ECE 602 Solution to Homework Assignment 4

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1 ECE 6 Lumpd Systms Thory Novmbr 8, 7 ECE 6 Solution to Homwork Assignmnt 4. Us two diffrnt mthods to comput th unit-stp rspons of th systm ẋ Assum zro initial conditions. x + u () y x. () Solution: First lt s us th frquncy domain calculation. W know that whr û(s) /s. W find that y(t) L {ĝ(s)û(s)} ĝ(s) C (si A) B + D () s s + s +. (4) W obtain th partial fraction xpansion of ŷ(s) to b y(s) ˆ / + s/ + 5 s s + s + / s + /(s + /) (s + ) ( /)( ) (s + ) + 4 (5) (6) which has invrs transform y(t) (/)u(t) + (/) t/ cos t/ + ( /) t/ sin t/ (7) Nxt, lt s outlin th procdur for th tim domain calculation. Th charactristic quation of A is s + s + so th ignvalus ar ± j. Taking f(λ) λt and h(λ) β + β λ and quating thm for λ + j yilds At h(a) β I + β A (8) ( t/ cos( t ) + sin( ( t )) ) t/ sin( t ) ( ) t/ sin( t ) ( t/ cos( t ) sin( t )) W can now calculat y(t) C A(t τ) Bu(τ)dτ (9) C A(t τ) Bu(τ)dτ. ()

2 ECE 6 Lumpd Systms Thory Novmbr 8, 7 Taking th constant matrix C insid th intgral mans that w can simplify th intgration by doing th matrix multiplications first and thn intgrating th rsulting scalar function of τ rathr than intgrating ach of th lmnts of th xponntial matrix. Our intgrand is thus ( t/ cos( t ) + sin( ( t )) ( ) t/ sin( t ) ( t/ cos( t t/ ( cos( ( t/ cos( t ) + 5 sin( t )) ( t/ cos( t ) sin( t )) t ) 4 sin( t ) ) ) t/ sin( t ) ) sin( t )) (). () W now nd to valuat ) t (t τ) (t τ) ( (t τ)/ cos( ) 4 sin( ) dτ which bcoms t (t τ) t/ τ/ cos( )dτ 4 t (t τ) t/ τ/ sin( )dτ. Nxt w us th appropriat trigonomtric idntitis to dcompos th sin and cosin of a sum, thn tak all functions of t outsid th intgral. Du to tim constraints, I ll stop hr.. Discrtiz th stat quation of Problm for sampling priods T and T π. Solution: For T w hav A d A and for T π w hav A d Aπ. Bcaus A is invrtibl (it had nonzro dtrminant) w can calculat B d using. Find th companion and modal forms of ẋ B d A (A d I) B. x + u (4) y x. (5) Solution: Th companion form is th rprsntation of th systm with rspct to th basis (b, Ab, A b). With Q b Ab A b 4 (6) 5 ()

3 ECE 6 Lumpd Systms Thory Novmbr 8, 7 our transformation matrix is P Q b, Ab, A b so with Ā PAP, b Pb, and C CP, th quivalnt systm is whr x Px. x 4 x + (7) u (8) ȳ x (9) 4. Find a ralization for th matrix Ĝ(s) s+ s s+ s (s+)(s+) s s+ () Solution: W follow th procdur on pag of th txtbook. Stp : In ordr for Ĝ(s) to b ralizabl, it must b a propr rational matrix, i.. all of its ntris must b propr rational transfr functions. Sinc this is satisfid hr, w conclud by Thorm 4. of th txt that Ĝ(s) is ralizabl. Stp : Dcompos Ĝ(s) into a sum of a constant matrix Ĝ( ) and a strictly propr rational matrix Ĝ sp (s). Ĝ(s) + (s + )(s + ) Stp Rwrit Ĝ sp (s) as a sum of matrics Ĝ sp (s) ( (s + )(s + ) s + s (s + ) (s + ) s + 4 ) (). () Stp 4 Our last common dnominator d(s) s + α s + α (s + )(s + ) s + s + so A ralization of Ĝ(s) is ẋ y 4 x + x + u () u (4)

4 ECE 6 Lumpd Systms Thory Novmbr 8, 7 4 Stp 5 Chcking our work, w find ˆ G(s) C (si A) B + D (5) s + s + 4 s +, s which, aftr som algbra, is qual to th original Ĝ(s). 5. Find fundamntal and stat transition matrics for t A(t). (6) Solution: Th coupld diffrntial quations ar ẋ x + t x (7) ẋ x. (8) Stp : Whr possibl, solv th associatd initial valu problm (IVP) to obtain quations for th two variabls in trms of arbitrary initial conditions. Trivially, Substituting this into th first quation yilds x (t) x () t. (9) ẋ x + x (). () Stp : Solv th IVP for ach of a pair of spcific linarly indpndnt initial condition vctors. Choosing th columns of th idntity matrix as our two indpndnt initial conditions w hav x I () lads to x I (t) t () and x II () lads to x II (t) t t, () whr () is drivd as follows. (), togthr with initial condition x II () yilds ẋ x +. ()

5 ECE 6 Lumpd Systms Thory Novmbr 8, 7 5 Now, dividing both sids by x + and multiplying both sids by dt w obtain Thn and bcaus ln, w hav x (t) x () dx t dt. (4) x (ln( x (t)) ln( x ())) t (5) so x (t) t and thus x (t) t. ln( x (t)) t (6) Stp : Construct a fundamntal matrix whos columns ar th two solutions obtaind in Stp. Thus th fundamntal matrix is X(t) t t t, (7) so X (t) t t + t t t t + t t (8) (9) (4) and X (t ) t t + t t. (4) Stp 4: Find th stat transition matrix, which is Φ(t, t ) X(t)X (t ) t t t ( t t ) t t. (4) 6. Find fundamntal and stat transition matrics for 5 A(t). (4) 4 Solution: Th matrix A is constant so th fundamntal matrix is At and th stat transition matrix is A(t τ). Sinc (si A) s +. (44) s 4s + 4 s 5

6 ECE 6 Lumpd Systms Thory Novmbr 8, 7 6 Th dnominator of th fraction is s 4s + (s )(s ) so finding th partial fraction xpansions and th invrs Laplac transform yilds, At t + t t t t + t t t (45) and τ t + (τ t) τ t (τ t) A(t τ) τ t + (τ t) τ t (τ t) (46)

7 ECE 6 Lumpd Systms Thory Novmbr 8, Find fundamntal and stat transition matrics for sin t A(t). (47) cos t Solution: First, w nd to find th solution to th stat quation. W s that th coupld diffrntial quations ar, in this cas, uncoupld, namly and hav solutions obtaind as follows ẋ sin t x (48) ẋ cos t x (49) so x (t) x () x (t) x () dx x dx x sin τdτ (cos t cos ) (5) cos τdτ (sin t sin ) (5) x (t) x () cos t (5) x (t) x () sin t. (5) Th columns of th fundamntal matrix ar obtaind by solving for x (t) and x (t) for a pair of linarly indpndnt initial conditions. W find that cos t x ( ) lads to x(t) and (54) x ( ) lads to x(t) sin t. (55) Thus X(t) cos t sin t and X (t) cos t+sin t sin t cos t (56) which lads us to X (t ) cos t +sin t sin t cos t. (57) Th stat transition matrix is thn Φ(t, t ) X(t)X cos t cos t (t ) sin t sin t (58) X(t)X cos t cos t (t ) sin t sin t. (59)