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I thought \$0\$ or a constant phase shift was necessary to avoid distortion of a square input.
How does linear phase shift help here? I'm familiar with some fourier series but I don't get my textbook explanation. Any help?

EDIT(My attempt):
Say the input signal is \$f(t) = \sin(t) + \sin(2t)\$.
Then the output will be \$g(t)=\sin(t+\phi) + \sin(2t + 2\phi)=f(t+\phi)\$
That's it?Wow! But this means a constant phase shift actually distorts a nonsinusoidal input? Doesn't this seem counterintuitive?

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How does linear phase shift help here? I'm familiar with some fourier series but I don't get my textbook explanation. Any help?

A linear phase shift (as produced by a Bessel filter) means a constant time delay for all the main frequency components of a signal. So, for instance, if the filter were a 2nd order low-pass type having a 3.01 dB cut-off point at (say) 12.64 kHz: -

  • 159.2 Hz would see a phase shift of about 0.99 degrees
  • 1.592 kHz would see a phase shift of about 9.86 degrees
  • 7.977 kHz would see a phase shift of about 49.02 degrees

Then, if you factored in the time shift you would get: -

  • 159.2 Hz has a time shift of 17.27 us
  • 1.592 kHz has a time shift of 17.20 us
  • 7.977 kHz has a time shift of 17.07 us

In other words, a 2nd order Bessel filter is good for constant delay up to around 50% of the cut-off frequency AND, that will largely maintain good shape with maybe about 0.5% overshoot in the time domain from a step change.

You can use this on-line tool to check out the numbers. Manuipulate the R value to give a \$\zeta\$ of 0.86. I used R = 172 ohms knowing that a 2nd order Bessel filter has a damping ratio of about 0.86.

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  • \$\begingroup\$ The period of \$\sin(t)\$ is \$k\$ times less than the period of \$\sin(kt)\$. So a phase shift of \$\phi\$ radians in \$\sin(t)\$ will have same "time delay" as the phase shift of \$k\phi\$ radians in \$\sin(kt)\$. Beautiful! Thanks again XD \$\endgroup\$ – beccaboo Mar 25 at 14:33
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    \$\begingroup\$ @beccabecca a lot of text books talk about group delay but, for simple low pass filters (a good majority of applications) simple phase delay is the key. Phase delay is the time delay converted to an equivalent phase angle at the natural resonant frequency of the LPF. \$\endgroup\$ – Andy aka Mar 25 at 14:35
  • \$\begingroup\$ But ....the definition of the Thomson-Bessel approximation requires a "maximally flat group delay" characteristic. \$\endgroup\$ – LvW Mar 25 at 18:05
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When we require that a frequency-dependent block (filter) should only delay any signal by the time T - without changing the form of the signal the output-to-input relation is:

v2(t)=v1(t-T) .

Applying the LAPLACE transformation to this equation the ratio (transfer function) is:

V2(s)/V1(s)=H(s)=exp(-sT).

That means: (1) Magnitude |H(w)|=1 and (2) phase function phi(w)=-wT.

Hence, we require a phase function that is a LINEAR function of the frequency.

Definition: The group delay D is defined as D=-d(phi)/d(w). We see that for a linear phase function the group delay is D=constant.

However, it is to be mentioned that a real lowpass can never realize a transfer function H(s)=exp(-sT). Hence, we need an approximation "as good as possible" within a limited frequency band. This requirement leads to the well-known Thomson-Bessel approximation.

In most cases, the cut-off frequency for a Bessel filter is defined in the time domain (and NOT in the frequency domain, as it is normal for all other filter types). The cut-off (end of the passband) is defined at a frequency where the deviation of the (nearly constant) group delay D has a certain percentage of the value for low frequencies - depending on the particular application.

Remember: The BUTTERWORTH approximation has a "maximally flat magnitude" within the passband - the BESSEL approximation is derived from a "maximally flat group delay" requirement.

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Consider a cable. That has constant time delay. This is the same as linear phase with frequency. Does a cable distort your signals?

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  • \$\begingroup\$ No a cable doesn't distort digital signals assuming it has constant time delay for all the frequencies. Earlier I was confusing phase shift with time delay. I get it now! For a beginner like me it's just amazing how filters are able to discriminate frequencies XD Thank you so much:) \$\endgroup\$ – beccaboo Mar 25 at 14:42

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