building a bandpass filter with ...

Discussion in 'Homework Help' started by 1233, Dec 26, 2009.

  1. 1233

    Thread Starter New Member

    Dec 26, 2009
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    I am asked to design a bandpass filter give the selectivity, the center frequency, and the passband gain.

    Q= 10
    fo = 5MHz
    Passband gain Ko=1

    Edit: using cascaded 1st order stages.

    I don't know where to start from.

    thanks in advance.
     
    Last edited: Dec 26, 2009
  2. ABoul

    Member

    Mar 30, 2009
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    i'm in a bit of a hurry so this is the best i can do: http://en.wikipedia.org/wiki/Q_factor

    go under the section titled "Physical interpretation of Q" (more specifically, the response equation, noting that s = jw) and the first part of the next section. if you play about with the equations, you should get values for R, L and C.

    EDIT: by inspection, LC = (2 * pi * 5000000)^2.
     
  3. 1233

    Thread Starter New Member

    Dec 26, 2009
    10
    0
    thx, I will have a look at that

    edit : forgot to mention that I should use cascaded 1st order stages.
     
    Last edited: Dec 26, 2009
  4. 1233

    Thread Starter New Member

    Dec 26, 2009
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    Is a simple LCR circuit considered a cascaded circuit ?

    [​IMG]
     
  5. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    With what parts may you make the filter ?
    Take a look at this PDF for filter design.
    sloa96_filter_design.pdf

    Greetings,
    Bertus
     
  6. 1233

    Thread Starter New Member

    Dec 26, 2009
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    I think I should build the filter using inductor, capacitor and resistors only. I am sure that it is not so difficult, but I don't know where to start from
     
  7. bertus

    Administrator

    Apr 5, 2008
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  8. Nanophotonics

    Active Member

    Apr 2, 2009
    365
    3
    It's OK to design filters using RLC, but I think it's not practical to build it. The range of values available for L is limited and there are various other disadvantages. I would probably go for opamps.
     
  9. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Start by considering the bandwidth: BW = fc/Q

    This, obviously is 500kHz
    So your f(low is 4.5MHz) and f(high)=5.5MHz.

    With the instructions to use "stages" I'd guess you're allowed to use transistors for amplification in order to offset the attenuation of the filter.

    A single common emmiter stage should do.
    Use the input capacitor to obtain f(low) and at the output put a capacitor in parallel with the load to get your f(high). To calculate the C use the input resistance in series with the source resistancefor fL and and the output resistance in parallel with the load resistance for fH. Put these in the equation C = 1/(2*Pi*R*f).

    At least, this is how I understand the problem. Better ask your instructor for more details.
     
  10. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    I'm betting not, it just means you have to predict the 1.5db points, and adjust accordingly.

    Generally filters use T attenuators to offset differences in terminations. A 6db attenuator will have a 12 db return loss.
     
  11. 1233

    Thread Starter New Member

    Dec 26, 2009
    10
    0

    if B.W is 500 kHz, then f(low) is 4.75MHz and f(high)=5.25MHz, am I right ?
     
  12. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Your right. I appologize for my sloppy math.
     
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