If group delay is a constant for all $\omega$, Does this system have a linear phase response?

I want to know if a system has a linear phase response using frequency response of this system.

So I got phase response of it and also group delay.

and after that I got to know that if group delay is a constant for all ω, this system has a linear phase response.

so questions are

1. Group delay = $1/(1+\omega^2)$ . so $\omega=0$, group delay is $1$ and $\omega=1$ group delay is $1/2$. This is not a constant for all $\omega$ so I think this system has not a linear phase response. Is it correct?

2. when $H \left( e^{j\omega} \right)$ = $-e^{-j\omega} + 2 e^{-2j\omega} - e^{-3j\omega}$. Can I also know if this system has a linear phase response with same logic in question 1?

This is a homework type question, so I will only give hints and no solutions. You should know that the group delay is the negative derivative of the phase (with respect to frequency), so if the phase is a linear function of frequency, the group delay must be a constant. This should answer your first question.

Concerning your second question, why don't you just compute the phase of the given frequency response? It may be helpful to rewrite $H(e^{j\omega})$ as

$$H(e^{j\omega})=2e^{-2j\omega}\left(1-\frac{e^{j\omega}+e^{-j\omega}}{2}\right)\tag{1}$$

and to realize that the term in parentheses is real-valued.

EDIT:

The frequency response $H(e^{j\omega})$ can be written as

$$H(e^{j\omega})=M(\omega)e^{j\phi(\omega)}\tag{2}$$

where $M(\omega)$ is the magnitude, and $\phi(\omega)$ is the phase. Now compare $(1)$ and $(2)$ and figure out what $\phi(\omega)$ must be.

• Thank you :) Can I ask you more? I think your hint is very important to solve it but I totally don't know how to use it... how can I get magnitude response and phase response by rewriting H(e^jω)? I think there is a simple way to get them by rewriting it.. Please help me.. Thank you. – allentando Jun 16 '18 at 16:09
• @allentando: I added some more information to my answer, I trust that you can take it from there. – Matt L. Jun 16 '18 at 16:17
• Thank you so much :) I almost understand but I still don't know what defines e^jϕ(ω) term. I think I can derive e^−jω or e^-2jω or e^-3jω from H(e^jω) = −e^−jω + 2e^−2jω − e^−3jω. but you derived only 2e^−2jω even it is 2e^−2jω not only e^−2jω. Are there something reasons? to make term in parentheses real-valued trigonometrical function? – allentando Jun 17 '18 at 1:58
• @allentando: So if you compare Equations (1) and (2), what do you get for $M(\omega)$ and $\phi(\omega)$? – Matt L. Jun 17 '18 at 9:51
• M(ω) is 1-cos(ω) and ϕ(ω) is -2ω. – allentando Jun 17 '18 at 9:58