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Homework Help: AC+DC signal equation=> Then Filter AC signal= output only DC signal visible

  1. Lowpass filter

    1 vote(s)
  2. FFT

    0 vote(s)
  1. Nov 12, 2009 #1
    Dear friends,

    I would like to know how can I remove an AC signal from a signal which is combination of AC+DC. AC signal 60V/50Hz DC=50V.

    Please suggest me a C code or any algoritham or a filter which can remove AC signal completely and I should be able to see only DC signal in the out put.

    with best regards,
    Sandhi :(

    Attached Files:

  2. jcsd
  3. Nov 14, 2009 #2


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    Science Advisor

    You would not be able to filter out the 60 Hz component.

    The AC input of 60 volts would overdrive the amplifier and the gain of 3.9 would make it worse if that was possible. The output would just be a 50 Hz square wave

    If you had an input of 0.6 volts, it may be possible to reduce the AC component in the output, but not remove it completely. You could put a large capacitor to ground after the 10 K resistor.
    Try 25 uF.
  4. Nov 16, 2009 #3
    I am sorry to tell that I have mentioned wrong value. It is 5V/1Hz Ac signal and 12 V DC signal.
  5. Nov 16, 2009 #4


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    The 1 Hz signal would make things difficult, but there is a bigger problem.

    12 volts DC into an amplifier with a gain of -3.9 will give an output of -46.8 volts.

    Now, unless you have a supply of +/- 50 volts this is going to overload your amplifier.

    Apart from that, filters for 1 Hz are not really practical. You can do calculations and design a filter, but the component values will become very large and expensive.
  6. Nov 26, 2009 #5
    I would like to know how can I post a new thread with attachment? I dont see option where I can send a new post with attachment.

    with best regards
  7. Nov 26, 2009 #6


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    Choose "new topic" at the top of the main page.

    Fill in the template, if it is on the homework page.

    When that opens go to the paperclip thing at the top of the edit screen.
    Browse and find the attachment on your hard drive. Upload it.

    Then put your cursor where you want the attachment to be. Go to the paperclip again and click on the name of the file you uploaded.
    It will put a link to the attachment and you can see it if you go to "preview post" at the bottom of the edit screen.

    This way, you can have various pictures or diagrams in the right place in your text.

    Click on "submit new thread".
  8. Nov 26, 2009 #7
    Thank you, I have found it. I have posted my new thread about origin of Capacitance formula.
  9. Dec 15, 2009 #8
    Dear Sir,

    I would like to know if you have recieved my mail few hours ago

    Prashant Kumar
  10. Dec 15, 2009 #9


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    Yes, I received your mail.

    At 0.5 Hz, the capacitor has a reactance of 106103 ohms. The net impedance of the capacitor and resistor combination is 9.956 Kohms. So, really the capacitor has no effect.

    In series with 1.01 Mohm, the output is 9.56 / 1010 or 0.00946 times the input voltage or 0.047 volts. Without the capacitor, it would be 0.0495 volts.

    0.5 Hz is 1 cycle every 2 seconds. This is a very low frequency. What are you actually doing?
  11. Dec 16, 2009 #10
    I am trying to calculate Rf by measuring the signal at Rm which I call it as Um. When there is no capacitance it is very simple to calculate Rf but with varying capacitance Ce and Rf I need to calculate du/dt to the signal measured at Rm i.e to Um volts. Um is almost like a squarewave.

    Dear Sir, I would like to know the exact equation for Rf when I include capacitance Ce.

    with regards,
  12. Dec 16, 2009 #11


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    This is the homework section, so the rules apply. This type of problem is solved by a process of steps, not really one formula.

    You might like to have a look at this:

    It should tell you how to work out the impedance of a parallel RC network. You can add currents as vectors if they are flowing in the same wire, but you can't add impedances if they are in parallel.

    You will also need to work out the reactance of a capacitor.
    This is given by:

    where F is in Hz and C is in Farads.
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