent
to addition,... ." That's not true, of course. We
don't simply add the two discrete sequences' samples,
we interleave the two sequences' samples.  Perhaps the
"addition" the author is referring to should be called

Sequences and signals are different objects; therefore, the
addition operator is (potentially) defined differently for the two.

It's like an object-oriented programming language - the " + " operator

Two sequences are added by pairwise addition of values with the
same index.

Whereas two signals are added by pairwise addition of values which
are at the same point along the time axis.

As you note the latter could be called "time aligned addtion", but
I propose this is the default notion of addition of signals.
Certainly real-life signals, wherein a summing node adds two
signals that exist in the same local time frame, are added in
a time-aligned fasion.

Steve
```
Reply by March 5, 20182018-03-05
```W dniu 03.03.2018 o&nbsp;19:51, Richard (Rick) Lyons pisze:
(...)

>> could you clarify why x(t), which spectrum is X(f) in Eq.(19), MUST BE a
>>    periodic signal ?
>
> I'll do my best. Let's say I have a ten-second duration analog audio recording of a pure sine wave. First I compute the spectrum, Y1(f), of the first five seconds of the audio signal. Then I compute the spectrum, Y2(f), of five seconds of the sine wave audio starting at the beginning of the 2nd second out to the end of the 6th second.  For this periodic sine wave input signal, Y1(f) and Y2(f) will be related to each other as shown in Eqs. (18) and (19) in the cited Razavi. That is, Y2(f) will be equal to Y1(f) multiplied by a phase angle that is a linear function of frequency. Also, in this periodic signal case, |Y1(f)| = |Y2(f)|.

both examples are inadequate.
If the Fs is correct sampling frequency (Fs > 2 x BW of signal), then we
are talking about situation where we have the first recording sampled
from the beginning, at Fs/2, and second, starting from the moment
delayed fom beginning by Ts, and also at Fs/2 rate.

Anyway, |Y1(f)| is not equal to |Y2(f)| generally. That is also true in
case of infinite periodic signal like, on example, sinusoid of Fs/4
frequency. The amplitude of this two sequences (+x1, -x1 for first, and
+x2, -x2 for second) will be generally different, but equal for moments
where sinusoid phase is pi/4+k*pi only.

As for mentioned paper, equation (19) has an error in exponent of e. The
correct exponent related to Tck/2 delay (Fck = Fs/2) is:

-j*2*pi*f*delta(t))= -j*2*pi*f*Tck/2 = -j*pi*f*Tck .

As f = k*Fck (exponent values for other f are multiplied by 0) we get:
-j*2*pi*k*Fck*Tck/2 = -j*pi*k,
what gives the sequence: +1, -1, +1, -1 ... .
Even spectral copies are added, odd - subtracted, so we should get
correct spectrum, as in Fs case, but because Y1(f) and Y2(f) are
different, all the more for non-periodic signals ... so, the Figure 6 is
not correct.

Where is the bug in this prove ?

Maybe the frequency analysis should be done for uspsampled sequences ?

Best regards,

Roman Rumian
```
Reply by March 5, 20182018-03-05
```W dniu 03.03.2018 o 23:43, Richard (Rick) Lyons pisze:
(...)
> Hi Roman. I forgot to ask you: Do you agree that the
> addition of the two sequences,
>
>      Seq# 1 = x(0), x(2), x(4), x(6)
>
>      Seq# 2 = x(1), x(3), x(5), x(7)
>
> produces the four-sample sequence:
>
>       x(0)+x(1), x(2)+x(3), x(4)+x(5), x(6)+x(7)
>
> ?
Hi Richard,

yes, of course.

Regards,

Roman Rumian
```
Reply by March 5, 20182018-03-05
```On Saturday, March 3, 2018 at 5:48:47 PM UTC-8, Steve Pope wrote:

[Snipped by Lyons]

> I also sort of disagree with your description that summing two
> discrete-time signals necessitates shifting one of them on the
> time axis.  That is not the natural way to add two signals.
> In hardware, there would be no such shift unless you deliberately
> added a delay to one signal.  And in this case, it does not give you
> the signal that you want.  I would venture to say it is incorrect
> to do that.
>
> Steve

Hi Steve.
I see what's causing the confusion here. On March 3rd I wrote:

"the addition of the two sequences,

Seq# 1 = x(0), x(2), x(4), x(6)

Seq# 2 = x(1), x(3), x(5), x(7)

produces the four-sample sequence:

x(0)+x(1), x(2)+x(3), x(4)+x(5), x(6)+x(7)."

Those above x(k) samples were meant to merely represent arbitrary sample values, nothing more. What I should have written is:

Do you agree that the addition of the two sequences,

Seq# 1 = 1, 2, 3, 4

Seq# 2 = 7, 9, -5, 2

produces the four-sample sequence:

8, 11, -2, 6

I wanted to reiterate that, unlike multiplexing two four-sample sequences, adding two four-sample sequences produces a four-sample result. Sorry for the confusion.

[-Rick-]
```
Reply by March 5, 20182018-03-05
```On Saturday, March 3, 2018 at 12:46:21 AM UTC-8, rr wrote:

Hi Roman. My use of indexing in my March 3rd might cause confusion.  I should have written:

Do you agree that the addition of the two sequences,

Seq# 1 = 1, 2, 3, 4

Seq# 2 = 7, 9, -5, 2

produces the four-sample sequence:

8, 11, -2, 6

Sorry for the possible confusion.

[-Rick-]
```
Reply by March 3, 20182018-03-03
```Richard (Rick) Lyons <r.lyons@ieee.org> wrote:

>On Saturday, March 3, 2018 at 2:54:36 PM UTC-8, Steve Pope wrote:
>> Richard (Rick) Lyons <r.lyons@ieee.org> wrote:
>>
>> >>> On Friday, March 2, 2018 at 1:19:22 PM UTC-8, Steve Pope wrote:
>> >>
>> >> >> Actually it's true.  A discrete-time signal is non-zero only at
>> >> >> discrete points in time.  So if for example,
>> >>
>> >> >>             X(t) is zero for t not in (0, 2, 4, 6...)
>> >> >>             Y(t) is zero for t not in (1, 3, 5, 7...)
>>
>> Okay my notation above was not the best.  I intended to write
>>
>>              X(t) is zero for t not in {0, 2, 4, 6...)
>>              Y(t) is zero for t not in (1, 3, 5, 7...)
>>
>> >> >> Then Z(t) = X(t) + Y(t) where
>> >>
>> >> >>        Z(t) = X(t) if t = 0, 2, 4, 6 ... , Y(t) otherwise.
>>
>> >> >> So I would say addition and multiplexing are the same, so long as
>> >> >> you haven't mangled the time axis.
>>
>> >> >Hi Steve.  I claim that multiplexing two sequences is not the same as
>> >> >adding two two sequences.  Multiplexing your above X(t) and Y(t)
>> >> >sequences yields
>> >> >
>> >> >  Seq# 1 --> 0, 1, 2, 3, 4, 5, ... .
>> >> >
>> >> >Adding your X(t) and Y(t) sequences produces:
>> >> >
>> >> >  Seq# 2 --> 1, 5, 9, 13, 17, ... .
>> >>
>> >> You've completely lost me.  I claim that you get a sequence that
>> >> is equal to X(t) for t = 0, 2, 4, 6 ... and equal to Y(t) for
>> >> t = 1, 3, 5, 7.
>>
>> I get:
>>
>> X(0) + Y(0) = X(0) + 0 = X(0)
>> X(1) + Y(1) = 0 + Y(1) = Y(1)
>> X(2) + Y(0) = X(2) + 0 = X(2)
>>
>> etc.
>>
>> >Yes, I agree that the sequence you describe is the sequence that results
>> >when you multiplex your X(t) and Y(t).  But I disagree with your claim
>> >that multiplexing X(t) and Y(t) and adding X(t) and Y(t) are equivalent
>> >operations.
>>
>> >I have an even simpler example of why I believe multiplexing and
>> >addition are not the same. If you multiplex two N-sample sequences
>> >you'll generate a 2N-sample sequence. If you add two N-sample sequences
>> >you'll produce an N-sample sequence. My claim is that the 2N-sample and
>> >the N-sample sequences are not equal to each other, thus multiplexing
>> >and addition are not equivalent operations.
>>
>> I think you are making sense, and if the author made a blanket
>> claim that multiplexing and adding are the same, then that is misleading.
>>
>
>[Snipped by Lyons]
>
>Hi Steve.

>On March 2 you claimed that addition and multiplexing are equivalent
>operations when you wrote: "So I would say addition and multiplexing are
>the same,... ."

What I actually wrote:

"So I would say addition and multiplexing are the same, so long as
you haven't mangled the time axis."

My qualifying language about not mangling the time axis is a little
loose, but I cleaned that up in the post you just responded to.

>And that statement of yours was what triggered my
>recent posts where I showed that addition and multiplexing are not
>equivalent operations. I agree with your above last sentence so long as
>we replace your word "misleading" with the word "incorrect."

Really I was just trying to explain why the original author used
this description, and why it is sometimes true.  I'm not asserting
anything beyond that.

I also sort of disagree with your description that summing two
discrete-time signals necessitates shifting one of them on the
time axis.  That is not the natural way to add two signals.
In hardware, there would be no such shift unless you deliberately
added a delay to one signal.  And in this case, it does not give you
the signal that you want.  I would venture to say it is incorrect
to do that.

Steve
```
Reply by March 3, 20182018-03-03
```On Saturday, March 3, 2018 at 2:54:36 PM UTC-8, Steve Pope wrote:
> Richard (Rick) Lyons <r.lyons@ieee.org> wrote:
>
> >>> On Friday, March 2, 2018 at 1:19:22 PM UTC-8, Steve Pope wrote:
> >>
> >> >> Actually it's true.  A discrete-time signal is non-zero only at
> >> >> discrete points in time.  So if for example,
> >>
> >> >>             X(t) is zero for t not in (0, 2, 4, 6...)
> >> >>             Y(t) is zero for t not in (1, 3, 5, 7...)
>
> Okay my notation above was not the best.  I intended to write
>
>              X(t) is zero for t not in {0, 2, 4, 6...)
>              Y(t) is zero for t not in (1, 3, 5, 7...)
>
> >> >> Then Z(t) = X(t) + Y(t) where
> >>
> >> >>        Z(t) = X(t) if t = 0, 2, 4, 6 ... , Y(t) otherwise.
>
> >> >> So I would say addition and multiplexing are the same, so long as
> >> >> you haven't mangled the time axis.
>
> >> >Hi Steve.  I claim that multiplexing two sequences is not the same as
> >> >adding two two sequences.  Multiplexing your above X(t) and Y(t)
> >> >sequences yields
> >> >
> >> >  Seq# 1 --> 0, 1, 2, 3, 4, 5, ... .
> >> >
> >> >Adding your X(t) and Y(t) sequences produces:
> >> >
> >> >  Seq# 2 --> 1, 5, 9, 13, 17, ... .
> >>
> >> You've completely lost me.  I claim that you get a sequence that
> >> is equal to X(t) for t = 0, 2, 4, 6 ... and equal to Y(t) for
> >> t = 1, 3, 5, 7.
>
> I get:
>
> X(0) + Y(0) = X(0) + 0 = X(0)
> X(1) + Y(1) = 0 + Y(1) = Y(1)
> X(2) + Y(0) = X(2) + 0 = X(2)
>
> etc.
>
> >Yes, I agree that the sequence you describe is the sequence that results
> >when you multiplex your X(t) and Y(t).  But I disagree with your claim
> >that multiplexing X(t) and Y(t) and adding X(t) and Y(t) are equivalent
> >operations.
>
> >I have an even simpler example of why I believe multiplexing and
> >addition are not the same. If you multiplex two N-sample sequences
> >you'll generate a 2N-sample sequence. If you add two N-sample sequences
> >you'll produce an N-sample sequence. My claim is that the 2N-sample and
> >the N-sample sequences are not equal to each other, thus multiplexing
> >and addition are not equivalent operations.
>
> I think you are making sense, and if the author made a blanket
> claim that multiplexing and adding are the same, then that is misleading.
>

[Snipped by Lyons]

Hi Steve.
On March 2 you claimed that addition and multiplexing are equivalent operations when you wrote: "So I would say addition and multiplexing are the same,... ."  And that statement of yours was what triggered my recent posts where I showed that addition and multiplexing are not equivalent operations. I agree with your above last sentence so long as we replace your word "misleading" with the word "incorrect."

[-Rick-]
```
Reply by March 3, 20182018-03-03
```Richard (Rick) Lyons <r.lyons@ieee.org> wrote:

>>> On Friday, March 2, 2018 at 1:19:22 PM UTC-8, Steve Pope wrote:
>>
>> >> Actually it's true.  A discrete-time signal is non-zero only at
>> >> discrete points in time.  So if for example,
>>
>> >>             X(t) is zero for t not in (0, 2, 4, 6...)
>> >>             Y(t) is zero for t not in (1, 3, 5, 7...)

Okay my notation above was not the best.  I intended to write

X(t) is zero for t not in {0, 2, 4, 6...)
Y(t) is zero for t not in (1, 3, 5, 7...)

>> >> Then Z(t) = X(t) + Y(t) where
>>
>> >>        Z(t) = X(t) if t = 0, 2, 4, 6 ... , Y(t) otherwise.

>> >> So I would say addition and multiplexing are the same, so long as
>> >> you haven't mangled the time axis.

>> >Hi Steve.  I claim that multiplexing two sequences is not the same as
>> >adding two two sequences.  Multiplexing your above X(t) and Y(t)
>> >sequences yields
>> >
>> >  Seq# 1 --> 0, 1, 2, 3, 4, 5, ... .
>> >
>> >Adding your X(t) and Y(t) sequences produces:
>> >
>> >  Seq# 2 --> 1, 5, 9, 13, 17, ... .
>>
>> You've completely lost me.  I claim that you get a sequence that
>> is equal to X(t) for t = 0, 2, 4, 6 ... and equal to Y(t) for
>> t = 1, 3, 5, 7.

I get:

X(0) + Y(0) = X(0) + 0 = X(0)
X(1) + Y(1) = 0 + Y(1) = Y(1)
X(2) + Y(0) = X(2) + 0 = X(2)

etc.

>Yes, I agree that the sequence you describe is the sequence that results
>when you multiplex your X(t) and Y(t).  But I disagree with your claim
>that multiplexing X(t) and Y(t) and adding X(t) and Y(t) are equivalent
>operations.

>I have an even simpler example of why I believe multiplexing and
>addition are not the same. If you multiplex two N-sample sequences
>you'll generate a 2N-sample sequence. If you add two N-sample sequences
>you'll produce an N-sample sequence. My claim is that the 2N-sample and
>the N-sample sequences are not equal to each other, thus multiplexing
>and addition are not equivalent operations.

I think you are making sense, and if the author made a blanket
claim that multiplexing and adding are the same, then that is misleading.

They only obtain the same result if:

(1) Both sequences are defined for all time indices of the result
(2) For any time index of the result, only one sequence is non-zero.

Under these conditions -- maybe it is much clearer in Verilog:

( x + y ) == (x ? x : y)

Steve
```
Reply by March 3, 20182018-03-03
```On Saturday, March 3, 2018 at 12:46:21 AM UTC-8, rr wrote:
> Hi Richard,
>
> W dniu 03.03.2018 o 00:57, Richard (Rick) Lyons pisze:
> (...)
>  > Hey Roman, It just "hit me". Isn't it true that the author's Eq. (19)
> only applies to periodic signals?  If that's true then his spectral
> example in Figure 6 would not be correct for real-world, unpredictable,
> information-carrying signals.  Am I missing something here?
>  > (...)
>
> could you clarify why x(t), which spectrum is X(f) in Eq.(19), MUST BE a
>   periodic signal ?
>
> Do you agree, that the addition of two upsampled, time-shifted sequences:
> x1u(n) = x(0),  0,     x(2),  0,    ... .
> x2u(n) = 0,     x(1),  0,     x(3), ... .
>
> is equal to multiplexing of two sequences:
>
> x1(n) = x(0), x(2), ... .
> x2(n) = x(1), x(3), ... .
>
> ?

Hi Roman. I forgot to ask you: Do you agree that the
addition of the two sequences,

Seq# 1 = x(0), x(2), x(4), x(6)

Seq# 2 = x(1), x(3), x(5), x(7)

produces the four-sample sequence:

x(0)+x(1), x(2)+x(3), x(4)+x(5), x(6)+x(7)

?

[-Rick-]
```
Reply by March 3, 20182018-03-03
```On Saturday, March 3, 2018 at 12:46:21 AM UTC-8, rr wrote:
> Hi Richard,
>
> W dniu 03.03.2018 o 00:57, Richard (Rick) Lyons pisze:
> (...)
>  > Hey Roman, It just "hit me". Isn't it true that the author's Eq. (19)
> only applies to periodic signals?  If that's true then his spectral
> example in Figure 6 would not be correct for real-world, unpredictable,
> information-carrying signals.  Am I missing something here?
>  > (...)

Hi Roman

> could you clarify why x(t), which spectrum is X(f) in Eq.(19), MUST BE a
>   periodic signal ?

I'll do my best. Let's say I have a ten-second duration analog audio recording of a pure sine wave. First I compute the spectrum, Y1(f), of the first five seconds of the audio signal. Then I compute the spectrum, Y2(f), of five seconds of the sine wave audio starting at the beginning of the 2nd second out to the end of the 6th second.  For this periodic sine wave input signal, Y1(f) and Y2(f) will be related to each other as shown in Eqs. (18) and (19) in the cited Razavi. That is, Y2(f) will be equal to Y1(f) multiplied by a phase angle that is a linear function of frequency. Also, in this periodic signal case, |Y1(f)| = |Y2(f)|.

Now replace the ten-second audio sine wave signal with a ten-second audio recording of Lady Gaga singing a song (which is not periodic in time) and compute the new Y1(f) and Y2(f) spectra. What I'm saying is the new Y1(f) and the new Y2(f) will be NOT be related to each other as shown in Eqs. (18) and (19) in the Razavi paper. In this non-periodic signal case, spec magnitude |Y1(f)| is not equal to spec magnitude |Y2(f)|.

>
> Do you agree, that the addition of two upsampled, time-shifted sequences:
> x1u(n) = x(0),  0,     x(2),  0,    ... .
> x2u(n) = 0,     x(1),  0,     x(3), ... .
>
> is equal to multiplexing of two sequences:
>
> x1(n) = x(0), x(2), ... .
> x2(n) = x(1), x(3), ... .

I definitely agree in principle with your multiplexing example.  But I would change the description of multiplexing, assuming that x1(n) and x2(n) are finite and equal in length, to following. Let's say that:

x1(n) = x(0), x(2), x(4), x(6) and
x2(n) = x(1), x(3), x(5), x(7).

1) Create Seq# 1 by following each sample in x1(n) with a zero-valued sample.
Seq# 1 = x(0), 0, x(2), 0, x(4), 0, x(6), 0.
(Seq# 1 is not simply stuffing zeros in between each x1(n) sample.
Seq# 1 must end with a zero-valued sample!)

2) Create Seq# 2 by following each sample in x2(n) with a zero-valued samples.
Seq# 2 = x(1), 0, x(3), 0, x(5), 0, x(7), 0.
(Seq# 2 must end with a zero-valued sample.)

3) Create Seq# 3 by circularly rotating Seq# 2 forward in time by one sample.
Seq# 3 = 0, x(1), 0, x(3), 0, x(5), 0, x(7).
(The 1st sample in Seq# 3 is the last sample in Seq# 2.)

I claim that multiplexing x(n) and x2(n) equals the sample-by-sample addition of Seq# 1 and Seq# 3 producing:

x(0), x(1), x(2), x(3), x(4), x(5), x(6), x(7).

Roman, sorry for my long-winded reply.

[-Rick-]
```