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I know that group delay is defined as :

$$\begin{align} \tau(\omega) = grd [H(e^{j\omega})] &=-\frac{d[\arg[H(e^{j\omega})]]}{d\omega}\end{align}$$

But if what is available is group delay itself, is it possible to recover the phase?

According to Discrete-Time Signal Processing 3rd edition by Alan Oppenheim equation 5.86 the expression would look like this:

$$\begin{align} \arg[H(e^{j\omega})] = -\int\limits_{0}^{\omega} \end{align} grd [ H(e^{j\phi})]\, d \phi + \arg[H(e^{j0})] $$

and my doubt is:

Where did the second term $$\arg[H(e^{j0}) ]$$ came from?

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    $\begingroup$ That's basic math – the antiderivative of a function $f(t)$ is always only certain up to a constant $C$ as in $\int f(t)\,dt +C$. That $C$ must fulfill the boundary conditions, and in this case, that condition is given by the the phase shift at DC. $\endgroup$ – Marcus Müller Aug 29 '17 at 20:03
  • $\begingroup$ Isn't the +C constant only for indefinite integrals? This one has integration limits. Why would there be a +C? $\endgroup$ – VMMF Aug 30 '17 at 13:23
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    $\begingroup$ no, it's not just for indefinite integrals. It's just that limited integrals usually inherently cancel that $C$. $\endgroup$ – Marcus Müller Aug 31 '17 at 1:04
  • $\begingroup$ If after integrating, when it comes to evaluate in 0 and W (integration limits) I get C - C then in the end there's no C constant remaining right? $\endgroup$ – VMMF Jan 29 '18 at 4:33
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Note that if $F(x)$ is an antiderivative of $f(x)$, i.e., if $F'(x)=f(x)$, then from the fundamental theorem of calculus we have

$$\int_a^bf(x)dx=F(b)-F(a)\tag{1}$$

Since the phase $\phi(\omega)$ is an antiderivative of the negative group delay $-\tau(\omega)$ we get from $(1)$

$$-\int_a^b\tau(\Omega)d\Omega=\phi(b)-\phi(a)\tag{2}$$

With $b=\omega$ and $a=0$ we get from $(2)$

$$-\int_0^\omega\tau(\Omega)d\Omega=\phi(\omega)-\phi(0)\tag{3}$$

By adding $\phi(0)$ on both sides of $(3)$ we obtain the given formula:

$$\phi(\omega)=-\int_0^\omega\tau(\Omega)d\Omega+\phi(0)\tag{4}$$

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When integrating any expression, there is a constant term that is added. This represents an offset that may exist. For example, if I take the derivative of the function $y=x^2+3$. The derivative $\frac{dy}{dx} = 2x$. If I integrate $\frac{dy}{dx}$, then I end up with the expression $y=x^2+c$, where $c$ is an unknown constant. In your question, $arg[H(e^{j0})]$ is the phase offset at DC. This is the missing constant offset when you integrate the group delay. Without it you, you can only know the phase at some frequency $\omega1$, relative to some other frequency $\omega2$.

I'll see if I can clarify (hopefully not confuse) and answer the comment below. Let's look at a different problem. Let's say I want to integrate velocity $v$ over time to find out my position $p$ on the highway. I started at time 0 and drove for $\omega$ seconds. My position relative to my starting position is $p=\int\limits_{0}^\omega vdt$. The key here is relative to my starting position. If I don't know my position at time 0, I do not know my absolute position. I can tell you I travelled 10 miles, but I don't know where from and therefore where to.

The same is true for integrating the group delay to produce a phase. If we wanted phase offset from DC, the $arg[H(e^{j0})]$ wouldn't be necessary. If we want absolute phase, we must add it.

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  • $\begingroup$ Thank you very much for your answer. I know that for indefinite integrals (no integration intervals) after integrating a +C constant is added. But this integral has integration intervals between 0 a w. Why would there be a +C? $\endgroup$ – VMMF Aug 30 '17 at 13:20
  • $\begingroup$ @VMMF: because the real phase response is not necessarily zero at zero. $\endgroup$ – sellibitze Sep 30 '17 at 17:16

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