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LTI systems- frequency response

  1. Apr 14, 2013 #1
    I have attached an image of a ''brick-wall'' band pass filter. Can someone please help me find the expression of frequency response for it?








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  3. Apr 14, 2013 #2

    rude man

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    Well, acctually, you're looking at it:

    H(f) = 0, f < f1
    H(f) = 1, f1 < f < f2
    H(f) = 0, f > f1
    where f2 - f1 = BW.
    This filter BTW is unrealizable since its input-output relationship is not causal ...

    From a web page:

    << Impulse Response and Causality
    • All the impulse responses of ideal filters are
    sinc functions, or related functions, which
    are infinite in extent
    • Therefore all ideal filter impulse responses
    begin before time, t = 0
    • This makes ideal filters non-causal
    • Ideal filters cannot be physically realized,
    but they can be closely approximated >>
     
  4. Apr 15, 2013 #3

    marcusl

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    Davidlong, in the future please fill out the question form completely. It is a rule of the Forum that you must show your attempt at a solution before receiving assistance.
     
  5. Apr 15, 2013 #4
    I have the following equation for finding the impulse response function. The equation is the inverse Fourier transform of the H(f) equation.(infinity and -infinity are the limits) h(τ)=∫e(2∏jfτ)H(f)df.

    I replaced the limits with f2 and f1 which represent F+BW/2 and F-BW/2 respectively.

    For the integral i get [H(f)/2∏jf]*[e(2∏jfF2)-e(2∏jfF1)].

    I don't understand what value i should use for H(f), should it just be 1? And would that be the full expression for the impulse response or can it be simplified further?


    Thanks
     
  6. Apr 15, 2013 #5

    marcusl

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    Yes, H(f) is 1 in that interval. You can factor the exponentials and then simplify (think about trig functions).
     
  7. Apr 16, 2013 #6
    This is the code I've written in Matlab;

    Fc=500; %filter frequency
    BW=80; %bandwidth frequency
    F2=Fc+BW/2; %upper limit
    F1=Fc-BW/2; %lower limit
    Fs=11025; %sampling frequency
    Ts=1/Fs; %sampling time

    IRF=(1/(2*pi*1j*Ts))*(exp(2*pi*1j*F2*Ts)-exp(2*pi*1j*F1*Ts)); %impulse response
    FRF=(1/(-2*pi*1j*Ts))*(exp(-2*pi*1j*F2*Ts)-exp(-2*pi*1j*F1*Ts)); %frequency response

    A1=abs(FRF);
    theta1=-angle(FRF);

    figure(1)
    plot(Fc,A1)
    title('System A')
    xlabel('frequency')
    ylabel('amplitude response')

    figure(2)
    plot(Fc,theta1)
    title('System A')
    xlabel('frequency in Hz')
    ylabel('phase response')


    I'm meant to shift the curve to t>0, truncate h(t) to some reasonable finite times and sample h(t) at a sampling time of Ts=1/Fs which is supposed to give a set of {h(n)} coefficients for the filter. I think I've done the sampling but I'm not sure on which variables to adjust to achieve the shift and truncating.
     
  8. Apr 16, 2013 #7

    marcusl

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    You have made some errors:

    1) Your code for FRF is incorrect. The FRF is H(f), which you were given at the start. (See rude man's post).
    2) You did not properly evaluate the integral that gives h(t). You can't put the limits in before you evaluate the integral, and, BTW, the answer should be a function of t, not f.

    Perhaps it would be helpful to reread the parts of your text that describe what frequency and impulse response functions are, and how they are related. Then look at rud man's post and mine, and try to work the problem again.
     
  9. Apr 16, 2013 #8

    rude man

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    The impulse reponse to an ideal bandpass filter is the sum of two sinc functions, one for each corner frequency, as you may be able to determine by taking the inverse Fourier of H(f). The impulse response has infinite delay and the output can build up to infinity, so it's not of much use except for theoretical work.

    Of course there are many realizable filter types that can approximate the ideal bandpass filter.
     
  10. Apr 16, 2013 #9

    marcusl

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    This is incorrect. It is a single sinc function, which the OP can determine by factoring the exponential terms and applying a trig identity as I suggested.

    To be precise, the theoretical impulse response has finite amplitude, but extends in time from -∞ to ∞.
     
  11. Apr 16, 2013 #10

    rude man

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    It is two sinc functions. You may be thinking of a low-pass.
     
  12. Apr 17, 2013 #11
    Yes it's a low pass filter. Do I use eulers formula to factor the exponentials and then apply the trig function?

    For the frequency response the equation ends in dt, so its a function of t. For impulse response it ends in df so the impulse response is a function of f. Am i correct on this?
     
  13. Apr 17, 2013 #12

    rude man

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    It's a bandpass, not a lowpass. Look at your figure more carefully.

    You have it backwards. ∫f(t)exp(-jwt)dt = F(f) and ∫F(f)exp(jwt)df = f(t).
     
  14. Apr 18, 2013 #13
    So if there's 2 sinc functions do i have 2 different integrals? So for example with limits Fc to fc-BW/2 and fc+BW/2 to Fc?

    For a low pass filter the sinc function is 2.Fc.sinc(2.Fc.t) when the limits are Fc and -Fc. But I'm not sure how to achieve this if it's a band-pass filter.
     
  15. Apr 18, 2013 #14

    rude man

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    EDIT:
    Can you express an ideal high-pass filter Hh(f) in terms of an ideal low-pass filter Hl(f) plus a constant, the cutoff frequency being the same in both cases? Then draw a picture of Hl and another of Hf where the low-pass cutoff frequency is f1 and the high-pass cutoff frequency is f2, f2 > f1.
     
    Last edited: Apr 18, 2013
  16. Apr 19, 2013 #15
    The H(f) you are given is a (frequency)-shifted version of a low-pass filter. Now, you already know the frequency response for a low-pass filter, and I bet your teacher covered a rule that relates shifting in the frequency domain to a simple operation in the time domain. (If not, look at the table in your textbook and I am sure you will find that property of the Fourier transform.)
     
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