Bode plot

Discussion in 'Homework Help' started by dilligaf, May 14, 2010.

  1. dilligaf

    Thread Starter New Member

    Dec 15, 2009
    I have a problem and need to handdraw a bode plot.
    3.16/[(1+2s)s] This is from the Jacobs Industrial Control Electronics.
    They do not explain it with zeros and poles.
    I am lost in their explanation.
  2. mik3

    Senior Member

    Feb 4, 2008
    To be able to plot it by hand correctly it has to be in pole-zero form.

    Your equation can be written as:


    Thus the DC gain is 1.58

    two zeros at infinity

    one pole at -0.5 and one pole at 0
    dilligaf likes this.
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
    A pole at the origin (DC) has infinite gain at DC.
  4. mel8030

    New Member

    Dec 24, 2009
    I hope the book is note using jw, w0, wn, absolutes and that squiggly notation?

    Write the terms in the form of (s+n), where n is a constant, including 0
    H(s) = 1.58 / [ (s+.5)(s+0)] {form (s+0) helps me avoid mistakes}

    BEEN MANY YEARS AGO since I worked with bodes. Could be errors that other should find

    Let s = 0 and evaluate function to get DC gain,---> DC gain = infinite
    (s+0) has no break point and its in the dominator, so the graph won't start out with a horizontal line (infinite gain was also indicating no horizontal). We do know any (s+n) in denominator will get -20 dB/dec slope .

    We have to start the graph somewhere so grab s=1 (20log1=0 , s=1 so dB=0) and then back up a tenth of s=1 for the start of graph----> s=.1

    (s+.5) has the breakpoint at s=.5 and and gets -20 dB/dec

    The graph starts out at .1 s, sloping down at 20 dB /dec, and would pass though (1,0) . Sloping line gets redirected at s=.5 when another downward 20 dB/dec kicks in for a total -40 dB /dec and continues to the edge of graph.

    note: ALL homework I have done use whole numbers greater than 1 (not .5) so (1,0) makes a good reference point for the line to pass though.

    poles are at -.5 and 0
    Last edited: May 16, 2010
  5. mik3

    Senior Member

    Feb 4, 2008
    That is right Ron, I wanted to say the nominator value is 1.58.