Topic #21 16.31 Feedback Control MIMO Systems Singular Value Decomposition Multivariable Frequency Response Plots Copyright 2001 by Jonathan How
Topic #21 16.31 Feedback Control MIMO Systems • Singular Value Decomposition • Multivariable Frequency Response Plots Copyright 2001 by Jonathan How. 1
Fal!2001 16.3121-1 Multivariable Frequency Response In the MIMO case, the system G(s)is described by a p m trans fer function matrix (TFM) Still have that =C(sI-A)B+D But G(s)A, B, C, D MUCH less obvious than in SISO case Discussion of poles and zeros of mimo systems also much more complicated e In siso case we use the bode plot to develop a measure of the system“size Given z=Gu, where G(jw)=G (jw) lejo(u) Then w=lejwit+)applied to G(jw)lejo(u)yields lullGGwn)lej(ant++o(w1)=I Amplification and phase shift of the input signal obvious in the SISO case ● MIMO extension? Is the response of the system large or small? /s0 010
Fall 2001 16.31 21—1 Multivariable Frequency Response • In the MIMO case, the system G(s) is described by a p×m transfer function matrix (TFM) — Still have that G(s) = C(sI − A) −1B + D — But G(s) → A, B, C, D MUCH less obvious than in SISO case. — Discussion of poles and zeros of MIMO systems also much more complicated. • In SISO case we use the Bode plot to develop a measure of the system “size”. — Given z = Gw, where G(jω) = |G(jω)|ejφ(w) — Then w = |w|ej(ω1t+ψ) applied to |G(jω)|ejφ(w) yields |w||G(jω1)|ej(ω1t+ψ+φ(ω1)) = |z|ej(ω1t+ψo) ≡ z — Amplification and phase shift of the input signal obvious in the SISO case. • MIMO extension? — Is the response of the system large or small? G(s) = ∙ 103/s 0 0 10−3/s ¸
Fal!2001 16.3121-2 For MIMO systems, cannot just plot all of the Gi; elements of G Ignores the coupling that might exist between them So not enlightening Basic MIMo frequency response Restrict all inputs to be at the same frequency Determine how the system responds at that frequenc See how this response changes with frequency So inputs are w= wceJut, where Wc Ec Then we get z=Gs0,→2= zceu and zo∈C We need only analyze ac=G(jw)c As in the siso case, we need a way to establish if the system re- sponse is large or small How much amplification we can get with a bounded input Consider 2c=G(w wc and set lwcll2=whC 1. What can we say about the - Answer depends on w and on the direction of the input wc Best found using singular values
Fall 2001 16.31 21—2 • For MIMO systems, cannot just plot all of the Gij elements of G — Ignores the coupling that might exist between them. — So not enlightening. • Basic MIMO frequency response: — Restrict all inputs to be at the same frequency — Determine how the system responds at that frequency — See how this response changes with frequency • So inputs are w = wcejωt , where wc ∈ Cm — Then we get z = G(s)|s=jω w, ⇒ z = zcejωt and zc ∈ Cp — We need only analyze zc = G(jω)wc • As in the SISO case, we need a way to establish if the system response is large or small. — How much amplification we can get with a bounded input. • Consider zc = G(jω)wc and set kwck2 = pwc Hwc ≤ 1. What can we say about the kzck2? — Answer depends on ω and on the direction of the input wc — Best found using singular values
Fal!2001 16.3121-3 Singular value decomposition Must perform the SVD of the matrix G(s)at each frequency s=ju G(ju)∈CpmU∈C∑∈RmV∈Cnxm G=U∑Vh UHU=IUUH=I VHV=I VVH=I and 2 is diagonal Diagonal elements ok 20 of 2 are the singular values of G 入(GG)oro i(GGH the positive ones are the same from both formulas The columns of the matrices U and V(; and v;)are the asso- ciated eigenvectors G Gu GGHa o2u 0U If the rank(G)=r0,k=1,,r Ok=0,k=r+1,., min(p, m An SVD gives a very detailed description of how a ma trix(the system G) acts on a vector (the input w) at a particular frequency
Fall 2001 16.31 21—3 Singular Value Decomposition • Must perform the SVD of the matrix G(s) at each frequency s = jω G(jω) ∈ Cp×m U ∈ Cp×p Σ ∈ Rp×m V ∈ Cm×m G = UΣV H — U HU = I, UU H = I, V HV = I, V V H = I, and Σ is diagonal. — Diagonal elements σk ≥ 0 of Σ are the singular values of G. σi = q λi(GHG) or σi = q λi(GGH) the positive ones are the same from both formulas. — The columns of the matrices U and V (ui and vj) are the associated eigenvectors GHGvj = σj 2 vj GGHui = σi 2 ui Gvi = σiui • If the rank(G) = r ≤ min(p, m), then — σk > 0, k = 1,..., r — σk = 0, k = r + 1,..., min(p, m) • An SVD gives a very detailed description of how a matrix (the system G) acts on a vector (the input w) at a particular frequency
Fal!2001 16.3121-4 So how can we use this result? Fix the size wcl2= 1 of the input, and see how large we can make the output Since we are working at a single frequency, we just analyze the relation ze=Guc,Gu≡G(s=ju) Define the maximum and minimum amplifications as 0≡maNx.‖zcll g≡min.‖ll Then we have that (let q=min(p, m)) qp≥mtal 0p<m“wide Can use O and g to determine the possible amplification and atten uation of the input signals Since G(s) changes with frequency, so will o and g
Fall 2001 16.31 21—4 • So how can we use this result? — Fix the size kwck2 = 1 of the input, and see how large we can make the output. — Since we are working at a single frequency, we just analyze the relation zc = Gwwc, Gw ≡ G(s = jω) • Define the maximum and minimum amplifications as: σ ≡ max kzck2 kwck2=1 σ ≡ min kzck2 kwck2=1 • Then we have that (let q = min(p, m)) σ = ½ σ1 σq p ≥ m “tall” σ = 0 p < m “wide” • Can use σ and σ to determine the possible amplification and attenuation of the input signals. • Since G(s) changes with frequency, so will σ and σ
Fal!2001 16.3121-5 SVD Example ●( Consider( wide case 100 500 101「500 010 00.50 0100.50 001 U∑L rh so that 01=5 and 0?=0.5 ≡max.‖Gal=5 mIn Gnl2=0≠ Jwc 2=1 But now consider(tall case) 50 100 10 00.5 01000.5 01 00 001」L00」 U∑VH o that 01=5 and 02=0.5 still G三max lwcl2= IGoWcl2=5=01 min.|Gl2=0.5=02
Fall 2001 16.31 21—5 SVD Example • Consider (wide case) G w = ∙ 500 0 0.5 0 ¸ = ∙ 1 0 0 1 ¸ ∙ 500 0 0.5 0 ¸ 100 010 001 = UΣV H so that σ1 = 5 and σ2 = 0.5 σ ≡ max kGwwck2 =5= σ1 kwck2=1 σ ≡ min kGwwck2 = 0 6= σ2 kwck2=1 • But now consider (tall case) G˜w = 5 0 0 0.5 0 0 = 100 010 001 5 0 0 0.5 0 0 ∙ 1 0 0 1 ¸ = UΣV H so that σ1 = 5 and σ2 = 0.5 still. σ ≡ max kGwwck2 =5= σ1 kwck2=1 σ ≡ min kwck2=1 kGwwck2 = 0.5 = σ2
Fal!2001 16.3121-6 For MIMO systems, the gains(or o's are only part of the story, as we must also consider the input direction To analyze this point further, note that Gn=U∑Vn=[u H H ilio Assumed tall case for simplicity, so p> m and q= m Can now analyze impact of various alternatives for the input Only looking at one frequency, so the basic signal is harmonic But, we are free to pick the relative sizes and phases of each of the components of the input vector wc 3 These define the input direction
Fall 2001 16.31 21—6 • For MIMO systems, the gains (or σ’s) are only part of the story, as we must also consider the input direction. • To analyze this point further, note that we can rewrite σ1 ... σm 0 vH 1 . . . vH m G w = UΣV H = £ u1 ... up ¤ = X m i=1 H σiuivi — Assumed tall case for simplicity, so p > m and q = m • Can now analyze impact of various alternatives for the input — Only looking at one frequency, so the basic signal is harmonic. — But, we are free to pick the relative sizes and phases of each of the components of the input vector wc. 3 These define the input direction
Fal!2001 16.3121-7 For example, we could pick wc =v1, then sInce v H Output amplified by o1. The relative sizes and phases of each of the components of the output are given by the vecto By selecting other input directions(at the same frequency), we can get quite different amplifications of the input signal e1 Gu is small if o(Gu<1 MO frequency response are plots of o jw) and g(jw Then use the singular value vectors to analyze the response at a particular frequency
Fall 2001 16.31 21—7 • For example, we could pick wc = v1, then à ! X m H zc = Gwwc = σiuivi v1 = σ1u1 i=1 since vi Hvj = δij. — Output amplified by σ1. The relative sizes and phases of each of the components of the output are given by the vector zc. • By selecting other input directions (at the same frequency), we can get quite different amplifications of the input signal kGwwck2 σ ≤ ≤ σ kwck2 • Thus we say that — Gw is large if σ(Gw) À 1 — Gw is small if σ(Gw) ¿ 1 • MIMO frequency response are plots of σ(jω) and σ(jω). — Then use the singular value vectors to analyze the response at a particular frequency
Fal!2001 16.3121-8 Example: Just picked a random 3x3 system -0.7500-2.0000 2.0000-0.7500 0 0 0-00 4.0000 0 1.0000 1.99946.4512 0989 3.45003.3430 7836 4.07815.9542-8.0204 3.5595-6.01231.2865 05650.5287-2.17070.6145 1.41510.2193-0.05920.5077 -0.8051-0.9219-1.01061.6924 0.00440.00920.0041 0.0062 740.009 0.00790.00180.0092 The singular value plot for this state space system is shown below SV Plots Freq(rad/sec) Figure 1: SVD of a typical 3x3 TFM
Fall 2001 16.31 21—8 • Example: Just picked a random 3x3 system a = -0.7500 -2.0000 0 0 2.0000 -0.7500 0 0 0 0 -1.0000 -4.0000 0 0 4.0000 -1.0000 b = -1.9994 6.4512 -0.0989 3.4500 3.3430 -0.7836 4.0781 5.9542 -8.0204 3.5595 -6.0123 1.2865 c = -1.0565 0.5287 -2.1707 0.6145 1.4151 0.2193 -0.0592 0.5077 -0.8051 -0.9219 -1.0106 1.6924 d = 0.0044 0.0092 0.0041 0.0062 0.0074 0.0094 0.0079 0.0018 0.0092 • The singular value plot for this state space system is shown below: SV Plots 102 101 100 10−1 10−2 10−1 100 101 102 Freq (rad/sec) SV Figure 1: SVD of a typical 3x3 TFM
Fal!2001 16.3121-9 Then applied a sinusoid input at about 3 rad / sec using the u di rection (see code on how this is done) aro w 0.4715 047150 0.2684+0.7121i0.76101.9313 0.2411-0.37460445-2.1427 so the three input sinusoids are scaled quite differently and phase shifted quite substantially Best input Gain 16.0751 Time sec) Figure 2 MIMO system response using the v1 input direction. Outputs converge to the expected output directions
Fall 2001 16.31 21—9 • Then applied a sinusoid input at about 3 rad/sec using the v1 direction (see code on how this is done) wc |wc| argwc 0.4715 0.4715 0 −0.2684 + 0.7121i 0.7610 1.9313 −0.2411 − 0.3746i 0.4455 −2.1427 so the three input sinusoids are scaled quite differently and phase shifted quite substantially. Best input. Gain 16.0751 15 10 5 lsim sin Yi 0 −5 −10 −15 0 1 2 3 4 5 6 7 8 9 10 Time sec) Figure 2: MIMO system response using the v1 input direction. Outputs converge to the expected output directions
