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Radio station tuned

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    You are building a simple passive AM radio using a series RLC circuit with an voltage source of V sin(omega*t) and the voltage drop V_R across the resistor as the system response.

    You would like to know the relations between the values of R,L,and C in order to tune to a station at 950kHz and with good enough reception. The bandwidth of the AM radio station is about 10 kHz. By good enough reception, you mean that the gain of V_R should drop to a small fraction, say 1/10 of the maximum gain at the two boundary frequencies v_1 kHz and v_2 kHz.

    Find C (with units microF) in terms of L (with units H) and v=950,000 Hz. (Remember omega has units rad/s, which are equivalent to units of 1/(2*pi)Hz.)

    (Substitute v=950,000 Hz, enter in terms of L)

    There is also the second question but perhaps I will be able to solve it on my own if I get this one right.


    2. The attempt at a solution

    I tried to use the formula for gain squared (with all information provided) but it gets very complicated. I am on the wrong track, I think.

    (95^2=(r^2*omega^2)/((1/c-l*omega^2)^2+r^2*omega^2))

    95^2=(r^2*(10+omega)^2)/((1/c-l*(10+omega)^2)^2+r^2*(10+omega)^2)

    where v_1=2*pi*omega
    v_2=2*pi*(omega+10)

     

    Attached Files:

  2. jcsd
  3. Apr 3, 2017 #2

    Baluncore

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    Consider the resonant frequency of the LC circuit, when XL = –XC.
    Consider the ratio XL/R and the Q of the circuit.
     
  4. Apr 3, 2017 #3
    Something like this: C= 1/((950000*2*pi)^2*l) ?
     
    Last edited: Apr 3, 2017
  5. Apr 3, 2017 #4
    Q=95? I am not sure.

    The second question: Find R (with units of kiloOhms) in terms of L (with units H) and v=950 kHz.

    1000*L ?
     
  6. Apr 3, 2017 #5

    Baluncore

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    Science Advisor

    Yes.

    Maybe, but first;
    Is that the bandwidth of the audio AF to be transmitted, or of the RF channel allocated in the radio spectrum?
    An AM RF signal has symmetrical upper and lower sidebands, so it needs twice the audio bandwidth in the RF channel.
     
  7. Apr 4, 2017 #6
    OK. I have to think about it. I have discovered that this is a different Q (full width at half maximum -1/sqrt(2) and not 1/10) so this formula doesn't fit here. Well, this is a course on differential equations. I don't know much about circuits. I will get to the bottom of it.
     
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