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 ?
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.