- #1

Jncik

- 103

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## Homework Statement

My problem is I want to know how to transform a low pass filter into a band pass and high pass filter.

Suppose that we are given the impulse response of the system which is h(n)

## The Attempt at a Solution

In order to transform this system into a band pass filter, I need to multiply by [tex]e^{j \omega_{0} n} [/tex] in the time domain, because shifting in the frequency domain means multiplication in the time domain.

with this multiplication I will shift H(ω) to the right, but since we have a band pass filter, I will need to shift it to the left as well and add this to the final result

so I multiply h(n) by [tex]e^{-j \omega_{0} n} [/tex]

and thus I have [tex]h'(n) = h(n)[e^{j \omega_{0} n} + e^{-j \omega_{0} n}] = 2h(n)cos(\omega_{0}n) [/tex]

my question is, why does my teacher find h(n)*cos(ω0*n) and not 2 times what I just said?

also, why do I need to multiply by cos(n*π) in order to transform my filter into a high pass filter? shouldn't it be just [tex]e^{j \omega_{0}} [/tex]? it means that we don't care if ω0 is π or not, it just is any ω in the frequency domain. I don't understand why it's cos either. I'm totally confused

thanks in advance