new equation of a function in semi-logarithm axis

Discussion in 'Math' started by mothermohammad, Oct 4, 2017.

  1. mothermohammad

    Thread Starter Member

    May 17, 2014
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    First, we plot the y=x, which is linear. now we want to plot this equation in semi-logarithm scale, in which the horizontal (x) is drawn logarithm. My question is what is the new equation of the y=x in the new plan. For more clarification please see these pictures.
    the y=x is
    upload_2017-10-4_11-35-14.png

    and the y=x in semi logarithm is
    upload_2017-10-4_11-35-35.png
    what is the equation of above curve?
     
  2. AlbertHall

    Distinguished Member

    Jun 4, 2014
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    The equation is still y=x. It is only the graph scaling that has changed. The data is the same, only the display of the data has changed.
     
  3. mothermohammad

    Thread Starter Member

    May 17, 2014
    30
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    Thank you. But, I want to know what is the new equation. For example in the new curve, I would like to calculate the slope of the curve in (1,1) point.
     
  4. AlbertHall

    Distinguished Member

    Jun 4, 2014
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    If you read the values from the scales on the graph the slope has not changed. It is still 1.
    To calculate the slope at (2,2) we get the rise and run between the points either side of that point - (1,1) and (3,3). Rise=2 and run=2, so rise/run=1.

    Only if you read values from the X-axis using some imposed linear scale - perhaps measuring the X values from the graph using a ruler - do you get anything else. In that case the curve will seem to be an exponential.
     
  5. mothermohammad

    Thread Starter Member

    May 17, 2014
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    yes. My mean is the slope of the exponential curve.
    Indeed, I have a problem with the phase of the filters like low pass filter in the bode diagram.
     
  6. AlbertHall

    Distinguished Member

    Jun 4, 2014
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    Then the equation of the line, and therefore the slope, depends on what scale you pretend the X-axis has.
     
  7. mothermohammad

    Thread Starter Member

    May 17, 2014
    30
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    OK. I understand.
    In the Erickson book, there is a low pass filter that its phase change from 0 to -90 degree. The low pass filter equation is G(jw)=1/(1+j(w/w0)), which w0 denotes the radian centre frequency of the filter in which the phase of the filter is -45 degree. The phase of the filter is -tang-1(w/w0).
    However, there are two points that their slop is equal to the slop of filter phase in w0. In the book, these two points are e^(-pi/2)w0 and e^(pi/2)w0. I want to know how these points are calculated. I sent the pic for clarification.
     
  8. AlbertHall

    Distinguished Member

    Jun 4, 2014
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    That's a question for someone else I'm afraid.
     
  9. mothermohammad

    Thread Starter Member

    May 17, 2014
    30
    1
  10. mothermohammad

    Thread Starter Member

    May 17, 2014
    30
    1
    how we can detect the fa and fb
    upload_2017-10-4_18-12-14.png
     
  11. MrAl

    Distinguished Member

    Jun 17, 2014
    3,754
    791
    Hi,

    I am not 100 percent sure what you want here, but if you use a conformal mapping of the space you get this:
    log(a)/log(10)+j*(a+atan2(0,a))/log(10)

    and what this gives you is a complete solution where you interpret the real part as units on the x axis and the imaginary part as the y axis. This spaces the x axis 1,10,100, 1000 as 0, 1, 2 ,3 which is just a long way of saying it's the log base 10.
    So for your straight line we for a=0.01 we get the point (-2,0) and for a=0.1 we get (-1,0.1) and for a=1 we get (0,1) and for a=10 we get (1,10), etc.
     
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  12. mothermohammad

    Thread Starter Member

    May 17, 2014
    30
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    Thank you. How you achieved this relation?
    I am thankful if you see my another question in this link.

    https://forum.allaboutcircuits.com/threads/the-value-of-fa-and-fb-in-the-book-of-erickson.140711/
     
  13. MrAl

    Distinguished Member

    Jun 17, 2014
    3,754
    791
    Hi,

    It was a trick of sorts. Just take the 'y' value to be the imaginary part and then take the log base 10 of the real part, then convert the whole thing into just anther complex number.
    BTW the function 'log()' in that previous post is the natural log function.
     
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