The given TFs are: 1 1() s = s s + 1 s + G p, () s ( )( ) >> Gp1=tf(1,ply([0-1 -])) Transfer functin: 1 ----------------- s^ + s^ + s Rt lcus G 1 = p ( s + 0.8 + j)( s + 0.8 j) >> Gp=tf(1,ply([-0.8-*i -0.8+*i])) Transfer functin: 1 ------------------ s^ + 1.6 s + 9.6 >> G=Gp1*Gp Transfer functin: 1 ----------------------------------------------- s^5 +.6 s^ + 16. s^ +.1 s^ + 19.8 s Using Matlab: >> rlcus(g) 8 Rt Lcus 6 Imaginary Axis 0 - - -6-8 -8-6 - - 0 6 Real Axis Figure 1 Rt lcus prduced using rlcus(.)
The graphical methd: The pen lps ples (there are n zers) are bviusly 0, -1, - frm the 1 st TF and - 0.8±j frm the nd TF but it can als be fund using Matlab: >> [p,z]=pzmap(g) p = 0-0.80000000000000 +.00000000000000i -0.80000000000000 -.00000000000000i -.00000000000001-1.00000000000000 z = Empty matrix: 0-by-1 Nte: The steps shwn here are at a different sequence that thse presented in yur handut (chapter, last page). Step 1: Draw the OL ples and zers: Figure OL ples and zers Step : Determine which parts f the real axis belng t the RL: Figure Real axis part dk Step : The Break in/ut pints can be fund frm = 0 ds k = s s + 1 s + s + 0.8 + j s + 0.8 j the CL CE: ( )( )( )( ) where k(s) is fund using
It will be easier if we first d the multiplicatin. This prduct is already calculated frm: >> G Transfer functin: 1 ----------------------------------------------- s^5 +.6 s^ + 16. s^ +.1 s^ + 19.8 s T islate the OL denminatr: >> [n, d]=tfdata(g,'v') n = 0 0 0 0 0 1 d = Clumns 1 thrugh 1.00000000000000.60000000000000 16.000000000000.1000000000001 Clumns 5 thrugh 6 19.8000000000000 0 T speed up the calculatins (fr ur system) we have that k=-ol Denminatr and hence: >> dd=[5*d(1) *d() *d() *d() d(5)] dd = Clumns 1 thrugh 5.00000000000000 18.0000000000000 9.000000000001 6.000000000001 Clumn 5 19.8000000000000 Nte: Be very careful when yu use the abve trick!!! dk The slutin f = 0 : ds >> rts(dd) ans = -0.85660095911 +.691796i -0.85660095911 -.691796i -1.59590681996-0.1798118 Hence the accepted slutin (break ut pint) is the last ne as the -1.59 is a pint n the real that is nt in the RL (use yur muse pinter n the Matlab figure t crsscheck yur answer).
Step : There are axis: n p n z = 5 Figure Break ut pint asympttes and the pint f intersectin with the real The ples are: >> pj=-p pj = 0 0.80000000000000 -.00000000000000i 0.80000000000000 +.00000000000000i.00000000000001 1.00000000000000 And hence: >> sum(pj)/5 ans = 0.9000000000000 Hence s=-0.9 The angles f the asympttes with the real axis are: # asymptte Angle 1 180 / 5 = 6 180 / 5 = 108 5 180 / 5 = 180 7 180 / 5 = 5 r 108 5 9 180 / 5 = r 6 The last can als be fund using the symmetry f the RL.
Figure 5 Asympttes Step 5: The angle f departure frm the cmplex ples is fund using the fllwing drawing: 0.8 + j θ θ θ 1 θ = tan θ = tan tan 1 1 1 0.8 = 68. 1. = 86. 0. = 75 θ = 180 Figure 6 Calculatin f departure angle 75 = 105 S the angle f departure is 180 68. 86. 105 90 = 169.
169. 169. Figure 7 Angle f departure Step 6: The last pint is the pint f intersectin with the imaginary axis. T d that we place at the CL CE s=jω: s 5 +. 6s 5 + 16. s + 1. s jω +. 6ω 16. jω 1. ω + 19.8 jω + k = 0 5 ω 16. ω + 19.8ω = 0. 6ω 1. ω + k ω 16. ω + 19.8 = 0. 6ω 1. ω + k >> rts([1 0-16. 0 19.8]) ans =.897818599817 -.897818599818 1.179770751-1.179770751 T find the accepted value: + 19.8s + k = 0 >> w=.897818599817; k=-.6*w^+.1*w^ k = -5.71568197e+00 >> w=1.179770751; k=-.6*w^+.1*w^ k =.9786819768 Hence the accepted value is 1.179770751 and hence the gain is. Nw we have all that we need t sketch the rt lcus:
169. 1.1 1.1 169. Figure 8
The given TFs are: 1 G p 1() s =, () ( )( ) s( s + 1)( s + ) G s + + 5 j s + 5 j s p = s + 0.8 + j s + 0.8 j Nte: This is a difficult ne >> gp1=tf(1,ply([0-1 -])) Transfer functin: 1 ----------------- s^ + s^ + s ( )( ) >> gp=tf(ply([-0.8-*i -0.8+*i]),ply([-0.8-*i -0.8+*i])) Transfer functin: s^ + 1.6 s + 9.6 ------------------ s^ + 1.6 s + 9.6 >> gp=tf(ply([--5*i -+5*i]),ply([-0.8-*i -0.8+*i])) Transfer functin: s^ + s + 9 ------------------ s^ + 1.6 s + 9.6 >> g=gp1*gp Transfer functin: s^ + s + 9 ----------------------------------------------- s^5 +.6 s^ + 16. s^ +.1 s^ + 19.8 s Using Matlab: >> rlcus(g) 8 Rt Lcus 6 Imaginary Axis 0 - - -6 The graphical methd: -8-10 -5 0 5 Real Axis Figure 9 Rt lcus prduced using rlcus(.)
The pen lps ples are bviusly 0, -1, - frm the 1 st TF and -0.8±j frm the nd TF. The zers are -±5j but it can als be fund using Matlab: >> [p,z]=pzmap(g) >> [p,z]=pzmap(g) p = 0-0.80000000000000 +.00000000000000i -0.80000000000000 -.00000000000000i -.00000000000001-1.00000000000000 z = -.00000000000000 + 5.00000000000000i -.00000000000000-5.00000000000000i Nte: The steps shwn here are at a different sequence that thse presented in yur handut (chapter, last page). Step 1: Draw the OL ples and zers: Figure 10 OL ples and zers Step : Determine which parts f the real axis belng t the RL:
Figure 11 Real axis part dk Step : The Break in/ut pints can be fund frm = 0 ds s( s + 1)( s + )( s + 0.8 + j)( s + 0.8 j) the CL CE: k = ( s + + 5 j)( s + 5 j) (nte: try t clearly understand the fllwing steps) >> [n, d]=tfdata(g,'v') dd=[5*d(1) *d() *d() *d() d(5)] nd=[5*d(1) *d() *d() *d() d(5)] where k(s) is fund using rts(cnv(dd,n)-cnv(d,nd)) n = 0 0 0 1 9 d = Clumns 1 thrugh 1.00000000000000.60000000000000 16.000000000000.1000000000001 Clumns 5 thrugh 6 19.8000000000000 0 dd = Clumns 1 thrugh 5.00000000000000 18.0000000000000 9.000000000001 6.000000000001 Clumn 5 19.8000000000000 nd = Clumns 1 thrugh 5.00000000000000 18.0000000000000 9.000000000001 6.000000000001 Clumn 5 19.8000000000000 ans = -0.786910810 +.06190965908i -0.786910810 -.06190965908i -0.85660095911 +.691797i
-0.85660095911 -.691797i -1.909506118 + 0.919967866i -1.909506118-0.919967866i -1.59590681997 0.655690980787-0.1798118 Hence the accepted slutin (break ut pint) is -0.1 (use yur muse pinter n the Matlab figure t crsscheck yur answer). Figure 1 Break ut pint Step : There are n n = 5 = asympttes and the pint f intersectin with the real axis: p z >> pj=-p pj = 0 0.80000000000000 -.00000000000000i 0.80000000000000 +.00000000000000i.00000000000001 1.00000000000000 >> zj=-z zj =.00000000000000-5.00000000000000i.00000000000000 + 5.00000000000000i >> -sum(pj)/sum(pi) ans = -1.65765 Hence s=-1.6
The angles f the asympttes with the real axis are: # asymptte Angle 180 / = 60 1 180 / = 180 5 180 / = 00 r 60 The last can als be fund using the symmetry f the RL. Figure 1 Asympttes Step 5: The angle f departure frm the cmplex ples is fund using the fllwing drawing:
φ 1 + 5 j 0.8 + j θ θ θ 1 0.8 j φ 5 j Figure 1 Calculatin f departure angle 1 θ = tan = 68. 1. 1 θ = tan = 86. 0. 1 tan = 75 θ = 180 75 = 105 0.8 1 + 5 φ = tan = 81.5 1. 1 5 tan = 59 φ1 = 70 + 59 = 9 1. S the angle f departure is 180 68. 86. 105 90 + 81.5 + 9 = 1 Figure 15 Angle f departure
Step 6: Angle f arrival at cmplex zer: + 5 j θ 0.8 + j θ θ 1 θ 0.8 j 5 j Figure 16 1 5 tan = 59 θ = 180 59 = 11 1. 1 5 + tan = 81.5 θ = 98.5 1. 1 5 tan = 78.7 θ = 101. 1 1 5 tan = 68. θ1 = 111.8 180-90 + 90 + 11 + 98. 5 + 101. + 1118. = 61.6 r 5.6
Figure 17 Angle f arrival Step 7: The last pint is the pint f intersectin with the imaginary axis. T d that we place at the CL CE s=jω: s 5 +. 6 s 5 + 16. s + 1. s + 19. 8s + K jω +. 6 ω 16. jω 1. ω + 19. 8 5 5 ω 16. ω + 19. 8ω + Kω = 0. 6ω 1. ω Kω + 9K = 0 ω 16. ω + 19. 8 + K = 0. 6ω 1. ω Kω + 9K = 0 ( s + s+ 9) = 0 jω + K( ω + jω+ 9) = 0 jω +. 6 ω 16. jω 1. ω + 19. 8 jω Kω + Kjω+ 9K = 0 T slve that I will use the cmmand slve(.) >> f=slve('w^-16.*w^+19.8+*k','.6*w^-.1*w^-w^*k+9*k') f = k: [6x1 sym] w: [6x1 sym] >> f.k ans = 1.505109555167589157 1.505109555167589157 8.0667857891865596+8.80111761950855096110*i 8.0667857891865596+8.80111761950855096110*i 8.0667857891865596-8.80111761950855096110*i 8.0667857891865596-8.80111761950855096110*i >> f.w ans = -1.19956855060155680770558 1.19956855060155680770558 -.91896819517900777067009119+1.6901167950099570160955*i.91896819517900777067009119-1.6901167950099570160955*i -.91896819517900777067009119-1.6901167950099570160955*i.91896819517900777067009119+1.6901167950099570160955*i
Hence the gain in 1.5 and the pint is 1.1 rad/s Nw we have all that we need t sketch the rt lcus: Figure 18 Further exercises 1) Repeat the riginal exercise but with G p ( s) = ( s + + 5 j)( s + 5 j). ) Repeat the riginal exercise but with G 1 () s p = ( s + 0.5 + 0. j)( s + 0.5 0. j). ) Repeat the riginal exercise but with G () ( s + 1. + j)( s + 1. j) s p = s + 0.8 + j s + 0.8. j ( )( ) Nte: it is EXTREMELY imprtant that yu d and fully UNDERSTAND these exercises.