On Fri, 25 Dec 2015 12:23:52 -0600, "Sharan123" <99077@DSPRelated>
wrote:

>Dear Eric, Steve, et al.
>
>I have few questions related to LTE/OFDM. I have summarized them below.
>I did not kept these questions separate from my previous above to avoid
>confusion ...
>
>1) In the base station downlink, does iFFT step handle 
>a) only frequency domain to time domain conversion
>b) it handles OFDM 
>c) just a part of OFDM. 
>
>I believe, it is a) and b) but want to just make sure ...

It's probably a point-of-view thing, but I think it's c).   That's a
fundamental part of what OFDM is, but a) and b) are essentially also
true.   Especially a).   ;)

>2) During the OFDM symbol period, Tu, the number of periods for each
>sub-carrier is an integer number. Is it 1/2/3 etc. for first/second/third
>etc. sub-carriers?

Just like any FFT, the bin number describes how many integer cycles
there are for that bin over the duration of the transform window.
It's the same in OFDM.   That being said, numbers or indices can be
arbitrarily assigned to bins/subcarriers for bookkeeping purposes that
may differ from the usualy signal processing indices.   I don't know
whether that is done much in LTE, but that is done in some systems.

>3) I have been thoroughly confused with TTI parameter. What role does this
>play on the PHY layer?

Not much.   It is a constant 1ms slice of physical resource that the
MAC can use to keep latency low.   The doc I referenced earlier, 

https://home.zhaw.ch/kunr/NTM1/literatur/LTE%20in%20a%20Nutshell%20-%20Physical%20Layer.pdf

shows how one TTI is split into resource block quanta, which are the
minimum chunks of assignable phy layer, i.e., 12 subcarrier for 14
symbols.   These quanta get filled by the MAC based on traffic demand
for each user, as controlled by the scheduler.

>Given a bandwidth, I believe that rate of data, symbols in a time slot,
>symbol period are all fixed. Hence I am trying to understand how TTI
>influences PHY parameters.

My understanding is that TTI is just a fixed 1ms interval, and is sort
of the basic time clock of the MAC.   Since the LTE frames and symbol
timing are locked with that, it makes translation from the MAC to PHY
resources simpler.

>4) why the transmit power of eNodeB is higher than power from terminals?
>is this to achieve higher downlink speed?

I usually think of it the other way, that the handset Tx power is
limited to manage battery life, which is a managed resource on the
handset.   It is not as much of a problem at the NodeB (Base Station),
so the power can be cranked up (or just not cranked down) to maintain
link reliability.   

There is a fair amount of asymmetry in the LTE system, not the least
of which is that the BS (NodeB) can make use of antenna spatial
diversity (or other diversity or even MIMO techniques), to enhance the
link, while the handset cannot as much due to its size.    This makes
it possible for the NodeB to achieve diversity or SIMO/MIMO gains at
the receiver in the Uplink, which allows the handset to transmit at
lower power (and save battery).

In the downlink, the BS (NodeB) can potentially do beamsteering or
transmit diversity encoding (e.g., Alamouti), but the handset is too
small to get much antenna diversity in the receiver.   LTE does use
MIMO, but the limitations in diversity due to the handset size limit
things on that end of the link.   

Basically, all of those add up to good reasons to use more power in
the downlink than uplink, but another good reason is just Becasue You
Can.   The BS isn't power limited like the handset, so it allows the
additional transmit power capability to increase DL reliability and
also increase the DL rate, as you mentioned.   Since the DL has to be
able to spray out a lot more data than the UL from a terminal
typically needs, this is just all part of the overall system
engineering. 

It's a very asymmetric system.

>I thank you all in advance ...
>---------------------------------------
>Posted through http://www.DSPRelated.com

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

On Fri, 25 Dec 2015 11:59:01 -0600, "Sharan123" <99077@DSPRelated>
wrote:

>Dear Eric, Steve, et al.
>
>I have been trying to work out OFDM in a little more detail.
>
>I am taking 20 & 15 MHz channels, as example, to explain the question.
>
>The bandwidth (or sample rate) = 30.72 MHz, 15.36 MHz for 20 and 10 MHz
>respectively.
>
>Sample time Ts (1/bandwidth) = 33 ns, 65 ns for 20 and 10 MHz
>respectively.
>
>Symbols time (excluding CP) = 1/15000 = 66 us
>
>Cyclic prefix period = for 20 MHz channel, 144 bits x Ts for 20 MHz 
>                 144 x 33 ns = 4.6 us
>
>Total symbol time including cyclic prefix for 20 MHz channel = 71 us
>
>In case of 10 MHz channel, cyclic prefix period = 144 x Ts for 10 MHz
>                                                  144 x 65 = 9.3 us
>
>Total symbol time including cyclic prefix for 10 MHz channel = 76 us
>
>However, I do understand that the total symbol time including the CP
>remains same for all the BW. So, where I am going wrong in my
>calculations?
>---------------------------------------
>Posted through http://www.DSPRelated.com

If you haven't seen it, this is a good description of the nuts and
bolts of the basics of the physical layer for LTE:

https://home.zhaw.ch/kunr/NTM1/literatur/LTE%20in%20a%20Nutshell%20-%20Physical%20Layer.pdf

The section labelled Physical Layer Pararmeters has some relevant
stuff.   I had to look this up again as I couldn't recall how the CP
lengths were managed.

I think the only thing you missed is that the time length of the CP is
not bandwidth dependent.   Remember that the length of the CP is
chosen based on the expected lengths of the channel impulse response,
which won't change with the signal bandwidth.

So the CP for the 10MHz case will have fewer samples.   It looks like
fs = 23.04MHz for the 10MHz BW case, so the CP would have around
23.04e6*4.7us = ~108 samples, for the 4.69us CP case.

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

Dear Eric, Steve, et al.

I have few questions related to LTE/OFDM. I have summarized them below.
I did not kept these questions separate from my previous above to avoid
confusion ...

1) In the base station downlink, does iFFT step handle 
a) only frequency domain to time domain conversion
b) it handles OFDM 
c) just a part of OFDM. 

I believe, it is a) and b) but want to just make sure ...

2) During the OFDM symbol period, Tu, the number of periods for each
sub-carrier is an integer number. Is it 1/2/3 etc. for first/second/third
etc. sub-carriers?

3) I have been thoroughly confused with TTI parameter. What role does this
play on the PHY layer?

Given a bandwidth, I believe that rate of data, symbols in a time slot,
symbol period are all fixed. Hence I am trying to understand how TTI
influences PHY parameters.

4) why the transmit power of eNodeB is higher than power from terminals?
is this to achieve higher downlink speed?

I thank you all in advance ...
---------------------------------------
Posted through http://www.DSPRelated.com

Dear Eric, Steve, et al.

I have been trying to work out OFDM in a little more detail.

I am taking 20 & 15 MHz channels, as example, to explain the question.

The bandwidth (or sample rate) = 30.72 MHz, 15.36 MHz for 20 and 10 MHz
respectively.

Sample time Ts (1/bandwidth) = 33 ns, 65 ns for 20 and 10 MHz
respectively.

Symbols time (excluding CP) = 1/15000 = 66 us

Cyclic prefix period = for 20 MHz channel, 144 bits x Ts for 20 MHz 
                 144 x 33 ns = 4.6 us

Total symbol time including cyclic prefix for 20 MHz channel = 71 us

In case of 10 MHz channel, cyclic prefix period = 144 x Ts for 10 MHz
                                                  144 x 65 = 9.3 us

Total symbol time including cyclic prefix for 10 MHz channel = 76 us

However, I do understand that the total symbol time including the CP
remains same for all the BW. So, where I am going wrong in my
calculations?
---------------------------------------
Posted through http://www.DSPRelated.com

On Tue, 01 Dec 2015 03:46:08 -0500, Randy Yates
<yates@digitalsignallabs.com> wrote:

>eric.jacobsen@ieee.org (Eric Jacobsen) writes:
>
>> On Mon, 30 Nov 2015 03:27:36 +0000 (UTC), spope33@speedymail.org
>> (Steve Pope) wrote:
>>
>>>Eric Jacobsen <eric.jacobsen@ieee.org> wrote:
>>>
>>>>The CP length or presence or absence or the processing thereof has no
>>>>affect on effective subcarrier width in an OFDM system, so I would
>>>>disagree with that statement in the context of answering the OPs
>>>>questions in that area so far.  There are lots of parts and pieces to
>>>>OFDM systems that have not been touched on here, appropriately so
>>>>because they're not germaine to the question.
>>>
>>>Well, the question was whether the subcarrier width is the same
>>>as the subcarrier spacing; so the possible differences between these 
>>>two was thereby "touched upon".
>>
>> And if the idea is to understand the workings of an OFDM system, which
>> I think was the case, an answer in that context is, IMHO, better than
>> one outside of it.
>
>Eric / Steve,
>
>I don't know what you mean by "inside/outside" but this question
>has perked my interest in the topic. Why IS the carrier bandwidth
>different than the carrier spacing?

I don't think it is different.   Steve seems to me to just be pointing
out a particular point of view whereby if you hold things up to the
light just right one can claim that the subcarrier bandwidths have
shrunk a little bit.

It's certainly not different to the receiver, though, since the
subcarriers *have* to maintain a high degree of orthogonality to get
reasonable performance.   That won't happen if their bandwidth
changes.

>I am woefully ignorant on the subject, but isn't the answer independent
>of OFDM? Isn't it that the FFT can viewed as a set of bandpass filters
>(a filter bank), and that each of these filters are not Fs / N wide but
>something less due to the mechanics of the FFT? 

This kind of begs for nitpicking on the definition of bandwidth,
sadly.  Since window functions change the 3dB width of the sinx/x
response of each bin, some will say that that alone will change their
bandwidth.   So what's the normal BW of an FFT bin?   I've always
liked the rectangular-window response where the 3dB width equals the
bin width as the starting reference, and from that perspective the
binwidth = the bin BW, but that's just me.

OFDM generally uses rectangular windows partly for this reason, so
that there is minimum interference between subcarriers in the presence
of synchronization errors.

>By the way, I've watched you guys argue for days (weeks?) now. I wish
>you would both acknowledge each other's maturity and capabilities, and
>spend all this energy for positive things. I, for one, respect you both
>immensely.

I don't think anybody's done any real mud-slinging or claimed anybody
else isn't capable or competent (that I recall, anyway).   I think
it's a fun dialogue.   Maybe I'm weird that way.

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

On Tue, 01 Dec 2015 12:09:14 -0600, "Sharan123" <99077@DSPRelated>
wrote:

>>>4) Considering the used sub-carriers of 1200, the used bandwidth is
>turns
>>>out to be - 1200*1500 = 18 MHz. But in any case 2048 sub-carriers are
>>>present (though not used). Hence the bandwidth is really - 2048*1500 =
>30
>>>MHz.
>>
>>Those "guard band" samples are zero, so no energy comes out at those
>>frequencies other than the sidelobe leakage from the populated
>>subcarriers.   The system is designed so that by the edge of the
>>channel that energy is low enough, and/or can be suppressed by an IF
>>filter, so that it is not an issue for the adjacent channels.
>
>Dear Eric, Steve, Kaz,
>
>I have couple of questions based on the response above.
>
>1. The guard band is 20-18 = 2 MHz or 30-18 = 12 MHz?

The 20MHz channels are adjacent to each other, so the guard band if
from the occupied part of the spectrum to the next adjacent channel,
or 20-18, split between the two sides (since there can be an adjacent
channel on each side), so 1MHz on each side. 

It is therefore imperative in the system that the additional 5MHz on
each side has significant filtering/suppression so that no energy
intrudes into the other channel.   This isn't difficult if those
subcarriers are zeros in the transmitter, and even easier if there is
an IF filter that helps out.   Having the extra transmit bandwidth
zeroed makes the design of the IF filter a bit easier.

>2. Also, if guard band samples are zero, I assume that in frequency
>domain, we still end up using 30 MHz and those unused frequencies are not
>available for usage from a spectrum perspective.

They adjacent channels are available for other transmitters in the
system to use, so they cannot be used by the original transmitter.

Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

>>4) Considering the used sub-carriers of 1200, the used bandwidth is
turns
>>out to be - 1200*1500 = 18 MHz. But in any case 2048 sub-carriers are
>>present (though not used). Hence the bandwidth is really - 2048*1500 =
30
>>MHz.
>
>Those "guard band" samples are zero, so no energy comes out at those
>frequencies other than the sidelobe leakage from the populated
>subcarriers.   The system is designed so that by the edge of the
>channel that energy is low enough, and/or can be suppressed by an IF
>filter, so that it is not an issue for the adjacent channels.

Dear Eric, Steve, Kaz,

I have couple of questions based on the response above.

1. The guard band is 20-18 = 2 MHz or 30-18 = 12 MHz?

2. Also, if guard band samples are zero, I assume that in frequency
domain, we still end up using 30 MHz and those unused frequencies are not
available for usage from a spectrum perspective.

Thanks a lot ...

---------------------------------------
Posted through http://www.DSPRelated.com

eric.jacobsen@ieee.org (Eric Jacobsen) writes:

> On Mon, 30 Nov 2015 03:27:36 +0000 (UTC), spope33@speedymail.org
> (Steve Pope) wrote:
>
>>Eric Jacobsen <eric.jacobsen@ieee.org> wrote:
>>
>>>The CP length or presence or absence or the processing thereof has no
>>>affect on effective subcarrier width in an OFDM system, so I would
>>>disagree with that statement in the context of answering the OPs
>>>questions in that area so far.  There are lots of parts and pieces to
>>>OFDM systems that have not been touched on here, appropriately so
>>>because they're not germaine to the question.
>>
>>Well, the question was whether the subcarrier width is the same
>>as the subcarrier spacing; so the possible differences between these 
>>two was thereby "touched upon".
>
> And if the idea is to understand the workings of an OFDM system, which
> I think was the case, an answer in that context is, IMHO, better than
> one outside of it.

Eric / Steve,

I don't know what you mean by "inside/outside" but this question
has perked my interest in the topic. Why IS the carrier bandwidth
different than the carrier spacing?

I am woefully ignorant on the subject, but isn't the answer independent
of OFDM? Isn't it that the FFT can viewed as a set of bandpass filters
(a filter bank), and that each of these filters are not Fs / N wide but
something less due to the mechanics of the FFT? 

By the way, I've watched you guys argue for days (weeks?) now. I wish
you would both acknowledge each other's maturity and capabilities, and
spend all this energy for positive things. I, for one, respect you both
immensely.
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

On Tue, 1 Dec 2015 00:26:28 +0000 (UTC), spope33@speedymail.org (Steve
Pope) wrote:

>Eric Jacobsen <eric.jacobsen@ieee.org> wrote:
>
>>On Mon, 30 Nov 2015 08:51:15 +0000 (UTC), spope33@speedymail.org
>
>>>I think the disconnect here is that you believe my reply to
>>>the OP's question, which was all of eight sentences, somehow
>>>derailed the preferred (by who?) context of the discussion, rather 
>>>than contributing to the discussion.
>
>>>I think this belief has gotta be baseless.
>
>>If a young primary-school student asks "Is 2+2 always 4?",  I think
>>the proper answer is "Yes, it is."     In my view your answer is akin
>>to replacing that with, "No, if the answer is in binary 2+2 is 100."
>>The answer is technically correct, but not appropriate.
>>
>>OFDM is often tricky even for experienced people to get their heads
>>around when they first start to learn it, so I think answers that
>>confuse rather than clarify are unhelpful.   I think your answer falls
>>in that category.
>>
>>It's not baseless.
>
>And I think that's totally whack.  I have not confused anyone.

How do you know?  Given that the full audience is unknown, I don't
think you have any way of knowing.  ;)


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com