Matlab function freqz gives wrong amplitude in dB

Hello guys,

i want to compute und plot the filter amplitude response and want to compare it with the filterDesigner function from Matlab.

For that i use this Code:

fs = 1200; Ts = 1/fs;

filterCoff = Filter1, %% FIR lOWPASS Filter

%% Plot of the amplitude response in db

figure(1);

plot(w/(2*pi*Ts),mag2db(abs(h)));

xlabel('Normalized Frequency (Hz)')

ylabel('Magnitude');

I tried everythink, but i dont get the same solution as the filterDesigner.

[hans petersen]

My Result:

Result of the FilterDesigner:

Answer 1

[Star Strider]

I do not see any call to freqz, or the designed filter, so I cannot test this.

What you are seeing may be a frequency-axis resolution problem. The freqz function has the option of choosing the length of the Fourier transform used to plot the Bode plot as the second (or third argument, depending on how the filter is defined), and I usually choose 2^16 for this. A higher value results in a more accurate plot with increased frequency resoution.

If the highest frequency in the passband is not the first frequency, the freqz function does not always scale the amplitude axis so that the highest amplitude is always 0 dB, so it may seem as though the filter is amplifying even though it is not. The relative amplitudes are nevertheless correct.

For freqz, also using:

set(subplot(2,1,1), 'YLim',[-130 0])

(in this instance) may make the plots more comparable by improving the amplitude resolution.

.

[Star Strider]

I do not see the problem —

Filter1 = [0.000178232054708589,
	 0.000709878618964491,
	 0.001585843072655027,
	 0.002791137421116118,
	 0.004305138743876396,
	 0.006101942058544247,
	 0.008150803562238728,
	 0.010416666666580653,
	 0.012860761825639317,
	 0.015441269893612304,
	 0.018114037662004680,
	 0.020833333333267912,
	 0.023552629004535204,
	 0.026225396772939561,
	 0.028805904840931838,
	 0.031250000000016098,
	 0.033515863104388588,
	 0.035564724608116896,
	 0.037361527922819888,
	 0.038875529245614424,
	 0.040080823594106516,
	 0.040956788047822303,
	 0.041488434612095185,
	 0.041666666666809973,
	 0.041488434612095185,
	 0.040956788047822303,
	 0.040080823594106516,
	 0.038875529245614424,
	 0.037361527922819888,
	 0.035564724608116896,
	 0.033515863104388588,
	 0.031250000000016098,
	 0.028805904840931838,
	 0.026225396772939561,
	 0.023552629004535204,
	 0.020833333333267912,
	 0.018114037662004680,
	 0.015441269893612304,
	 0.012860761825639317,
	 0.010416666666580653,
	 0.008150803562238728,
	 0.006101942058544247,
	 0.004305138743876396,
	 0.002791137421116118,
	 0.001585843072655027,
	 0.000709878618964491,
	 0.000178232054708589];


fs = 1200; Ts = 1/fs;
filterCoeff = Filter1;
[h,w] = freqz(filterCoeff,1, 2^16, fs);
figure(1);
plot(w,mag2db(abs(h)));
xlabel('Normalized Frequency (Hz)');
ylabel('Magnitude (dB)');
grid
ylim([-130 0])

[pks, locs] = findpeaks(mag2db(abs(h)));
Results = table(w(locs),pks, 'VariableNames',{'Freq (Hz)','Amplitude (dB)'})

Results = 22×2 table
    Freq (Hz)    Amplitude (dB)
    _________    ______________

     59.052         -31.468    
     85.199          -41.48    
     110.75         -48.489    
     136.08         -53.944    
     161.31         -58.432    
     186.47         -62.254    
     211.59         -65.591    
     236.68         -68.561    
     261.76         -71.248    
     286.81         -73.711    
     311.86         -76.001    
     336.89         -78.157    
     361.91         -80.216    
     386.92         -82.213    
     411.92         -84.188    
     436.91         -86.186    

Consider doing this same sort of analysis with the filterDesigner results. I suspect they will be the same.

.

[hans petersen]

Thank you for your help. But if you compare the graphs you see the differences.

For example at 100 Hz you see that the line in the filterDesigner ends between -80 and -100 db.

But in your plot the graph ends between -100 and - 120 db.

Why is there such a gap?

[Star Strider]

That is likely just the way the plots are drawn, and that has to do with the frequency resolution of the magnitude plot, and that is determined by the length of the Fourier transform. Higher frequency resolution will yield a more precise plot. You can experiment with that with respect to freqz as:

[h1,w1] = freqz(filterCoeff,1, 2^8, fs);

[h2,w2] = freqz(filterCoeff,1, 2^12, fs);

[h3,w3] = freqz(filterCoeff,1, 2^16, fs);

to see how the plots differ.

.

[hans petersen]

Okey with [h,w] = freqz(filterCoeff,1, 2^13, fs); i get a similar solution as the filterDesigner. But wich solution is right? The one with the highest resolution?

[Star Strider]

They are both correct. The only difference is the frequency resolution in the plots, and that has nothing to do with the actual filter performance.

The actual performance of the filter is important only in the context of the result of the signal filtered with it.

[hans petersen]

I am trying to understand this. But if the frequency resolution increases, shouldn't the absolute value of the gain be almost identical?

[Star Strider]

The values at any specific frequency will always be the same (within floating-point precision limits). The only difference is in the display and the specific frequencies that are presented in it. That is determined by the length of the fft used to calculate it.

The plot is simply for information to determine that the filter design matches the requirements, and does not at all affect the underlying filter values or how the filter will operate with respect to a specific signal.

[hans petersen]

First of all thank you for your help. I think i understood. I plot the same filter with 2^12 and 2^16 and the plots are the same. The only Difference is in the length of the vertikal lines.

Is this explanation correct?

The reason Why the vertical lines dont have the same length is just a mathematical issue. Somethink like rounding and decimal point operations. For The performance of the digital filter we have to look on the high points of the individual parabolas.

[Star Strider]

My pleasure.

‘Is this explanation correct?’

Essentially, yes. The difference is in the fft frequency resolution and therefore the plot resolution. The underlying data is the same.

‘For The performance of the digital filter we have to look on the high points of the individual parabolas.’

The entire filter is important with respect to the actual filtered signal. The filter is not linear in the stopband, however the variations do not matter because the stopband frequencies are significantly attenuated.

.

Answer 2

[hans petersen]

Hi,

sorry. This is the whole Code:

fs = 1200; Ts = 1/fs;

filterCoeff = Filter1

[h,w] = freqz(filterCoeff,1);

figure(1);

plot(w/(2*pi*Ts),mag2db(abs(h)));

xlabel('Normalized Frequency (Hz)');

ylabel('Magnitude (dB)');

% The FIR Filter Coefficient:

Filter1 = [

0.000178232054708589,

0.000709878618964491,

0.001585843072655027,

0.002791137421116118,

0.004305138743876396,

0.006101942058544247,

0.008150803562238728,

0.010416666666580653,

0.012860761825639317,

0.015441269893612304,

0.018114037662004680,

0.020833333333267912,

0.023552629004535204,

0.026225396772939561,

0.028805904840931838,

0.031250000000016098,

0.033515863104388588,

0.035564724608116896,

0.037361527922819888,

0.038875529245614424,

0.040080823594106516,

0.040956788047822303,

0.041488434612095185,

0.041666666666809973,

0.041488434612095185,

0.040956788047822303,

0.040080823594106516,

0.038875529245614424,

0.037361527922819888,

0.035564724608116896,

0.033515863104388588,

0.031250000000016098,

0.028805904840931838,

0.026225396772939561,

0.023552629004535204,

0.020833333333267912,

0.018114037662004680,

0.015441269893612304,

0.012860761825639317,

0.010416666666580653,

0.008150803562238728,

0.006101942058544247,

0.004305138743876396,

0.002791137421116118,

0.001585843072655027,

0.000709878618964491,

0.000178232054708589

];

With this Code my frequencies are near to the frequencies of the filterDesigner Tool, but my Amplitude is absolutely different.

Thank you for your help.