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Is IF Data real or complex

Started by b2508 September 15, 2015
Well, probably sounds stupid but I get quite confused on this. In my
specification I get Intermediate Frequency data as real from third party
RF Front end.

Isn't RF Data suppose to be real? Then I assume it was mixed with some LO
to downconvert it from RF to IF. How this this result in a real signal?

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Posted through http://www.DSPRelated.com
On 15.9.15 16:06, b2508 wrote:
> Well, probably sounds stupid but I get quite confused on this. In my > specification I get Intermediate Frequency data as real from third party > RF Front end. > > Isn't RF Data suppose to be real? Then I assume it was mixed with some LO > to downconvert it from RF to IF. How this this result in a real signal?
You may have either, depending on the mixer. If nothing is said, it is pretty safe to assume that IF is a real, band-limited signal. If the mixer has two channels with local oscillators at 90 degree phase difference, you can have complex IF signal pair at the output. -- -TV
On Tuesday, September 15, 2015 at 9:35:36 AM UTC-4, Tauno Voipio wrote:
> On 15.9.15 16:06, b2508 wrote: > > Well, probably sounds stupid but I get quite confused on this. In my > > specification I get Intermediate Frequency data as real from third party > > RF Front end. > > > > Isn't RF Data suppose to be real? Then I assume it was mixed with some LO > > to downconvert it from RF to IF. How this this result in a real signal? > > > You may have either, depending on the mixer. If nothing is said, it is > pretty safe to assume that IF is a real, band-limited signal. > > If the mixer has two channels with local oscillators at 90 degree > phase difference, you can have complex IF signal pair at the output. > > -- > > -TV
as a simple minded hardware guy... my rule is if the signal takes one cable to transport, it is real.. if the signal takes two cables to transport i.e I and Q. , it is complex, not sure if this is technically correct. Mark
On Tue, 15 Sep 2015 08:06:22 -0500, "b2508" <108118@DSPRelated> wrote:

>Well, probably sounds stupid but I get quite confused on this. In my >specification I get Intermediate Frequency data as real from third party >RF Front end. > >Isn't RF Data suppose to be real? Then I assume it was mixed with some LO >to downconvert it from RF to IF. How this this result in a real signal? > >--------------------------------------- >Posted through http://www.DSPRelated.com
If it is mixed with a real-valued sinusoid the result will be real-valued mixing products on both the high side and low side of the RF signal. This is due to the product of two sinusoids being two separate sinusoids at the sum and difference frequencies. There's a trig identity for this. If it is mixed with a complex-valued local oscillator, then the entire spectrum is shifted by the frequency of the LO. The product of the real-valued rf input and the complex-valued LO will be a complex-valued result. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
On Tue, 15 Sep 2015 08:06:22 -0500, b2508 wrote:

> Well, probably sounds stupid but I get quite confused on this. In my > specification I get Intermediate Frequency data as real from third party > RF Front end. > > Isn't RF Data suppose to be real? Then I assume it was mixed with some > LO to downconvert it from RF to IF. How this this result in a real > signal?
Well, RF _data_ is whatever you specify it to be, but real live signals out in the real live aether are really real. It has something to do with the physical world not containing any imaginary numbers -- odd, that. So perhaps this comes about because the result of multiplying two real numbers is a real number? Or, to get even more distressingly physical, because in a REAL radio you can't have IMAGINARY numbers running around? So the circuit designers are constrained -- yes, constrained!, tied down!, chained, even! -- to either having a single LO signal or trying to simulate a complex signal by making a pair of them that are very accurately 90 degrees apart (which isn't at all trivial, even in an IC)? There's a very high probability that your 3rd-party RF front end is a plain old superheterodyne radio of the sort that's been around for 90 years or so. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
On Tue, 15 Sep 2015 07:21:05 -0700, makolber wrote:

> On Tuesday, September 15, 2015 at 9:35:36 AM UTC-4, Tauno Voipio wrote: >> On 15.9.15 16:06, b2508 wrote: >> > Well, probably sounds stupid but I get quite confused on this. In my >> > specification I get Intermediate Frequency data as real from third >> > party RF Front end. >> > >> > Isn't RF Data suppose to be real? Then I assume it was mixed with >> > some LO to downconvert it from RF to IF. How this this result in a >> > real signal? >> >> >> You may have either, depending on the mixer. If nothing is said, it is >> pretty safe to assume that IF is a real, band-limited signal. >> >> If the mixer has two channels with local oscillators at 90 degree phase >> difference, you can have complex IF signal pair at the output. >> >> -- >> >> -TV > > as a simple minded hardware guy... my rule is > > if the signal takes one cable to transport, it is real.. > if the signal takes two cables to transport i.e I and Q. , it is > complex, > > not sure if this is technically correct.
It's not an IIF proposition. You can turn it around: If the signal is real it only takes one cable to transport. If the signal is quadrature it must have two cables to transport. But there are other ways you can put a group of related signals onto multiple cables that aren't I and Q. Being a stickler about this, I wouldn't even say "I and Q is complex" -- I would say "I and Q is _pretending_ to be complex". Because -- there ain't no imaginary numbers in this real world of ours. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
<makolber@yahoo.com> wrote:

>as a simple minded hardware guy... my rule is
>if the signal takes one cable to transport, it is real..
Yep
>if the signal takes two cables to transport i.e I and Q. , it is complex,
Depends. A signal comprised of two real-valued signals is only a "complex" signal if has some quality about it, such that it makes sense to view it on the complex plane. For example, if one signal is an FM-modulated carrier, and the second signal is a voltage level corresponding to how sunny it is in Mexico City, that is not a complex signal. Steve
On 9/15/15 1:09 PM, Tim Wescott wrote:
> ... > Being a stickler about this, I wouldn't even say "I and Q is complex" -- > I would say "I and Q is _pretending_ to be complex".
agree with everything up to here...
> Because -- there ain't no imaginary numbers in this real world of ours.
i would not agree with that. i might state it as "there ain't no imaginary physical quantities in this real world of ours." "number" != "physical quantity" -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Tim Wescott <seemywebsite@myfooter.really> wrote:
> On Tue, 15 Sep 2015 07:21:05 -0700, makolber wrote:
(snip)
>> as a simple minded hardware guy... my rule is
>> if the signal takes one cable to transport, it is real.. >> if the signal takes two cables to transport i.e I and Q. , it is >> complex,
>> not sure if this is technically correct.
> It's not an IIF proposition. You can turn it around:
> If the signal is real it only takes one cable to transport. > If the signal is quadrature it must have two cables to transport.
> But there are other ways you can put a group of related signals onto > multiple cables that aren't I and Q.
> Being a stickler about this, I wouldn't even say "I and Q is complex" -- > I would say "I and Q is _pretending_ to be complex". Because -- there > ain't no imaginary numbers in this real world of ours.
I suppose, but in an analog system, the signal in the cable is analogous to some physicial quantity, not the actual quantity, so it doesn't seem so hard to call it imaginary. Also, there are some physical quantities that really are complex, such as index of refraction: https://en.wikipedia.org/wiki/Refractive_index#Complex_refractive_index The index of refraction depends on the motion of electrons in a material subject to an electromagnetic field. They don't know that one motion contributes to changing the phase velocity, and another to absorption. If that isn't obvious enough, consider that the index of refraction is the complex square root of the dielectric constant. (That is, relative pemittivity.) https://en.wikipedia.org/wiki/Refractive_index#Relative_permittivity_and_permeability Again, the electrons don't know the difference, but note also how the square root mixes the real and imaginary parts. Phasors, on the other hand, are an artificial use of complex math to measure quantities that are basically real, but might have different phases. -- glen
Tim Wescott wrote:
> On Tue, 15 Sep 2015 07:21:05 -0700, makolber wrote: > >> On Tuesday, September 15, 2015 at 9:35:36 AM UTC-4, Tauno Voipio wrote: >>> On 15.9.15 16:06, b2508 wrote: >>>> Well, probably sounds stupid but I get quite confused on this. In my >>>> specification I get Intermediate Frequency data as real from third >>>> party RF Front end. >>>> >>>> Isn't RF Data suppose to be real? Then I assume it was mixed with >>>> some LO to downconvert it from RF to IF. How this this result in a >>>> real signal? >>> >>> >>> You may have either, depending on the mixer. If nothing is said, it is >>> pretty safe to assume that IF is a real, band-limited signal. >>> >>> If the mixer has two channels with local oscillators at 90 degree phase >>> difference, you can have complex IF signal pair at the output. >>> >>> -- >>> >>> -TV >> >> as a simple minded hardware guy... my rule is >> >> if the signal takes one cable to transport, it is real.. >> if the signal takes two cables to transport i.e I and Q. , it is >> complex, >> >> not sure if this is technically correct. > > It's not an IIF proposition. You can turn it around: > > If the signal is real it only takes one cable to transport. > If the signal is quadrature it must have two cables to transport. > > But there are other ways you can put a group of related signals onto > multiple cables that aren't I and Q. > > Being a stickler about this, I wouldn't even say "I and Q is complex" -- > I would say "I and Q is _pretending_ to be complex". Because -- there > ain't no imaginary numbers in this real world of ours. >
Two wrongs don't make a right, but two reals might make an imaginary. -- Les Cargill