EECE 301 Signals & Systems Prof. Mark Fowler
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1 EECE 301 Signals & Systems Prof. Mark Fowler C-T Systems: Bode Plots Note Set #36 1/14
2 What are Bode Plots? Bode Plot = Freq. Resp. plotted with H() in db on a log frequency axis. Its easy to use computers to make Bode plots we already saw that! Use MATLAB s freqs command But good engineers need insight to: - understand the results of an analysis - make decisions for design They do also include phase plots but we ll focus here on the magnitude part Learning how to make by hand a Bode magnitude plot will help give you the needed insight! /14
3 Key Idea of the Method Consider a Freq Resp Magntiude factored into terms of its poles and zeros K jz1 jz ( j) ( 1 n,1 ) jn,1 H ( ) j p j p ( j) ( ) j 1 n, n, Complex conjugate roots are kept together Convert to db form and use log property: (db converts multiplications to additions!) log(ab) = log(a) + log(b) log(a/b) = log(a) log(b) 0log H ( ) 0log K 0log j z 0log ( j) ( ) j 1 1 n,1 n,1 0log j p 0log ( j) ( ) j 1 n, n, So Key Idea is that we have a sum of terms and only need to figure out two things: 1. What does each of these terms look like?. How do we add them together? 3/14
4 One Little Trick First After factoring: H() s Note zero at origin (might not have one... Or could be pole at origin) Ks s s s s ( s5)( s0)( s 6s6,400) ( )( )( 60 40,000) Complex conjugate roots are kept together Our trick is to pull out constants for each term (except a pole or origin) like this: H() s 40,000 ss ( 1)( s 1) s00 (60 40,000) s 1 K 5 0 6,400 ( s 5 1)( s 0 1) s 80 (6 6,400) s1 H ( ) K 40,000 j( j 1)( j 1) j 00 (60 40,000) j , 400 ( j 51)( j 01) j 80 (6 6,400) j1 The whole point of doing this is so that each term (once in db form) is at 0 db at low frequecies this will make it easy to add the db terms together! 4/14
5 What the Terms Look Like Plotted 0log K Just a constant Flat Plot 0log ( j) n An n th order Origin (Plot shown for n = ) Cross 0 = 1 n 0log ( j) 0 db/dec 0log ( j) (rad/sec) 5/14
6 0log j a1 1 st order a log j a1 Straight-Line Approx Flat at 0 db until = a Then break to a slope of 0 db/dec 0-0 0log j a = a 4 5 (rad/sec) 0log ( j) ( ) j1 n 1 nd order n Straight-Line Approx Flat at 0 db until = n Then break to a slope of 40 db/dec n Different values cause different spike levels 0 db/dec (rad/sec) = n 0log 40 db/dec 0log 6/14
7 What This Does For Us Notice that the 1 st and nd Order Non-Origin terms contribute nothing (i.e., they add 0 db) at frequencies below their break points So if we start below all break points it is as if these terms don t exist So if we start plotting at a low enough frequency all we have to do is plot the effect of the constant term and any origin-located poles/zeros (if any exist). Then as we go up in frequency at each break point we change the slope by the amount of slope the new term provides! Sum of two Lines has Slope equal to Sum of Slopes! 7/14
8 General Steps to Sequentially Build Bode Plots 1. Factor H(s) leave complex-root terms as quadratics. Convert to j form 3. Pull out constants into a gain term 4. Combine constant term with j terms (if any) 5. Identify break points and put in ascending order 6. Plot constant term with j terms at values below the lowest break point 7. At break point, change slope by 0dB/decade or 40dB/decade for 1 st order or nd order terms, repectively. Repeat this step through ordered list of breakpoints. 8. Make resonant corrections for under damped nd order terms (i.e. when < 0.5). 8/14
9 Example 1. Factor: H() s H() s s s s s s s s 5s 00s 0.1 s( s ) 3 3 s 4s 4s00 s 4s 4s ss ( 50)( s 00) ( s)( s s0) >> roots([ ]) ans = Convert to j: >> roots([ ]) ans = i i Leave as nd order term 0.1 j( j 50)( j 00) H ( ) ( j )(( j) j 0) s n n n s n Pull Out Constants : H ( ) Gain Term j( j 50)( j 00) H ( ) ( j )(( j) j 0) j1 j / 501 j / 00 1 j / 1 j /0 j / 9/14
10 4. Combine gain term with j term : H ( ) (5 j) 1 j / 501 j / 00 1 j / 1 j /0 j / 5. Identify Breakpoints and List in Ascending Order: H ( ) (5 j) 1 j / 501 j / 00 1 j / 1 j /0 j / List breakpoints in ascending order: Break Points Change in slope -0dB/decade 1 st order term in denominator -40dB/decade nd order term in denominator 50 +0dB/decade 1 st order term in numerator 00 +0dB/decade 1 st order term in numerator /14
11 6. Plot constant term with j terms at values below the lowest break point : H ( ) (5 j) 1 j / 501 j / 00 1 j / 1 j /0 j / -Evaluate 5j in db at value that is (at least) 1 decade below the lowest BP (Here used = 0.1 which is more than a decade below ): 0 log(5 0.1) 0log (0.5) 6dB -Plot a point at -6 db at = 0.1 -Draw a line of slope +0dB/decade from this point up to the first BP It is + slope because here the j term is in numerator It would be slope if it were in the denomator It is 0 db slope because here there is a first order j term It would be px0 slope if it were (j) p 11/14
12 7. At break point, change slope by 0dB/decade or 40dB/decade for 1st order or nd order terms, repectively. : Break Points Change in slope -0dB/decade 1 st order term in denominator -40dB/decade nd order term in denominator 50 +0dB/decade 1 st order term in numerator 00 +0dB/decade 1 st order term in numerator 8. Make resonant corrections for under damped nd order terms (i.e. when < 0.5). : Finally: Make adjustment for the value from the plot of the nd order term: = 0.1 gives peak 14dB up value Adjustment db 0. 8 db db db db 1/14
13 Approximate Bode Plot for Example in Notes 60 db 40 db Adjust for = 0.1 H() (db) 0 db 0 db -0 db up 1 decade up 0 db up 1 decade down 40 db down 0 db up 1 decade -40 db ,000 (rad/sec) 13/14
14 60 Exact Bode Plot for Example H() (db) Approximate Exact (rad/sec) 14/14
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