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What is Image Rejection Ratio (IRR)

  1. Jun 26, 2011 #1
    What is Image Rejection Ratio (IRR)
    and How to calculate an Image Rejection Ratio (IRR)?
     
  2. jcsd
  3. Jun 26, 2011 #2

    vk6kro

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    This is from Wikipedia:

    In radio, the image rejection ratio, or image frequency rejection ratio, is the ratio of (a) the intermediate-frequency (IF) signal level produced by the desired input frequency to (b) that produced by the image frequency.

    The image rejection ratio is usually expressed in dB.

    When the image rejection ratio is measured, the input signal levels of the desired and image frequencies must be equal for the measurement to be meaningful.
     
  4. Jun 27, 2011 #3
    What is an formulae for Image Rejection Ratio?
    Can anyone solve this for clarification...
    -----------------------------------------------------------------------------------------------------------------------------
    Suppose, if a superheterodyne receiver with an IF of 3.1 MHz is tuned to receive a carrier of 78 MHz and the radio frequency tuning stage has a Q-factor of 100. What is the Image Rejection Ratio (IRR) in dB?

    Given is |Vo/Vi| = 1/ Sqrt [1+Q2 (wo/w -w/wo)2]
     
  5. Jun 27, 2011 #4

    vk6kro

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  6. Jun 28, 2011 #5
    After substituting all values I got -23.7 dB, Kindly let me know was it the same for you or anything else is your answer varun. Thank you so much for your links and support.
     
  7. Jun 28, 2011 #6
    Where can I get an eBook of Communication Engineering By A.P.Godse U.A.Bakshi, It seems it is very good one's and alare so clearly written in it. Thanks and Thank you so much...
     
  8. Jun 28, 2011 #7

    vk6kro

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    Yes, I got 23.7 dB too. Not sure about the negative sign. The definition seems to say that it should be a the log of a number greater than 1. In this case the book formula gives 15.33 as the voltage ratio.

    I don't know about that book. It is on Google books.

    You could try Amazon.

    I hope you noticed the typo in the first example. They had a "*" instead of a "+" sign. The formula was right but the calculation was wrong, although the answer was right. So, it was a simple typo.
     
  9. Jun 29, 2011 #8
    Are you sure of converting something in dB is about 20xlog10(0.065) or 10xlog10(0.065). Also I'm confused slightly in converting in dB and dBm of any value so can you please clarify me in what should we use when converting to dB and dBm
     
  10. Jun 29, 2011 #9

    vk6kro

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    Yes, 20 log10(voltage ratio) is correct.

    Decibels are only ever concerned with comparing ratios of power.
    If you have two powers, you can compare them in dB by using the formula:

    10 log10(power ratio)

    For the special case where you know two voltages across the same impedance, you can use the 20 log10(voltage ratio) formula to get the power ratio between them in dB.
    The 20 in the formula takes care of the squaring of voltages you would normally do to work out the powers.

    There are a lot of special units where you compare the power you have with some reference.
    It might be milliwatts, microwatts or something else.
    There is a list of them here:
    http://en.wikipedia.org/wiki/Decibel

    You just use them the same way as above except that one of the measurements is already decided for you and you provide the one you have measured.
     
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