A simplified two-path fading radio channel can be modelled as a first-order FIR filter, with delay T, and the strength of the direct signal being b0 and the strength of the indirect signal is b1. The channel can be equalised by the use of an IIR filter, which has the opposite frequency response. The frequency response is given by: G/(1+az^-1) where G = 1/b0 and a = b1/b0 However, due to the possibility of instability with an IIR filter, an equivalent FIR filter can be used. By taking the first 7 IIR impulse response coefficients, a 7th order FIR filter can be designed. Seeing as the FIR impulse response IS its filter coefficients, and the impulse response has to be the same for the FIR as the IIR to make it an accurate approximation (!) then you use the IIR impulse response as the FIR coefficients. (I hope I have made this clear!!) The pole zero plot of the FIR equaliser shows that there are 7 poles at the centre, and 7 zeros, each at magnitude of |a| where a = b1/b0. Why is this? My reasoning is this: This is because of the nature of stability and the Region of Convergence. For an IIR filter to be stable, the poles need to be on or inside this circle with radius |a|. If the poles are outside, then the impulse response does not converge, and the filter is unstable. As the IIR coefficients are being used in the FIR filter, so the rule stays the same. If you are confused by anything, which I'm sure you are, then please ask questions! I hope you can help me! Phil

# pole-zero plot with zeros |a|

Started by ●March 31, 2004

Reply by ●March 31, 20042004-03-31

"Philip Newman" <nojunkmail@ntlworld.com> writes:> [...] > The pole zero plot of the FIR equaliser shows that there are 7 poles at the > centre, and 7 zeros, each at magnitude of |a| where a = b1/b0. > > Why is this?It depends on what "this" is. Do you mean "Why are there 7 poles at the centre"? If so, then that's easy: An FIR is a sum of delayed versions of the signal, so its z-transform has the form h[z] = a0 + a1*z^{-1) + ... + aN*z^(-N). If you rearrange this in terms of positive powers of z instead of negative powers, you get a rational function with z^N in the denominator, just by simple algebra. -- % Randy Yates % "Remember the good old 1980's, when %% Fuquay-Varina, NC % things were so uncomplicated?" %%% 919-577-9882 % 'Ticket To The Moon' %%%% <yates@ieee.org> % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr

Reply by ●March 31, 20042004-03-31

"Randy Yates" <yates@ieee.org> wrote in message news:smfoljrc.fsf@ieee.org...> "Philip Newman" <nojunkmail@ntlworld.com> writes: > > [...] > > The pole zero plot of the FIR equaliser shows that there are 7 poles atthe> > centre, and 7 zeros, each at magnitude of |a| where a = b1/b0. > > > > Why is this? > > It depends on what "this" is. Do you mean "Why are there 7 poles at > the centre"? >Oops, sorry, no. I meant why are there 7 zeros each at magnitude |a|. I can understand why there are 7 poles and zeros, because it is a 7th order FIR filter, but I wish to know why they are all at |a|. Phil

Reply by ●March 31, 20042004-03-31

"Philip Newman" <nojunkmail@ntlworld.com> writes:> "Randy Yates" <yates@ieee.org> wrote in message > news:smfoljrc.fsf@ieee.org... >> "Philip Newman" <nojunkmail@ntlworld.com> writes: >> > [...] >> > The pole zero plot of the FIR equaliser shows that there are 7 poles at > the >> > centre, and 7 zeros, each at magnitude of |a| where a = b1/b0. >> > >> > Why is this? >> >> It depends on what "this" is. Do you mean "Why are there 7 poles at >> the centre"? >> > > > Oops, sorry, no. I meant why are there 7 zeros each at magnitude |a|. I > can understand why there are 7 poles and zeros, because it is a 7th order > FIR filter, but I wish to know why they are all at |a|.Who said they were? I almost just sent the sentence above alone, but that wouldn't be nice. You've discussed how an equalizer *could* be formed to equalize the channel, but you haven't said *how* it actually was formed. Is this the output of a simulation and you are examining the EQ output after it had converged? -- % Randy Yates % "Midnight, on the water... %% Fuquay-Varina, NC % I saw... the ocean's daughter." %%% 919-577-9882 % 'Can't Get It Out Of My Head' %%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra http://home.earthlink.net/~yatescr

Reply by ●March 31, 20042004-03-31

> > Who said they were?I did :-) I have simulated this, and I know they are, and I know they should be!> > I almost just sent the sentence above alone, but that wouldn't be nice.You've> discussed how an equalizer *could* be formed to equalize the channel, butyou> haven't said *how* it actually was formed. Is this the output of asimulation and you> are examining the EQ output after it had converged?I am just examining the equaliser output, not the equalised output. I originally used an IIR filter as an equaliser, but for reasons of instability under certain conditions, I decided to change to an FIR filter. I did this by truncating the IIR impulse response at 7T and then using those numbers printed off by MATLAB as the FIR filter coefficients, the FIR filter could be used to do a decent job. I plotted the magnitude response of the FIR and it is a good approximation of the IIR response, with only a 2dB ripple due to the finite number of poles. I then took the impulse response of the FIR equaliser, and observed the pole-zero plot, using zplane(hFIR) where hFIR is the impulse response of the FIR equaliser. The pole zero plot showed 7 poles at the centre, which is obvious seeing as there is no feedback, and it also showed 4 zeros, each at radius of magnitude |a| from the centre. why do they all have radius of mag |a|? the original channel had FIR coefficients of [b0 b1] and the IIR equaliser had coefficients [1 a] and [G] for feedback, where a = b1/bo and G = 1/bo I can send you my report if it will help.... cheers Phil

Reply by ●April 1, 20042004-04-01

>>>>> "Philip" == Philip Newman <nojunkmail@ntlworld.com> writes:Philip> A simplified two-path fading radio channel can be modelled as a first-order Philip> FIR filter, with delay T, and the strength of the direct signal being b0 and Philip> the strength of the indirect signal is b1. The channel can be equalised by Philip> the use of an IIR filter, which has the opposite frequency response. Philip> The frequency response is given by: Philip> G/(1+az^-1) Philip> where G = 1/b0 and a = b1/b0 [snip] Philip> The pole zero plot of the FIR equaliser shows that there are 7 poles at the Philip> centre, and 7 zeros, each at magnitude of |a| where a = b1/b0. Philip> Why is this? Let see if I can do this right. Let H(z) = (1+a*z^(-1)). This is basically your frequency response, but with the G scalar removed. This is unimportant when we're looking for the poles and zeroes.n Divide that out to get: H(z) = 1 - a/z + (a/z)^2 - (a/z)^3 + ... You truncated this to 7th order: H'(z) = 1 - a/z + (a/z)^2 - (a/z)^3 + ... - (a/z)^7. But this is just a geometric series, whose sum is H'(z) = (1-(a/z)^8)/(1-a/z) = (z^8-a^8)/z^7/(z-a) So we can see the 7 poles at zero, and also the 7 zeroes at a. Ray, who hopes he got this right.

Reply by ●April 5, 20042004-04-05

>>>>> "Raymond" == Raymond Toy <toy@rtp.ericsson.se> writes:Raymond> H'(z) = (1-(a/z)^8)/(1-a/z) Raymond> = (z^8-a^8)/z^7/(z-a) Raymond> So we can see the 7 poles at zero, and also the 7 zeroes at a. Raymond> Ray, who hopes he got this right. Oops, I didn't get it quite right. There are not 7 zeros at a. But there are 7 zeroes with magnitude |a|. From (z^8-a^8)/(z-a), the numerator has the 8 roots of a^8. The denominator removes the real root at a, leaving the 7 other roots. Ray

Reply by ●April 6, 20042004-04-06

"Raymond Toy" <toy@rtp.ericsson.se> wrote in message news:4ny8pfn3l2.fsf@edgedsp4.rtp.ericsson.se...> >>>>> "Philip" == Philip Newman <nojunkmail@ntlworld.com> writes: > > Philip> A simplified two-path fading radio channel can be modelled asa first-order> Philip> FIR filter, with delay T, and the strength of the directsignal being b0 and> Philip> the strength of the indirect signal is b1. The channel can beequalised by> Philip> the use of an IIR filter, which has the opposite frequencyresponse.> > Philip> The frequency response is given by: > > Philip> G/(1+az^-1) > > Philip> where G = 1/b0 and a = b1/b0 > > [snip] > > Philip> The pole zero plot of the FIR equaliser shows that there are 7poles at the> Philip> centre, and 7 zeros, each at magnitude of |a| where a = b1/b0. > > Philip> Why is this? > > Let see if I can do this right. > > Let H(z) = (1+a*z^(-1)). This is basically your frequency > response, but with the G scalar removed. This is unimportant when > we're looking for the poles and zeroes.n > > Divide that out to get: > > H(z) = 1 - a/z + (a/z)^2 - (a/z)^3 + ... > > You truncated this to 7th order: > > H'(z) = 1 - a/z + (a/z)^2 - (a/z)^3 + ... - (a/z)^7. > > But this is just a geometric series, whose sum is > > H'(z) = (1-(a/z)^8)/(1-a/z) > > = (z^8-a^8)/z^7/(z-a) > > So we can see the 7 poles at zero, and also the 7 zeroes at a. > > Ray, who hopes he got this right. >That looks great Ray, cheers. I am unsure how you arrived at the penultimate line though, and which bits divide through by which, could you go through with brackets so that it is more clear - sorry. I thought this issue might have something to do with stability, but it seems not! Cheers Phil

Reply by ●April 6, 20042004-04-06

>>>>> "Philip" == Philip Newman <nojunkmail@ntlworld.com> writes:>> >> H'(z) = (1-(a/z)^8)/(1-a/z) >> >> = (z^8-a^8)/z^7/(z-a) >> >> So we can see the 7 poles at zero, and also the 7 zeroes at a. >> >> Ray, who hopes he got this right. >> Philip> That looks great Ray, cheers. I am unsure how you arrived at the Philip> penultimate line though, and which bits divide through by which, could you Philip> go through with brackets so that it is more clear - sorry. Sure. (1-(a/z)^8) ---------- (1-(a/z) Multiply top and bottom by z^8, and we have z^8 - a^8 ----------- z^8 - a*z^7 So we have (z^8-a^8)/z^7/(z-a) = 1/z^7*(z^8-a^8)/(z-a). So the first term gives the 7 poles at 0. Now look at (z^8 - a^8)/(z-a). Clearly the numerator has 8 zeroes at the 8 roots of a^8 (which all have magnitude |a|), one of which is a. This is cancelled by the 1/(z-a) term, so we have 7 zeroes left, all with magnitude |a|. Does that make sense? Ray

Reply by ●April 6, 20042004-04-06