SanjKrish
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Hello all,
While seeing a lecture about designing a simple digital high pass filter, I came to know its frequency response will of the form
H(ej\[\omega\]) = je-j\[\omega\]/2sin(\[\omega\]/2)
The author plotted the phase response graph as below.. :?:
View attachment phase.bmp
He explained that the phase at 0 will be \[\pi\]/2 because of the 'j' term and
beyond \[\omega\]=0 it will be \[\pi\]/2 - \[\omega\]/2 so it will linearly come down to 0 at \[\omega\] = \[\pi\]/2..
Please explain me how did he find out the phase response to be \[\pi\]/2 - \[\omega\]/2
I know tat the phase is given by tan-1(img part/real part) and for me, for the given frequency response, it came out to be \[\omega\]/2
and not \[\pi\]/2 - \[\omega\]/2..:-(
While seeing a lecture about designing a simple digital high pass filter, I came to know its frequency response will of the form
H(ej\[\omega\]) = je-j\[\omega\]/2sin(\[\omega\]/2)
The author plotted the phase response graph as below.. :?:
View attachment phase.bmp
He explained that the phase at 0 will be \[\pi\]/2 because of the 'j' term and
beyond \[\omega\]=0 it will be \[\pi\]/2 - \[\omega\]/2 so it will linearly come down to 0 at \[\omega\] = \[\pi\]/2..
Please explain me how did he find out the phase response to be \[\pi\]/2 - \[\omega\]/2
I know tat the phase is given by tan-1(img part/real part) and for me, for the given frequency response, it came out to be \[\omega\]/2
and not \[\pi\]/2 - \[\omega\]/2..:-(
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