hello, what is the difference between channel gain and path loss? is both are same? 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?
what is difference between pathloss and channel gain?
Started by ●October 10, 2012
Reply by ●October 10, 20122012-10-10
On Wed, 10 Oct 2012 06:16:16 -0500, "shaishavgandhi" <62238@dsprelated> wrote:>hello, >what is the difference between channel gain and path loss? >is both are same? > >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?They're generally not the same, but be careful to make sure that you understand the definitions as they are used in your particular application, as they can vary. Usually path loss is a component of channel gain, with channel gain also including other effects such as multipath fading, shadowing, and sometimes interference and other impairments. How this is treated in detail may depend on a lot of things that are application dependent, so the definition of channel gain may be different from application to application. Eric Jacobsen Anchor Hill Communications www.anchorhill.com
Reply by ●October 10, 20122012-10-10
Let me answer with a question: what is the physical meaning of Pt*H^2? what is the physical meaning of Pr? That's the best I've got to offer, being unwilling to give straight answers to homework-style problems :-)
Reply by ●October 10, 20122012-10-10
On Wed, 10 Oct 2012 15:50:00 +0000, Eric Jacobsen wrote:> On Wed, 10 Oct 2012 06:16:16 -0500, "shaishavgandhi" <62238@dsprelated> > wrote: > >>hello, >>what is the difference between channel gain and path loss? is both are >>same? >> >>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? > > They're generally not the same, but be careful to make sure that you > understand the definitions as they are used in your particular > application, as they can vary. > > Usually path loss is a component of channel gain, with channel gain also > including other effects such as multipath fading, shadowing, and > sometimes interference and other impairments. > > How this is treated in detail may depend on a lot of things that are > application dependent, so the definition of channel gain may be > different from application to application.Isn't pathloss mostly about the path in the eather, while the channel gain includes such things as corroded connections on your coax and dicky amplifiers? Or am I confused? -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
Reply by ●October 10, 20122012-10-10
I think the complete question is something like "a transmitter with a transmit power Pt is located at a distance d from a receiver. Transmitter and receiver have antenna gain Gt and Gr, respectively. Given a bandwidth W and spectral noise power density N at the receiver, calculate the maximum throughput". I once documented a little measurement [1] to illustrate the path loss equation and the exponent of 2 (because power conservation implies that all the radiated power must pass through the surface of a sphere, and its surface is the radius (antenna distance) squared. If the sphere isn't empty, the power will decay faster (exponent > 2). If both antennas are inside a metal tunnel, the exponent can go down to 0, as distance doesn't matter at all if the power can't escape the waveguide. Those things seem much more easy to grasp with a little hands-on work. Sadly, radio engineering seems very theoretical and abstract nowadays. [1] http://www.dsprelated.com/showarticle/32.php PS: Kids don't try this at home, not if the national communications authority is located on the other side of the road. Which happened to be the case, but they never found out who was jamming :o)
Reply by ●October 10, 20122012-10-10
On Wed, 10 Oct 2012 15:51:44 -0500, mnentwig wrote: <snip>> If both > antennas are inside a metal tunnel, the exponent can go down to 0, as > distance doesn't matter at all if the power can't escape the waveguide.Hmm. I would expect the exponent to go down to no less than 1, but perhaps with coefficient adjusted for very small loss. Unless it's a superconducting tunnel, that's perfect, and has no material inside of it to absorb energy, that is. Yes, I know: picky, picky. -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
Reply by ●October 10, 20122012-10-10
On Wed, 10 Oct 2012 18:00:46 -0500, Tim Wescott wrote:> On Wed, 10 Oct 2012 15:51:44 -0500, mnentwig wrote: > > <snip> > >> If both >> antennas are inside a metal tunnel, the exponent can go down to 0, as >> distance doesn't matter at all if the power can't escape the waveguide. > > Hmm. I would expect the exponent to go down to no less than 1, but > perhaps with coefficient adjusted for very small loss. Unless it's a > superconducting tunnel, that's perfect, and has no material inside of it > to absorb energy, that is. > > Yes, I know: picky, picky.Come to think of it, it won't be 1/x, will it -- it'll be e^(something negative * distance). D'oh. -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
Reply by ●October 10, 20122012-10-10
On Wed, 10 Oct 2012 19:42:30 -0500, Tim Wescott <tim@seemywebsite.com> wrote:>On Wed, 10 Oct 2012 18:00:46 -0500, Tim Wescott wrote: > >> On Wed, 10 Oct 2012 15:51:44 -0500, mnentwig wrote: >> >> <snip> >> >>> If both >>> antennas are inside a metal tunnel, the exponent can go down to 0, as >>> distance doesn't matter at all if the power can't escape the waveguide. >> >> Hmm. I would expect the exponent to go down to no less than 1, but >> perhaps with coefficient adjusted for very small loss. Unless it's a >> superconducting tunnel, that's perfect, and has no material inside of it >> to absorb energy, that is. >> >> Yes, I know: picky, picky. > >Come to think of it, it won't be 1/x, will it -- it'll be e^(something >negative * distance). > >D'oh. >Generaly pathloss is proportional to d^n where d is the range (distance) and, as Markus mentioned, n = 2 for free space, and n>2 for environments with clutter/reflectors/barriers/etc., and <2 for environments that function like waveguides, or for LOS with rich multipath. Exponents of <2 are not unusual in practice for LOS with multipath. This effect is often referred to as "multipath gain". Eric Jacobsen Anchor Hill Communications www.anchorhill.com
Reply by ●October 11, 20122012-10-11
>> Hmm. I would expect the exponent to go down to no less than 1, butperhaps with coefficient adjusted for very small loss. Unless it's a superconducting tunnel, that's perfect, and has no material inside of it to absorb energy, that is. ..>> Unless it's a superconducting tunnel, that's perfectThat's exactly what I meant. You can buy them in a shop as "waveguides" (or fiber optic cable, here the "tunnel" is filled with glass). Not superconducting, but extremely low-loss nonetheless, something like 0.2 dB/km (kilometer!) for micrometer wavelengths. There may be even material inside the "tunnel", but the trick is that the wave travels in a non-metallic medium. The reason I brought this up is simply to establish a range for the path loss exponent (0 for lossless guided wave, 2 for free space, higher when it's raining cats and dogs). You'll rarely find free space propagation and waveguides in the same book chapter, and what I'm trying to show is that they have a lot in common. With some intuitive understanding, one can look at an equation and see immediately, whether or not the units match, for example.
Reply by ●October 11, 20122012-10-11
On 11.10.12 9:42 , mnentwig wrote:>>> Hmm. I would expect the exponent to go down to no less than 1, but > perhaps with coefficient adjusted for very small loss. Unless it's a > superconducting tunnel, that's perfect, and has no material inside of it > to absorb energy, that is. > .. >>> Unless it's a superconducting tunnel, that's perfect > > That's exactly what I meant. > You can buy them in a shop as "waveguides" (or fiber optic cable, here the > "tunnel" is filled with glass). > Not superconducting, but extremely low-loss nonetheless, something like 0.2 > dB/km (kilometer!) for micrometer wavelengths. There may be even material > inside the "tunnel", but the trick is that the wave travels in a > non-metallic medium. > > The reason I brought this up is simply to establish a range for the path > loss exponent (0 for lossless guided wave, 2 for free space, higher when > it's raining cats and dogs). > > You'll rarely find free space propagation and waveguides in the same book > chapter, and what I'm trying to show is that they have a lot in common. > With some intuitive understanding, one can look at an equation and see > immediately, whether or not the units match, for example.There is a very principal difference between waveguides and free space: In free space the power gets distributed to a surface proportional to the square of distance, but in waveguides the cross-section stays the same. -- Tauno Voipio
