Pre-emphasis filter design

Question

I am using the standard MATLAB filter function to design pre-emphasis filter. However, I am getting the wrong results back! I have the ground truth so I can check it.

For example I wrote this one Matlab code:

 B = ones(1,6);
 B=B*(-0.97);
 B(1)=1;
 z = 1; 
 xd = filter(B,z,data,[], 2);

However, I only getting back correctly only the first two columns.

Ground truth:

-604.981536865234   -98.1470046997071   -105.546778869629   -114.096517944336   -113.826526184082   -112.466567687988
-615.981201171875   -17.4794665527344   -22.4493148803712   -27.5691586303711   -19.8393945312500   -21.8693325805664
-448.986297607422   4.52986175537109    6.06981475830077    1.63994995117184    -0.939971313476576  -7.60976776123050
-10.9996643066406   -18.3294406127930   -18.8694241333008   -16.4094992065430   -16.8594854736328   -15.3095327758789
207.993652343750    -3.75988525390625   5.93981872558595    3.93987976074220    2.87991210937500    2.78991485595705
105.996765136719    11.1796588134766    6.41980407714844    6.50980133056640    10.5996765136719    3.80988372802734

What I recover:

-604.981536865234   -98.1470046997071   481.285311889649    1137.16529541016    1884.31249359131    2720.81696472168
-615.981201171875   -17.4794665527344   575.052450256348    1166.46440124512    1774.60584167480    2381.71731353760
-448.986297607422   4.52986175537109    441.586523437500    855.213900146484    1252.26178283691    1631.64020477295
-10.9996643066406   -18.3294406127930   -8.19974975585937   22.3893167114258    67.5279391479492    129.216056518555
207.993652343750    -3.75988525390625   -195.814024047852   -389.868101806641   -582.982208251953   -773.186403503418
105.996765136719    11.1796588134766    -96.3970581054687   -206.883686218262   -316.280347595215   -439.466588134766

My coeff is supposed to be 0.97, I don't get the partial correct results to be honest.

cheers for any input!

Answer 1

I don't really understand what are you trying to achieve with vector B that has 6 elements. Pre-emphasis filter is defined as:

$$y[n]=x[n]-\alpha x[n-1]$$

Where $\alpha$ is $0.97$ in your case.

You can see that there are only two elements in vector B, right? So the code should be something like that:

B=[1, -0.97];
xd = filter(B, 1, data, [], 2);

You can verify the frequency response:

freqz(B, 1)

Answer 2

Only two columns of your output match the expected results, because the first $N$ columns only depend on the first $N$ filter coefficients and only the first two coefficients are consistent with the desired result.

Working back from your original data (which I recreated back from the provided "recovered" result):

-604.9815368652340   -684.9790954589840   -769.9765014648426   -860.9737243652289   -948.9710388183510  -1032.9684753417814
-615.9812011718750   -614.9812316894531   -618.9811096191402   -627.9808349609343   -628.9808044433603   -631.9807128906195
-448.9862976074220   -430.9868469238282   -411.9874267578127   -397.9878540039070   -386.9881896972709   -382.9883117675836
 -10.9996643066406    -28.9991149902344    -46.9985656738281    -61.9981079101562    -76.9976501464843    -89.9972534179683
 207.9936523437500    197.9939575195312    197.9939575195308    195.9940185546867    192.9941101074209    189.9942016601541
 105.9967651367190    113.9965209960940    116.9964294433599    119.9963378906258    126.9961242675798    126.9961242675811

You can compute what the coefficients should be to get your expected results with:

row = 1; % choose any row, results should be the same within numerical errors
B_expected = zeros(size(B));
for col=1:size(data,2)
  sum = 0;
  for idx=2:col
    if col <= length(B)
      sum = sum + B_expected(col-idx+1)*data(row,idx);
    end
  end
  B_expected(col) = (Ground_Truth(row,col)-sum)./data(row,1);
end
B_expected

Which gives you:

1.0000e+000  -9.7000e-001  1.3859e-015  5.7785e-015  -3.2181e-015  6.0134e-015

Asides from numerical errors, this essentially corresponds to using only two non-zero coefficients of [1 -0.97] and is consistent with the pre-emphasis filter definition as indicated in @jojek's answer:

B=[1, -0.97];
xd = filter(B, 1, data, [], 2);