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If symmetry conditions are met, FIR filters have a linear phase. This is not true for IIR filters.

However, for what applications is it bad to apply filters that do not have this property and what would be the negative effect?

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6 Answers 6

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A linear phase filter will preserve the waveshape of the signal or component of the input signal (to the extent that's possible, given that some frequencies will be changed in amplitude by the action of the filter).

This could be important in several domains:

  1. coherent signal processing and demodulation, where the waveshape is important because a thresholding decision must be made on the waveshape (possibly in quadrature space, and with many thresholds, e.g. 128 QAM modulation), in order to decide whether a received signal represented a "1" or "0". Therefore, preserving or recovering the originally transmitted waveshape is of utmost importance, else wrong thresholding decisions will be made, which would represent a bit error in the communications system.

  2. radar signal processing, where the waveshape of a returned radar signal might contain important information about the target's properties

  3. audio processing, where some believe (although many dispute the importance) that "time aligning" the different components of a complex waveshape is important for reproducing or maintaining subtle qualities of the listening experience (like the "stereo image", and the like)

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    $\begingroup$ (I've done ABX listening tests and was able to distinguish between simulated 8th-order Linkwitz-Riley crossover vs without. Impulsive sounds become "chirpy" as the high frequencies arrive a little sooner than the low. So #3 isn't totally far-fetched.) $\endgroup$
    – endolith
    Commented Apr 28, 2017 at 19:31
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    $\begingroup$ Needless to say the waveform preservation property is only applicable for narrowband signals... Othewise (for general wideband signals) the filter (whether linear phase or not) will be changing the signal shape as much as the impulse response convolves with the signal... $\endgroup$
    – Fat32
    Commented Nov 10, 2018 at 15:44
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    $\begingroup$ @Fat32 but we could have a constant magnitude with non linear phase that would completely disrupt the shape (as in my example below)- so that can’t be true? $\endgroup$ Commented Mar 18, 2020 at 18:20
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    $\begingroup$ @DanBoschen nonlinear phase will be distorting in any case. But linear phase will be preserving only if |H(w)| is (approximately) constant or sigal is (sufficiently) narrowband. $\endgroup$
    – Fat32
    Commented Mar 18, 2020 at 18:25
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    $\begingroup$ So reading your original comment I understand your intent which is the bandwidth of the signal must be less than the bandwidth of the filter- otherwise whether linear phase or not, distortion will result— this was your intention I believe! $\endgroup$ Commented Jul 11, 2023 at 22:26
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Let me add the following graphic to the great answers already given, with the intention of a specific and clear answer to the question posed. The other answers detail what linear phase is, this details why it is important in one graphic:

Comparison of linear and non-linear phase signals

When a filter has linear phase, then all the frequencies within that signal will be delayed the same amount in time (as described mathematically in Fat32's answer). When a filter has non-linear phase, individual frequencies or bands of frequencies within the spectrum of the signal are delayed different amounts in time.

Any signal can be decomposed (via Fourier Series) into separate frequency components. When the signal gets delayed through any channel (such as a filter), as long as all of those frequency components get delayed the same amount, the same signal (signal of interest, within the passband of the channel) will be recreated after the delay.

Consider a square wave, which through the Fourier Series Expansion is shown to be made up of an infinite number of odd harmonic frequencies.

In the graphic above I show the summation of the first three components. If these components are all delayed the same amount, the waveform of interest is intact when these components are summed. However, significant group delay distortion will result if each frequency component gets delayed a different amount in time.

The following may help give additional intuitive insight for those with some RF or analog background.

Consider an ideal lossless broadband delay line (such as approximated by a length of coaxial cable), which can pass wideband signals without distortion.

The transfer function of such a cable is shown in the graphic below, having a magnitude of 1 for all frequencies (given it is lossless) and a phase negatively increasing in direct linear proportion to frequency. The longer the cable, the steeper the slope of the phase, but in all cases "linear phase". This is also consistent with the equation for Group Delay, which is the negative derivative of phase with respect to frequency.

This makes sense; the phase delay of 1 Hz signal passing through a cable with a 1 second delay will be 360°, while a 2 Hz signal with the same delay will be 720°, etc...

Bringing this back to the digital world, $z^{-1}$ is the z-transform of a 1 sample delay (therefore a delay line), with a similar frequency response to what is shown, just in terms of H(z); a constant magnitude = 1 and a phase that goes linearly from $0$ to $-2\pi$ from f = 0 Hz to f = fs (the sampling rate).

a cable with its frequency delays graphed out

The simplest mathematical explanation is that the a phase that is linear with frequency and a constant delay are Fourier Transform pairs. This is the shift property of the Fourier Transform. A constant time delay in time of $\tau$ seconds results in a linear phase in frequency $-\omega \tau$, where $\omega$ is the angular frequency axis in radians/sec:

$$\mathscr{F}\{g(t-\tau)\} = \int_{-\infty}^{\infty}g(t-\tau)e^{j\omega t}dt$$ $$u = t - \tau$$ $$\mathscr{F}\{g(u)\} = \int_{-\infty}^{\infty}g(u)e^{-j\omega (u+\tau)}du$$ $$ = e^{-j\omega \tau}\int_{-\infty}^{\infty}g(u)e^{-j\omega u}du$$ $$ = e^{-j\omega \tau}G(j\omega)$$

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    $\begingroup$ Dan, your happy- and sad-face graph made me laugh out loud at how simply informative it is! Nicely done! $\endgroup$
    – Oreo
    Commented Nov 29, 2017 at 0:18
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Just to add to what's already been said, you can see this intuitively by looking at the following sinusoid with monotonically increasing frequency.

enter image description here

Shifting this signal to the right or left will change its phase. But note also that the phase change will be larger for higher frequencies, and smaller for lower frequencies. Or in other words, the phase increases linearly with frequency. Thus a constant time shift corresponds to a linear phase change in the frequency domain.

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  • $\begingroup$ Best answer imo. $\endgroup$ Commented Nov 28, 2017 at 22:20
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The essence and importance of linear phase property lies in the definition and the effect of group delay $$\tau(\omega) = - \frac {d\phi(\omega)}{d\omega}$$ on the applied signal $x[n]$, where $\phi(\omega)$ is the phase response of the filter; (phase of its frequency response).

Assume that a filter, with a fixed group delay of $n_0$ samples, is applied a narrowband input signal $x[n]$ (a signal having sufficiently narrow bandwidth across which any filter's arbitrary frequency-response magnitude remains approximately constant). Then the output signal will be (approximately) of the form $y[n] = K x[n-n_0]$ where $K$ is the filter gain evaluated at the center frequency of the narrowband input signal $x[n]$. If a filter has a flat passband of arbitrary large bandwidth, then the signal's bandwidth remaining inside the passband of filter is sufficient, and narrowband condition, of generality, is not necessary then. Eventually, for a linear phase filter, the input signal will be weighted and shifted intact as a whole by the group delay of the filter. Indeed, this can happen only when the filter's group delay is independent of the frequency $\omega$. And it's so if the underlying filter has linear phase (or generalized linear phase). Note that if the input signal is of broadband type, then the approximation of constant filter gain may not be valid in general, and eventhough filter may still be a linear-phase one, the output amplitudes will be weighted by frequency dependent gain $K(w)$ of filter, hence $y[n] \neq K x[n-n_0]$ anymore.

Then what's the effect of a filter with nonlinear phase (or frequency dependent group delay) on the input signal? A simple example would be a complicated input signal considered as a sum of multiple wavepackets at different center frequencies. After the filtering, each packet with a particular center frequency will be shifted (delayed) differently due to frequency dependent group delay. And this will be resulting in a change in the time-order (or space order) of those wave packets, sometimes drastically, depending on how nonlinear the phase is, which is called as dispersion in communications terminalology. Not only the composite waveshape, but also some event orders may be lost. This kind of dispersive channels have severe effects such as ISI (inter symbol interference) on transmitted data.

This property of linear phase filters, therefore, is also known as waveform-preserving property, which is applicable to narrowband signals in particular. An example where waveform is important, other than ISI as mentioned above, is in processing of images, where the Fourier transform phase information is of paramount importance compared to magnitude of Fourier transform, for intelligibility of the image. The same, however, cannot be said for perception of sound signals due to a different kind of sensitivity of the ear to the stimulus.

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  • $\begingroup$ What does generalized linear phase mean in this context? $\endgroup$
    – user33568
    Commented Feb 25, 2018 at 3:45
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    $\begingroup$ @0MW I suppose it means that constant phase shift is also allowed, as in Hilbert transform. $\endgroup$ Commented Dec 26, 2018 at 10:33
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The answer to this question is already been explained clearly in the previous replies. Yet I wish to give it a try to present a mathematical interpretation of the same

Consider a linear time invariant System whose frequency response is governed by $H(w)$.

i.e if the input to this system is $e^{jw_{0}t}$ the output will be $H(w_{0})e^{jw_{0}t}$

Here $H(w_{0})$ is a complex number which has a phase component denoted by $arg(H(w))$ and magnitude component denoted by $|H(w)|$

if the system has linear phase response then $$arg(H(w)) = Kw$$ where $K$ is a constant

If the phase is linear the output of the system for the input $e^{jw_{0}t}$ will be $$y(t) = |H(w)|*e^{jw_{0}t + jKw_{0}}$$ $$ = |H(w)|*e^{jw_{0}(t + K)}$$ which is nothing but a delayed version of input with some scaling applied.

So if the phase is linear then all the frequency components of the signal will undergo the same amount of delay in time-domain which results in shape preservation.

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I will just put a summary for these great answers mentioned above:

  • shifting the signal in the time domain will results in phase shift proportional to the frequency so f(t+dt) would be F(f)e(j2πfdt)
  • When a filter with a liner phase response all the frequencies of the input signal to this filter will be shifted with the same amount in the time domain so this will lead to the feasibility of recreation of the input signal.
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