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In VSB filter we have the Transfer function condition as $H(F-F_c)+H(F+F_c)=1$ So can we accept that for the range of $F$ in $(F-df,F_c)$ we have $H(F)$ and for $(F_c,F+dF)$ we have $A-H(-F)$ where $A=H(F_c)$. I am learning Latex so need some time:(

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For $(F_c,F+dF)$, it needs to be $2A-H(-F)$, when $H(F_c)=A$. See the picture below taken from (http://163.152.6.72/contents/vod/21/20121129093403/20121129093403.pdf). Here $A$ corrsponds to $1/2$ as per below figure, that is the value at $H(F_c)$. In order for the sum to become $1$, $A$ needs to $1/2$. Also, $\epsilon$ is value at any frequency in the range $(F-dF,F_c)$. So the value in $(F_C,F+dF)$ needs to be $2A-H(-F)$ so that sum $H(F) + 2A-H(-F) = 1$, as $H(F)=H(-F)$ along $F=0$.

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  • $\begingroup$ Hey Thanks..i just made an error while translating the graph while deriving it!! and i posted..Thanks again for making it right.. i was following the NPTEL lectures where the prof just mentions the condition for a straight line! $\endgroup$ Apr 11, 2020 at 17:00

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