Facebook
Google
GitHub
Linkedin
Bode plot
- Thread starter dilligaf
- Start date
To be able to plot it by hand correctly it has to be in pole-zero form.
Your equation can be written as:
3.16/[2(0.5+s)s]=1.58/[(0.5+s)s]
Thus the DC gain is 1.58
two zeros at infinity
one pole at -0.5 and one pole at 0
A pole at the origin (DC) has infinite gain at DC.
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
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:
That is right Ron, I wanted to say the nominator value is 1.58.
