# express pass band filter as sum of low pass filter

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 🙈

• "...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? – A_A Jan 30 at 11:14
• 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 $$H_{BP}(f) = \frac{1}{2} 2 H_{LP} (f-f_0) + \frac{1}{2} 2 H_{LP }(f+f_0)$$ – Elena Martini Jan 30 at 12:53