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You are right. Even though the phase jumps at the zeros of the frequency response, such a phase response is usually still called "linear". For a frequency selective filter with frequency response zeros in the stopband, the phase always has jumpsdiscontinuities at the zeros. A purely linear phase response (without jumps) is only possible for filters with no zeros in their frequency response.

The frequency response of your filter can be written as

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$

where $\phi(\omega)$ is the phase response with jumps at the zeros, as shown in your figure. For such a linear phase system, the frequency response (1) can equivalently be written as

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

where $A(e^{j\omega})$ is a real-valued but bipolar (i.e. positive and possibly negative) function satisfying $|A(e^{j\omega})|=|H(e^{j\omega})|$. The phase $\hat{\phi}(\omega)$ in (2) is now a purely linear function without any jumps.

You are right. Even though the phase jumps at the zeros of the frequency response, such a phase response is usually still called "linear". For a frequency selective filter with frequency response zeros in the stopband, the phase always has jumps at the zeros. A purely linear phase response (without jumps) is only possible for filters with no zeros in their frequency response.

The frequency response of your filter can be written as

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$

where $\phi(\omega)$ is the phase response with jumps at the zeros, as shown in your figure. For such a linear phase system, the frequency response (1) can equivalently be written as

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

where $A(e^{j\omega})$ is a real-valued but bipolar (i.e. positive and possibly negative) function satisfying $|A(e^{j\omega})|=|H(e^{j\omega})|$. The phase $\hat{\phi}(\omega)$ in (2) is now a purely linear function without any jumps.

You are right. Even though the phase jumps at the zeros of the frequency response, such a phase response is usually still called "linear". For a frequency selective filter with frequency response zeros in the stopband, the phase always has discontinuities at the zeros. A purely linear phase response (without jumps) is only possible for filters with no zeros in their frequency response.

The frequency response of your filter can be written as

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$

where $\phi(\omega)$ is the phase response with jumps at the zeros, as shown in your figure. For such a linear phase system, the frequency response (1) can equivalently be written as

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

where $A(e^{j\omega})$ is a real-valued but bipolar (i.e. positive and possibly negative) function satisfying $|A(e^{j\omega})|=|H(e^{j\omega})|$. The phase $\hat{\phi}(\omega)$ in (2) is now a purely linear function without any jumps.

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source | link

You are right. Even though the phase jumps at the zeros of the frequency response, such a phase response is usually still called "linear". For a frequency selective filter with frequency response zeros in the stopband, the phase always has jumps at the zeros. A purely linear phase response (without jumps) is only possible for filters with no zeros in their frequency response.

The frequency response of your filter can be written as

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$

where $\phi(\omega)$ is the phase response with jumps at the zeros, as shown in your figure. For such a linear phase system, the frequency response (1) can equivalently be written as

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

where $A(e^{j\omega})$ is a real-valued but bipolar (i.e. positive and possibly negative) function satisfying $|A(e^{j\omega})|=|H(e^{j\omega})|$. The phase $\hat{\phi}(\omega)$ in (2) is now a purely linear function without any jumps.