Re: how do you compute the channel gain from path loss?

On Sun, 3 Aug 2008 21:31:33 -0700 (PDT), Ginu <osheikh81@xxxxxxxxx>
wrote:

On Aug 3, 8:48 pm, Eric Jacobsen <eric.jacob...@xxxxxxxx> wrote:
On Sun, 3 Aug 2008 14:52:15 -0700 (PDT), Ginu <osheik...@xxxxxxxxx>
wrote:





On Aug 3, 5:36 pm, Eric Jacobsen <eric.jacob...@xxxxxxxx> wrote:
On Sun, 3 Aug 2008 10:02:28 -0700 (PDT), Ginu <osheik...@xxxxxxxxx>
wrote:

On Aug 2, 7:00 pm, Eric Jacobsen <eric.jacob...@xxxxxxxx> wrote:
On Sat, 2 Aug 2008 14:04:14 -0700 (PDT), Ginu <osheik...@xxxxxxxxx>
wrote:

On Aug 2, 2:26 pm, Ginu <osheik...@xxxxxxxxx> wrote:
On Aug 1, 8:17 pm, Ginu <osheik...@xxxxxxxxx> wrote:

Hello, I had a really simple question. If I've calculated the path
loss between two nodes as Pr = Pt*Gt*Gr*(lambda/(4*pi*d)^2, what is
the channel gain H^2 in the Shannon Capacity formula R = w*log2(1 +
(Pt*H^2)/(w*N))?

Does H^2 = Gt*Gr*(lambda/(4*pi*d)^2?

anyone?

Once I calculate the path loss G = Gt*Gr*(lambda/(4*pi*d))^2, does G =
H? If so, do I have to then square G to get the corresponding power to
use in the Shannon Capacity?

Or do you compute channel gain completely separate from the path loss?
What is the relationship?

One problem might be your terminology.   I don't know what you mean by
"channel gain".    If you mean the channel response, that's often
described in the form of either an impulse response (often denoted
'h') or the channel frequency response (the FT of h).   The average
power in that impulse response is typically estimated with a pathloss
model, but that's usually separate from describing h.   i.e., h is
scaled by the estimated pathloss.

"Channel gain" isn't a term that's familiar to me as anything in
common use.

Eric Jacobsen
Minister of Algorithms
Abineau Communicationshttp://www.ericjacobsen.org

Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php-Hidequoted text -

- Show quoted text -

Thanks for your response. I am confused but I'm reading directly from
a textbook - Fundamentals of Wireless Communication by Tse and
Viswanath. "h" is the "fixed complex channel gain from the transmit
antenna to the receive antenna" where the received signal is y = h*x +
m, x is the transmitted signal and m is the additive white Gaussian
noise. The capacity of the channel is C = w*log2 (1 + P*|h|^2/(w*N0 +
I)), where w is the bandwidth, P is the average transmit power, N0 =
kT is the variance of the additive white Gaussian noise, and I is the
interference.

I assume that |h|^2 is the corresponding power?

This is where my confusion lies. "h" is described as the (complex)
channel gain, but SHOULD have some relation to the path loss between a
transmitter and receiver because clearly the transmitted signal goes
through some degradation along the way. You suggest that the average
power in the impulse response is typically estimated with a pathloss
model. Are we talking about "h"? How is the average power in the
impulse response estimated with the path loss? You also suggest that
the path loss model is usually separate from describing h?

I'm trying to figure out what to use in the Shannon Capacity C =
w*log2 (1 + P*|h|^2/(w*N0 + I)) when I know the path loss G =
Gt*Gr*(lambda/(4*pi*d))^2 between the two nodes. How do I get from one
to the other? If I have to scale the h by the estimated path loss, how
do I calculate h?

Thanks. As you can see I'm extremely confused lol

Pathloss and channel response are usually treated separately because
if the signal power is within capture range of a properly designed
receiver the Automatic Gain Control will normalize the received power.
In other words, after synchronization and AGC, the demodulator only
has to deal with the resulting SNR and channel response (independent
of the actual pathloss).   The signal magnitude will have been
normalized by the AGC so that the slicer can properly demodulate the
constellation(s).

I have a copy of Tse and Viswanath but I don't know where you're
reading the part you've quoted.   If you could give specifics (i.e.,
chapter, page) I might be able to help sort out the context of what
they're saying.

Eric Jacobsen
Minister of Algorithms
Abineau Communicationshttp://www.ericjacobsen.org

Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php-Hide quoted text -

- Show quoted text -

I'm looking at section 5.3.1 Single input multiple output (SIMO)
channel, page 179 of the 2005 edition.

I'm trying to simulate a multiple technology network of wireless node
transmissions in the 2-5 GHz band. I'm using SIMO because I'm using
omni-directional antennas and thus while the transmitter sends one
transmission, multiple nodes receive it. Since the operating frequency
bands overlap in the 2-5 GHz range, each node in the network will
receive a transmission which will cause interference on that channel
based on distance, frequency, transmit power, etc. (I assume no co-
channel interference, not that it matters for my question).

So my issue is how to model the channel gain (the "h") between a
transmitter and receiver to calculate the Shannon Capacity. A second
part of the question then is, does the path loss formula only give me
the received power? A third part of the question is, isn't the
numerator of the fraction inside the logarithm of the Shannon capacity
supposed to represent the received signal power? That is why I assumed
that |h|^2 and path loss G are identical.

If not, how would I go about modeling it?

Looking at that section in Tse and Viswanath it looks like your
initial intuition is correct:  ||h||^2 is just the fixed pathloss so
that the received SNR = (Pt ||h||^2)/No for each h at each receiver.

IMHO, it's a bit confusing because h is usually the channel impulse
response.   If the assumption is that the fading is flat, then h
becomes a scalar or a single complex value (depending on how one wants
to account for things), which is essentially the pathloss or link
loss.

Also, be careful about accounting for "pathloss", as that term may
mean different things to different people as well.   Whether or not to
include the antenna gains may make a difference, and the appropriate
exponent may change depending on the environmental assumptions (e.g.,
the exponent may be 2 for free space, it may be less than two for an
indoor system, it may be much more than 2 for an NLOS outdoor system.

Eric Jacobsen
Minister of Algorithms
Abineau Communicationshttp://www.ericjacobsen.org

Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php- Hide quoted text -

- Show quoted text -

Thanks very much for your clarification. From your initial post, you
seemed pretty certain that the path loss and "channel gain" were
separate. Is your intuition telling you that my initial thinking was
correct, or are you certain after reading the text? i'm sure that this
channel gain/path loss relationship has been used thousands of times,
so it should be described somewhere. Have you seen any technical
material that explains this thoroughly?

For a flat-faded channel there is no delay spread and no frequency
selectivity. This means that a single coefficient (sometimes scalar,
sometimes complex, if the phase matters), is used for channel
attenuation. Since the described channel is static, i.e., the
coefficient is fixed, the only attenuation present is the pathloss. So
in this particular case h is strictly the channel attenuation (or
gain).

In the presence of things like multipath fading or other channel
memory effects h includes the channel impulse response which can then
be, and often is, treated separately from the channel fading. Usually
the approach is that the AGC takes care of the pathloss and the
Equalizer takes care of the channel memory effects. Since those are
generally developed independently the effects can be analyzed
independently.

So, yes, it can be confusing when authors lump them together, but it
isn't really wrong to do so.


Also, do you think that ||h||^2 is the fixed path loss, or the square
of the fixed path loss? since the path loss is dimensionless (as it is
Pr/Pt), would I have to square the path loss to get the related
channel "power"?

I haven't worked through the math, but that didn't look obvious to me,
either. I suspect it may be due to converting the channel
coefficient to a power attenuation, but I'm not sure.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
.

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