Why does hilbert tranformer give a phase-shifted but amplitude-reduced signal?

I have performed hilbert transform by using a FIR hilbert filter rather than the matlab built-in hilbert function. An example code is as follows:

fx = 10;
t = linspace(0,1,512);
x = 0.5*cos(2*pi*fx*t); %sampled signal
h = firpm(26,[0.1 0.8],[1 1],'h');   % hilbert filter
xh = conv(x,h,'same'); %hilbert transform of x
figure;
plot(t, [x;xh]); legend('Original','Transformed');

The resulting figure is given as follows:

How to address the problem to obtain the result equivalent to that by using hilbert function ?

Answer 1

[Honglei Chen]

The phase shift is expected so I assume you are talking about the amplitude? This is because the filter is not ideal. If you do a

fvtool(h)

You can see that the magnitude response is not unity when the frequency is close to DC or Nyquest. In your case, the signal frequency is 10 and the sampling rate is 512, so that is close to DC so you see the attenuation. if you change your sampling rate to, say, 128, then the result would be similar to what you get from hilbert.

HTH

[Yuzhen Lu]

Thanks for your answer! You are right on the sampling rate. But I don't see clearly how reducing the sampling rate leads to the improved result. If I fix the frequency and sampling rate (I want to use a high sampling rate to ensure a smooth signal), is there any way to modify the Hilbert filter to obtain a phase-shifted and amplitude-unchanged signal?

[Honglei Chen]

Hilbert transform is non-causal, so it is not realizable. Therefore, some distortion is introduced in the filter. In this case, I think there is a good tradeoff between the transition band and the passband ripple. Unfortunately your sample rate makes the interested frequency fall in the transition band. But note that the mismatch is really on magnitude, not the phase and phase is actually what you are interested. The mismatch on magnitude is easier to fix compared to the phase. For example, in your example, if you know this is a single frequency, you can also do something like

xh = xh/rms(xh)*rms(x)

to preserve the power once you get xh as the filter output.

HTH

[Yuzhen Lu]

Your suggestion is a simple yet effective way for obtaining the signal amplitude! But can you expand further on how the sampling rate (e.g., N=512) makes the frequency (e.g., fx=10) fall in the transition band. In designing the Hilbert filter by using firpm, I set the passband region of [0.1 0.8] that is amenable to futher adjustments:

h = firpm(26,[0.1 0.8],[1 1],'h');   % hilbert filter

I am not clear of how the sampling rate get involved here. Thank you very much!

[Honglei Chen]

So in your design, you specify that the resulting filter should have a unity response between normalized frequency [0.1 0.8]. Now in your script, you have

t = linspace(0,1,512)

This is roughly 512 points within one sec, which means that your sampling rate is 512, and your Nyquest rate is 256. So combining the two together, you are asking the filter to have a unity response between

[0.1 0.8]*256

which is approximately 25 Hz through 200 Hz. In your signal definition, you have

fx=10;

so your signal is 10 Hz, and it is not in the band where the response is unity.

HTH

[Yuzhen Lu]

Thanks!

Based on such analysis, if the passband is set such that the frequency of interest is covered, an amplitude-preserved signal should be recovered. But in practice it does not, as you see below. The FIR hilbert filter seems incomparable to the hiltert function based on Fourier transform anyway, and both cannot shed the Gibbs artifacts.

fx = 10;
t = linspace(0,1,512);
x = 0.5*cos(2*pi*fx*t); %sampled signal
h = firpm(26,[0.01 0.99],[1 1],'h'); % hilbert filter with a modifed passband
xh = conv(x,h,'same'); %hilbert transform of x
xh2 = (rms(x)/rms(xh))*xh; %amplitude corrrection based on RMS
figure;
plot(t,xh,'b-',t, xh2,'r-',t,imag(hilbert(x)),'c'); 
legend('Hilbert filter','Amplitude Corrected','Hilbert transform','Location','best');
figure;
plot(t,sqrt(xh.^2+x.^2),'b-',t,sqrt(xh2.^2+x.^2),'r-',t,abs(hilbert(x)),'c'); 
legend('Hilbert filter','Amplitude Corrected','Hilbert transform','Location','best');

[Honglei Chen]

This is true because you just cannot get a causal FIR filter that does a perfect unity response across the entire band, the same reason as a boxcar filter is not realizable. That's why you need to make those tradeoffs between different characteristics. HTH

[Yuzhen Lu]

My ultimate goal is to recover the signal amplitude without causing noticeable(Gibbs) boundary oscillations. It seems that both FIR filter and Hilbert transform cannot fulfill the goal.

Do you have any improvement suggestions?

[Honglei Chen]

I don't have good suggestion on that. I'm not aware of a way to avoid Gibbs phenomenon.

[Yuzhen Lu]

Thank you all the same!!

[叶]

Do you have any idea to solve this problem?I mean if signal amplitude decreased,how can I get the normal ampliude?what should I do ?