Negative group delay and envelope advance

Question

I am having a doubt reading about delays in signal processing.

Let there be an input to a LTI system with frequency response $H(f)$, given signal $x(t) = a(t)\cos(2\pi f_ot)$, where $a(t)$ is a narrowband envelope and of bandwidth $B$ Hz, such that $f_o \gg B$.

The output $y(t)$ is given by $$y(t) =\lvert H(f_o) \rvert a\left(t - \tau_g\right)\cos\left(2\pi f_o\left(t - \tau_p\right)\right)$$ here $\tau_g$ is the group delay and $\tau_p$ is the phase delay at frequency $f_o$.

My question

It is not unusual to have positive gradients in the phase response and hence have a negative group delay which seems to suggest that the envelope is advanced in time, so the input appears at the output prior to being applied!! This of course should not happen in practice so what are we missing here. Can somebody explain this. Is there an issue with the derivation? But this is a well known equation.

Note:I have read the research article and other questions that conclude with the notion of "filter being able to predict values from past" , I am not convinced by those. In a causal LTI practical system I am sure, an input will appear at output only after it is applied at input.

Answer 1

Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies.

Yes, you are right that it is not uncommon to have practical LTI systems with $+ve$ gradient of $\angle H(f)$ in parts of the response and hence making group delay $-ve$ around those parts of $H(f)$. And, if we can give a Narrow-Band input $x(t)$ such that the bandwidth of $x(t)$ is restricted in that part of $H(f)$, then you would have a time-advanced output. So, are we able to look into the future?

NO!!!! We are not. My point will get clear in a minute.

Let me take an example of a very common and practically realizable IIR filter in equivalent discrete time scenario : the Leaky Integrator.

The $H(e^{j\omega})$ of a leaky integrator is given by the following: $$H(e^{j\omega}) = \frac{1-\lambda}{1-\lambda e^{-j\omega}},$$So, $$|H(e^{j\omega})|^2 = \frac{(1-\lambda)^2}{1 + \lambda^2 -2\lambda cos(\omega)}, \angle{H(e^{j\omega})} = arctan \{\frac{-\lambda sin(\omega)}{1-\lambda cos(\omega)} \}$$

The shape can be plotted in MATLAB by following:

freqz(0.1, [1 -0.9], (-pi:0.001:pi));

Now, if we can give a very narrow-band input $x[n]$ centered around $\omega = 0.6\pi$ and bandlimited within a very small $\Delta \omega$, then we would get a response as follows:

$x[n] = s[n]cos[\omega_o n]$, where s[n] is a narrowband baseband signal and $\omega_o = 0.6\pi$ and group delay of the filter is $g_d$ around $\omega_o$ $$Y(e^{j\omega}) = X(e^{j\omega}).e^{-j.g_d(\omega-\omega_o)},$$ You can work this out to get $y[n] = s[n - g_d]cos[\omega_o n]$

According to the equation above, leaky integrator is basically producing an output which is having a delayed envelope of input by $g_d$ samples. And, what happens if this $g_d$ is negative!

Check out that $g_d$ is indeed negative around $\omega_o = 0.6\pi$. Does that mean that the leaky integrator is able to produce the $s[n]$ envelope $g_d$ samples in advance?

No, it is not. The caveat is that we cannot have a perfectly bandlimited narrowband causal input $x[n]$. We cannot have a $x[n]$ which has an absolute start in time and yet it is having a bandlimited narrowband response in frequency domain.

Because we cannot have such input $x[n]$, hence we cannot have a "future seeing time machine".

In order to produce a causal input, which has an identifiable absolute start in time, the frequency response of the input will spread in frequency domain and the input $X(e^{j\omega})$ will be present at all frequencies with non-zero spectral components, and this will make the overall delay to be positive.

Indeed, if you plot the group delay response of the leaky integrator, you get the following, and check that even though the group delay is small negative number away from $\omega = 0$, it is taking high $+ve$ values around $\omega = 0$:

Hope that answers your question.

Answer 2

Here is a actual example with negative group delay that will provide further insight:

Below is a plot of the output and input of a pulse through a realizable filter that has negative group delay:

It seems like a complete violation of causality, but it is just a clever DSP magic trick. Let's explore further:

The filter above that did this had the following transfer function with the normalized frequency of the carrier within the pulse envelope was 0.1 radians/sample:

$$H(z) = \frac{42.7(z-.9)^2}{(z-.1)^2}$$

Notice that a scaled derivative of the input would almost provide this, but there are other features in the memory of this filter that cause the peak of the envelope to go down based on the previous cycles. In either case, as with the derivative the pulse can lead without having started before the input.

This filter can be factored into a cascade of filters including the transfer function below.

$$G(z) = \frac{z-0.9}{z-0.1} $$

H(z) as a more complex filter has a larger delay offset so was more obvious for the plot, but G(z) is simpler and will be easier to see what is occurring since it is given by this equation:

$$y[n] = x[n] - 0.9x[n-1] + 0.1y[n-1]$$

A plot of the same input pulse through $G(z)$ is shown below.

Below shows the initial conditions out of the filter with a constant envelope input signal of the same frequency:

A zoom in of the very first samples shows how the leading pulse shape can develop. Answering how the output knows how to turn up before the input will further help explain how the pulse envelope can do similar things while still being causal. Working through the equation above manually for the first 40 samples can help further illustrate how the memory of the past samples in the filter can help predict the future, given that the pulse occupies a narrow band of frequencies.

Extending this further and it gets really fascinating: consider an extended pulse by using a Tukey (cosine taper) window where the pulse is of very long duration:

And we zoom in at the beginning and end of the pulse and see that the envelope of the output is indeed advanced in time, but here it really gives the illusion that the output could predict the input since it appears that the envelope begins to decay at the output before the first sample of the input!

Start of Pulse

End of Pulse -- How can this be??

The secret to the DSP Magic Trick revealed!

At the scale of the entire pulse it appears that the change in the input is somehow predicted before it even occurs. However if we zoom in and look carefully at the 5 peaks that at whole scale appear unchanged, we see that indeed the input starts to change and it is this change that gets captured and amplified in the memory of the filter in the creating of the next output, it is completely causal. We are essentially seeing the DSP equivalent of economic "leading indicators" applicable to conditions when this can occur (when the derivative of phase with respect to frequency is positive for the "group" of frequencies within our signal). We see the bandwidth constraint in that this occurs with very small changes that start to occur over many cycles, even before it is immediately visible to us such as on this plot. A sudden unannounced larger change from one sample to the next would require high bandwidth, while low bandwidth implies memory over multiple samples.

For futher details please refer to: What is meant by "Group delay"?in simple words?