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I have to find impulsive response of an ideal pass band filter, but I have a problem to express $$ H_{BP} (f) $$ as a sum of $$ H_{LP} (f) $$. I mean that $$ H_{BP} (f) = rect ( \frac{f-f_0}{B} ) + rect ( \frac{f+f_0}{B} ) $$ but $$ H_{LP }= rect (\frac{(f- f_0)}{2B} ) $$ so I wrote $$ H_{BP} (f) = rect ( \frac{2(f-f_0)}{2B} ) + rect ( \frac{2(f+f_0)}{2B} ) $$. Now probably it’s wrong but I wrote $$ H_{BP} (f) = 2 H_{LP} (f-f_0) + 2 H_{LP} (f+f_0)$$ Now I should apply wave demodulation $$ h_{BP}(t)= \frac{1}{2} h_{LP} + \frac{1}{2} h_{LP} cos(2 \pi 2 f_0 t) $$. I hope i wrote at least one right thing 🙈

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    $\begingroup$ "...as a sum of low pass filters" or as a "..sum of a high-pass and a low pass filter" or as a "...shifted low pass filter prototype" (?)... The $h_{BP}$ expression brings together two overlapping low pass filters which amount to 1 filter in the end. What you are expressing is strictly wrong and along the right track if you make a decision as to what you had in mind when approaching this problem. Can you please clarify what the thinking was? $\endgroup$ – A_A Jan 30 at 11:14
  • $\begingroup$ To solve this problem my book write as suggestion that $$ H_{BP} (f) $$ can be expressed as the transfer function of a low pass filter and write $$ H_{BP}(f) = \frac{1}{2} 2 H_{LP }(f-f_0) + \frac{1}{2} 2 H_{LP} (f+f_0) $$ assuming this for true I think now I had to apply modulation property to find $$ h_{BP}(t) $$. This because i know $$ h_{LP}(t)$$ from past exercises. But I don’t know how my book obtained $$ H_{BP}(f) = \frac{1}{2} 2 H_{LP} (f-f_0) + \frac{1}{2} 2 H_{LP }(f+f_0) $$ $\endgroup$ – Elena Martini Jan 30 at 12:53

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