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I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n-\tau_g]\cos[\omega_0 (n -\tau_p)] $$ (if $\tau_g$ and $\tau_p$ are integers) where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related questionrelated question I could find here did not have a complete answer to my question.

I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n-\tau_g]\cos[\omega_0 (n -\tau_p)] $$ (if $\tau_g$ and $\tau_p$ are integers) where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n-\tau_g]\cos[\omega_0 (n -\tau_p)] $$ (if $\tau_g$ and $\tau_p$ are integers) where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

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I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n+\tau_g]\cos[\omega_0 (n +\tau_p)] $$$$y[n] \approx |H(e^{j\omega_0})|s[n-\tau_g]\cos[\omega_0 (n -\tau_p)] $$ where(if $\tau_g$ and $\tau_p$ are integers) where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n+\tau_g]\cos[\omega_0 (n +\tau_p)] $$ where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n-\tau_g]\cos[\omega_0 (n -\tau_p)] $$ (if $\tau_g$ and $\tau_p$ are integers) where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

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I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n+\tau_g]\cos[\omega_0 (n +\tau_p)] $$ where $$\tau_g =-\frac{d\phi(\omega_0)}{d\omega_0} $$$$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =-\frac{\phi(\omega_0)}{\omega_0} $$$$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n+\tau_g]\cos[\omega_0 (n +\tau_p)] $$ where $$\tau_g =-\frac{d\phi(\omega_0)}{d\omega_0} $$ $$\tau_p =-\frac{\phi(\omega_0)}{\omega_0} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

I know that for a linear-phase filter with frequency response given by $$H(e^{j\omega}) = |H(e^{j\omega})|e^{j\phi(\omega)} $$ if the input of the system is $$x[n] = s[n]\cos[\omega_0 n] $$ where $s[n] $ is a narrowband signal with bandwidth $W\ll\omega_0$, then the output of the system is approximately $$y[n] \approx |H(e^{j\omega_0})|s[n+\tau_g]\cos[\omega_0 (n +\tau_p)] $$ where $$\tau_g =\left.-\frac{d\phi(\omega)}{d\omega}\right|_{\omega=\omega_o} $$ $$\tau_p =\left.-\frac{\phi(\omega)}{\omega}\right|_{\omega=\omega_o} $$ A broadband signal can be thought as a superposition of narrowband signals, this means that if we have a filter with linear phase (i.e. $\phi(\omega)= \tau\omega$) both group delay and phase delay are constant and equal, therefore all the narrowband signal packets will be shifted in the same amount. But what about generalized linear phase filters? In this case $$\phi(\omega) = n_d \omega + \phi_o $$ so the output of a narrowband input centered at $\omega_0$ will now be $$y[n]= |H(e^{j\omega_0})|s[n-n_d]\cos\left[\omega_0\left(n -n_d-\frac{\phi_0}{\omega_0}\right)\right] $$ and the term $\frac{\phi_0}{\omega_0} $ will differ from one signal packet to another as the center frequency ($\omega_0$) of each packet changes. So why are generalized linear phase filters as important as linear phase ones? Isn't the term $\frac{\phi_0}{\omega_0} $ important?

The only related question I could find here did not have a complete answer to my question.

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