Simple digital Bandpass Filter confusions

In summary: If I kind of get this right, your saying create this system: sin(2*pi*10hz) + sin(2*pi*30hz)*(some lowpass filter) = sin(2*pi*10hz)?
  • #1
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Lets assume I have a signal that exists in time and it has an unit impulse at t=5, I only want to filter out that portion of the signal. So I plan to design a passband filter to filter the signal out at t=5.

This is where I get confused, my passband filter exist in frequency so how is that going to filter the signal out? I'm not sure what's the range for example 50 hz to 100 hz. If I want to have the filter exist at t = 4 through t=6 is it simply going to be 1/4 hz to 1/6 hz?

Assuming I get the range right, will my output after getting passed through the filter just be a unit impulse? Will it still exist at t=5? This isn't homework I'm just trying to understand so I can start designing.
 
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  • #2
Filters work in the frequency domain, not the time domain. If you want to just leave out some samples, design a sequential circuit (finite state machine) that copies (passing through) all samples but the ones you don't want to.

Maybe if you shared more about what you want to achieve (we're pretty sure your signal doesn't contain only one sample at t=5) we could give you a better advice.
 
  • #3
FailedLaunch said:
Filters work in the frequency domain, not the time domain. If you want to just leave out some samples, design a sequential circuit (finite state machine) that copies (passing through) all samples but the ones you don't want to.

Maybe if you shared more about what you want to achieve (we're pretty sure your signal doesn't contain only one sample at t=5) we could give you a better advice.

It has to be a filter, pretty much my goal is to demonstrate some form of filtering and demonstrate it in the real world. I can choose any signal I want, it's my choice and I have to build the physical circuit to prove the concept. I haven't taken Signals and Systems in years.

I've spent some time reading online and I'm kind of getting the idea that I need to multiply two signals in the frequency domain then go back to time domain. So in the case of the unit impulse at t=5, it'll become (in z-domain using z transforms) z^-5.

So couldn't I just use a 3 unit impulses in frequency domain (z), at z=4, z=5,z=6 for a filter? So that'll be derac(z-4)+derac(z-5)+derac(z-6) then go back to time. So my system can be described by this F(z) = z^-5 and G(z) = derac(z-4)+derac(z-5)+derac(z-6).

Using z transform properties of convolution I do the inverse z-transform{F(z)xG(z)}= f(t)*g(t) (convolution not multiplication) to get my filtered signal. Darn I made it too hard now.
 
  • #4
Now you got completely lost in it :tongue2:

  1. you can not determine frequency properties (sprectrum) from a single sample. You need more than only one lonely sample
  2. filter with "unit pulses at z=4,5,6" is pretty hopeless. You need to have a deeper look at Signals and Systems

As for your project, I think the easiest way would be to generate 2 sines with different frequencies (something like 10Hz and 30Hz - fairly apart, yet still close enough),add them, then do a simple low pass (or high pass depending on what you want to filter out), and have one of the sines on the output.

Instead of designing a circuit you can program it in a processor (or MATLAB) and then plot input and output signals.
 
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  • #5
FailedLaunch said:
Now you got completely lost in it :tongue2:

  1. you can not determine frequency properties (sprectrum) from a single sample. You need more than only one lonely sample
  2. filter with "dirac at z=4,5,6" is pretty hopeless. You need to have a deeper look at Signals and Systems

As for your project, I think the easiest way would be to generate 2 sines with different frequencies (something like 10Hz and 30Hz - fairly apart, yet still close enough), then do a simple low pass (or high pass depending on what you want to filter out), and have one of the sines on the output.

I think I do need a lot more reading.
 
  • #6
Actually, my advice about using Matlab is probably the best idea for you. You can model there whatever you want and see how it interacts and it will help you understand how it all plays together.
 
  • #7
FailedLaunch said:
Actually, my advice about using Matlab is probably the best idea for you. You can model there whatever you want and see how it interacts and it will help you understand how it all plays together.

If I kind of get this right, your saying create this system: sin(2*pi*10hz) + sin(2*pi*30hz)*(some lowpass filter) = sin(2*pi*10hz) ?
 

1. What is a simple digital bandpass filter?

A simple digital bandpass filter is an electronic circuit that allows a specific range of frequencies to pass through while reducing or eliminating all other frequencies. It is commonly used in signal processing applications to filter out unwanted noise or isolate a specific frequency range.

2. How does a bandpass filter work?

A bandpass filter works by using a combination of capacitors and inductors to create a circuit that resonates at a specific frequency. This resonance allows only signals within a certain frequency range to pass through, while attenuating signals outside of that range.

3. What types of signals can be filtered using a bandpass filter?

A bandpass filter can be used to filter a wide range of signals, including audio signals, radio waves, and digital signals. It is commonly used in audio equipment, telecommunications, and digital signal processing applications.

4. Can a bandpass filter be adjusted to different frequency ranges?

Yes, a bandpass filter can be adjusted to filter different frequency ranges by changing the values of the capacitors and inductors in the circuit. This allows for flexibility in filtering different types of signals.

5. What are the advantages of using a digital bandpass filter over an analog one?

One advantage of using a digital bandpass filter is that it can be easily programmed and adjusted using a microcontroller or digital signal processor. It also tends to have a more precise and stable frequency response compared to an analog bandpass filter.

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