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Simple digital Bandpass Filter confusions

  1. Oct 13, 2012 #1
    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 whats 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.
     
  2. jcsd
  3. Oct 13, 2012 #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.
     
  4. Oct 13, 2012 #3
    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.
     
  5. Oct 13, 2012 #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.
     
    Last edited: Oct 13, 2012
  6. Oct 13, 2012 #5
    I think I do need a lot more reading.
     
  7. Oct 13, 2012 #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.
     
  8. Oct 13, 2012 #7
    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) ?
     
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