Transforming a low pass filter to a band pass and low pass filter

[Jncik]

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 $$e^{j \omega_{0} n}$$ 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 $$e^{-j \omega_{0} n}$$

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

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 $$e^{j \omega_{0}}$$? 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

 

[KataKoniK]

Hello,

To transform a low pass filter into a band pass filter, you are correct in multiplying the impulse response by e^{j \omega_{0} n} in the time domain. This will shift the frequency response to the right, but as you mentioned, for a band pass filter we also need to shift it to the left. This can be achieved by multiplying by e^{-j \omega_{0} n} as well, resulting in a final frequency response of H'(ω) = H(ω)e^{j \omega_{0}} + H(ω)e^{-j \omega_{0}}.

Now, for your question about why your teacher used h(n)*cos(ω0*n) instead of 2h(n)cos(ω0*n), it is because the frequency response of the band pass filter will have two peaks at ω0 and -ω0. So, your teacher's solution takes into account the fact that the frequency response will have a magnitude of 2 at ω0 and -ω0, while your solution only considers the peak at ω0.

As for transforming a low pass filter into a high pass filter, you are correct in multiplying by e^{j \omega_{0}} in the time domain. This will shift the frequency response to the right, but to achieve a high pass filter, we need to shift it to the left as well. This can be achieved by multiplying by cos(n*π) in the time domain. This is because cos(n*π) has a value of 1 when n is even and -1 when n is odd, which will result in a frequency response that is shifted to the left.

I hope this helps clarify your confusion. Please let me know if you have any further questions.