Transforming a low pass filer to a high pass filter

In summary, in order to make a low pass filter into a high pass filter, a shift in the frequency domain by π is needed, which results in a multiplication of the impulse response by cos(nπ). This causes the new impulse response to be [-1 1 -1].
  • #1
Jncik
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0
Hi, I have trouble understanding how to achieve this transformation

Suppose for example that we have a low pass filter which impulse response is the following

h(n) = 1 for 0<=n<=2 where n is an integer

In the frequency domain we would have the following frequency response:

[tex] H(e^{j\omega}) = \sum_{n=0}^{2} h(n) e^{-j \omega n} = 1 + e^{-j \omega} + e^{- 2j\omega} [/tex]

now, this is a low pass filter, and in order to make it high pass I have to make a shift in the frequency domain, suppose by π

So we would have in the time domain a multiplication by cos(n*π)

so the new impulse response is h(n) = cos(nπ) h(n)

for n = 0 we have hn(n) = 1
for n=1 we have hn(n) = -1
for n=2 we have hn(n) = 1

so the NEW impulse response will be the following vector [1 -1 1]

the problem is that my professor finds the result [-1 1 -1]

and I'm not sure why

here's his explanation:

[tex] h_{hp}(n) = [1 1 1] .* [cos(-1 \pi) \;\; cos(0 \pi)\; \; cos(1 \pi) ] = [1 1 1] .* [-1 \;1 \;-1] = [-1 \;1 -\;1][/tex]

isn't this wrong? I mean where does this cos(-1 π) come from?

thanks in advance
 
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  • #2
for your help.No, your professor is correct. The frequency shift by π that you need to do will cause the impulse response to be multiplied by the cosine function with the argument of nπ. This means that when n=0, the argument is 0π, when n=1, the argument is 1π, and when n=2, the argument is 2π. This is why the new impulse response is [-1 1 -1].
 

1. How do you transform a low pass filter to a high pass filter?

To transform a low pass filter to a high pass filter, you can use a process called frequency inversion. This involves swapping the positions of the resistors and capacitors in the circuit, effectively flipping the filter's frequency response curve.

2. What is the purpose of transforming a low pass filter to a high pass filter?

The purpose of transforming a low pass filter to a high pass filter is to change the frequency response of the filter. A low pass filter allows low frequencies to pass through while attenuating high frequencies, whereas a high pass filter does the opposite. This can be useful in different applications, such as audio equalization or signal processing.

3. Can any low pass filter be transformed into a high pass filter?

Yes, any low pass filter can be transformed into a high pass filter using the frequency inversion method. However, the resulting high pass filter may not have the same characteristics or performance as a dedicated high pass filter, so it is important to consider the specific requirements of your application.

4. What are the advantages of transforming a low pass filter to a high pass filter?

One advantage of transforming a low pass filter to a high pass filter is that it allows you to use a single circuit for both low and high frequency filtering. This can save space and reduce the number of components needed. It also allows for more flexibility in adjusting the filter's frequency response.

5. Are there any limitations or drawbacks to transforming a low pass filter to a high pass filter?

One limitation of transforming a low pass filter to a high pass filter is that it can introduce phase shifts and distortions in the frequency response. This may not be desirable in certain applications that require precise filtering. Additionally, the resulting high pass filter may not have the same performance or characteristics as a dedicated high pass filter, so it is important to carefully consider the trade-offs before making the transformation.

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