2 added 10 characters in body edited May 17 '15 at 14:40 Matt L. 56k22 gold badges4040 silver badges105105 bronze badges 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. 1 answered May 17 '15 at 13:10 Matt L. 56k22 gold badges4040 silver badges105105 bronze badges 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.