Root locus ( )( ) The given TFs are: 1. Using Matlab: >> rlocus(g) >> Gp1=tf(1,poly([0-1 -2])) Transfer function: s^3 + 3 s^2 + 2 s

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1 The given TFs are: 1 1() s = s s + 1 s + G p, () s ( )( ) >> Gp1=tf(1,ply([0-1 -])) Transfer functin: s^ + s^ + s Rt lcus G 1 = p ( s j)( s j) >> Gp=tf(1,ply([-0.8-*i -0.8+*i])) Transfer functin: s^ s >> G=Gp1*Gp Transfer functin: s^5 +.6 s^ s^ +.1 s^ s Using Matlab: >> rlcus(g) 8 Rt Lcus 6 Imaginary Axis Real Axis Figure 1 Rt lcus prduced using rlcus(.)

2 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 = i i 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 j s j the CL CE: ( )( )( )( ) where k(s) is fund using

3 It will be easier if we first d the multiplicatin. This prduct is already calculated frm: >> G Transfer functin: s^5 +.6 s^ s^ +.1 s^ s T islate the OL denminatr: >> [n, d]=tfdata(g,'v') n = d = Clumns 1 thrugh Clumns 5 thrugh 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 Clumn Nte: Be very careful when yu use the abve trick!!! dk The slutin f = 0 : ds >> rts(dd) ans = i i Hence the accepted slutin (break ut pint) is the last ne as the is a pint n the real that is nt in the RL (use yur muse pinter n the Matlab figure t crsscheck yur answer).

4 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 = i i And hence: >> sum(pj)/5 ans = Hence s=-0.9 The angles f the asympttes with the real axis are: # asymptte Angle / 5 = / 5 = / 5 = / 5 = 5 r / 5 = r 6 The last can als be fund using the symmetry f the RL.

5 Figure 5 Asympttes Step 5: The angle f departure frm the cmplex ples is fund using the fllwing drawing: j θ θ θ 1 θ = tan θ = tan tan = = = 75 θ = 180 Figure 6 Calculatin f departure angle 75 = 105 S the angle f departure is = 169.

6 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 s s + 1. s jω +. 6ω 16. jω 1. ω jω + k = 0 5 ω 16. ω ω = 0. 6ω 1. ω + k ω 16. ω = 0. 6ω 1. ω + k >> rts([ ]) ans = T find the accepted value: s + k = 0 >> w= ; k=-.6*w^+.1*w^ k = e+00 >> w= ; k=-.6*w^+.1*w^ k = Hence the accepted value is and hence the gain is. Nw we have all that we need t sketch the rt lcus:

7 Figure 8

8 The given TFs are: 1 G p 1() s =, () ( )( ) s( s + 1)( s + ) G s j s + 5 j s p = s j s j Nte: This is a difficult ne >> gp1=tf(1,ply([0-1 -])) Transfer functin: s^ + s^ + s ( )( ) >> gp=tf(ply([-0.8-*i -0.8+*i]),ply([-0.8-*i -0.8+*i])) Transfer functin: s^ s s^ s >> gp=tf(ply([--5*i -+5*i]),ply([-0.8-*i -0.8+*i])) Transfer functin: s^ + s s^ s >> g=gp1*gp Transfer functin: s^ + s s^5 +.6 s^ s^ +.1 s^ s Using Matlab: >> rlcus(g) 8 Rt Lcus 6 Imaginary Axis The graphical methd: Real Axis Figure 9 Rt lcus prduced using rlcus(.)

9 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 = i i z = i i 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:

10 Figure 11 Real axis part dk Step : The Break in/ut pints can be fund frm = 0 ds s( s + 1)( s + )( s j)( s j) the CL CE: k = ( s 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 = d = Clumns 1 thrugh Clumns 5 thrugh dd = Clumns 1 thrugh Clumn nd = Clumns 1 thrugh Clumn ans = i i i

11 i i i 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 = i i >> zj=-z zj = i i >> -sum(pj)/sum(pi) ans = Hence s=-1.6

12 The angles f the asympttes with the real axis are: # asymptte Angle 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:

13 φ j j θ θ θ j φ 5 j Figure 1 Calculatin f departure angle 1 θ = tan = θ = tan = tan = 75 θ = = φ = tan = tan = 59 φ1 = = 9 1. S the angle f departure is = 1 Figure 15 Angle f departure

14 Step 6: Angle f arrival at cmplex zer: + 5 j θ j θ θ 1 θ 0.8 j 5 j Figure tan = 59 θ = = tan = 81.5 θ = tan = 78.7 θ = tan = 68. θ1 = = 61.6 r 5.6

15 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 s s + 1. s s + K jω +. 6 ω 16. jω 1. ω ω 16. ω ω + Kω = 0. 6ω 1. ω Kω + 9K = 0 ω 16. ω 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 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 = *i *i *i *i >> f.w ans = *i *i *i *i

16 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 j)( s + 5 j). ) Repeat the riginal exercise but with G 1 () s p = ( s j)( s j). ) Repeat the riginal exercise but with G () ( s j)( s + 1. j) s p = s j s j ( )( ) Nte: it is EXTREMELY imprtant that yu d and fully UNDERSTAND these exercises.