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

• Jncik
In summary, to transform a low pass filter into a band pass filter, you need to multiply the impulse response by e^{j \omega_{0} n} and e^{-j \omega_{0} n} in the time domain. To transform a low pass filter into a high pass filter, you need to multiply by e^{j \omega_{0}} and cos(n*π) in the time domain.
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

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.

## 1. What is a low pass filter and how does it work?

A low pass filter is an electronic circuit that allows low frequency signals to pass through while attenuating or blocking high frequency signals. It works by using a combination of capacitors and resistors to create a frequency-dependent voltage divider. Low frequency signals are able to pass through while high frequency signals are filtered out.

## 2. How can a low pass filter be transformed into a band pass and low pass filter?

A low pass filter can be transformed into a band pass and low pass filter by adding an additional circuit element, such as an inductor or a second capacitor, to create a band pass filter. This additional element allows a specific range of frequencies to pass through, creating a band pass filter, while the original low pass filter component still filters out high frequencies.

## 3. What is the benefit of using a band pass and low pass filter?

The benefit of using a band pass and low pass filter is that it allows for a more precise control over the frequencies that are allowed to pass through. This is useful in applications where specific frequency ranges need to be isolated, such as in audio signal processing or radio communication.

## 4. Can a low pass filter be transformed into other types of filters?

Yes, a low pass filter can also be transformed into a high pass or band stop (notch) filter by using different combinations of circuit elements. A high pass filter allows high frequency signals to pass through while blocking low frequency signals, and a band stop filter blocks a specific range of frequencies while allowing all others to pass through.

## 5. Are there any limitations to transforming a low pass filter into a band pass and low pass filter?

One limitation is that it can be more difficult to achieve a precise and accurate band pass filter compared to a standalone low pass filter. This is because the additional circuit elements can introduce more variables and complexities. Additionally, the transformed filter may have a narrower bandwidth and may not be able to attenuate high frequencies as effectively as a standalone low pass filter.

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