Background Review• Elementary functionsElementary functionsxex, eexa) (x y x yx y x (xy)eeeeeeeeeexxxaxxyyxyxyxyxyx10)1ln(,lnlnlnlnlnlnlnln Function Logrithmic1,)(/ ,eFunction lExponentia1lnlnln0x=====−=+=====−−+y x y x y) (x y x y x y) (x x x x y x y x y) (x y x y x y) (xxx x(-x)x, x)(xf(x)f(-x)xf(x)f(-x)sinsincoscoscossinsincoscoscoscossin22sinsincoscossinsinsincoscossinsin1cossincoscossinsin, :Function Odd,:FunctionEven 22+=−−=+=−=−+=+=+=−=−∀−=∀=Elementary functionsElementary functionsx x, xx, xxx, xxx x x) (π x x) (π-x)π ()π( xx-x)π ()π( x-xx)(xx)(xxxxxx sin1csccos1secsincoscotcossintancoscossinsin2sin2sincos2cos2cossin2cos121sin2cos121cos1cos2sin21sincos2cos222222====−=−=−=+===−=+=−=−=−=Elementary functions2sin2sin2coscos2cos2cos2coscos2cos2sin2sinsin22then Substitutesinsin21cossincoscos21coscoscoscos21sinsinvuvuvuvuvuvuvuvuvuvu,yvuxx-y, y, vxuy)](xy)(x[yxy)](xy)(x[yxy)](xy)(x[yx−+=−−+=+−+=+−=+==+=−++=−++=−++−=Elementary functions phase, δ magnitudeCδ)(xBAxBxABAδ, BAδ, then CCδ, BCAxBxAxδCxδCδ)(xCδ)(xBAxBxAABδ, BACδ, Cδ, BCAxBxAxδCxδCδ)(xC xB x A ::sinsincostancossinsincoscossinsincossinAlsocossincostanthen sincossincossinsincoscoscossincos gSimplifyin22222222−+=+−=+==−=+=−=−−+=+=+===+=+=−+Elementary functions•F(t)=3sin 3t +4cos 3t• F(t)=Asin(3t-δ)=Acosδ sin3t –Asin δcos3t• Acos δ =3• Asin δ =-4•A2=25, A=5 • tan δ =-4/3, δ=−53.13ο• F(t)=5sin(3t+53.13ο)Complex Numbers•X2+1=0 Æ x=i where i2=-1•X2+4=0, then x=2i, or 2j• If z1=x1+iy1, z2=x2+iy2• Then z1+ z2= (x1+ x2)+i(y1 +y2)•z1z2=(x1+iy1)(x2+iy2)=(x1x2 -y1y2) +i(x1y2 +x2y1)222221122222212122222211221121))(())((yxyxyxiyxyyxxziyxiyxiyxiyxiyxiyxzzz+−+++=−+−+=++==Polar form of Complex Numbers• z=x+iy, let’s put x=rcosθ, y= rsinθ• Then z = r(cosθ+i sinθ) = r cisθ= r∠θ• Absolute value (modulus) r2=x2+y2• Argument θ= tan-1(y/x)• Example z=1+i,...2,1,0,24arg2=±==nnzzππEuler Formula•z=x+iy•ez=ex+iy= ex eiy= ex (cos y+i sin y)•eix=cos x+i sin x = cis x•| eix| = sqrt(cos2x+ sin2 x) = 1•z=r(cosθ+i sinθ)=r eiθ• Find e1+i• Find e-3iTest waveforms used in control systems1storder differential equations• y’ + a y = 0; y(0)=C, and zero input• Solution: y(t) = Ce-at• y’ + a y = δ(t); y(0)=0, input = unit impulse• Unit impulse response: h(t) = e-at• y’ + a y = f(t); y(0)=C, non zero input• Total response: y(t) = zero input response + zero state response = Ce-at + h(t) * f(t)• Higher order LODE: use LaplaceLaplace Transform• Definition and examplesssFesesesdteLsFtutfFLtfdttfefLsFsststst1)(1111)1()(1)()(ExampleLapalce inverse );()()()()(00010=∴+−=−========∞−∞−∞−−∞−∫∫Unit Step Function u(t)Laplace TransformaseLssFsesdtedteeeLsFetfattstststtt−=−=∴−=−−=====∞−−∞−−∞−∫∫1)(31)(3131 )()()(Example0)3(0)3(0333Laplace Transform32000202220200002122 211)(111 11)()(ssstdtesdttesetsdttetLsesdtesdtestestdtetLttfststststststststst===−−−===−=+=−−−===∫∫∫∫∫∫∞−∞−∞−∞−∞−∞−∞−∞−∞−Laplace Transform{}{})(1)'(1)(1ass'Let )(1)()()(zatLet ?)}({about what ),()}({ If0'00asFasFadzzfeadzzfeaatfLdtatfeatfLatfLsFtfLzsazsst========∫∫∫∞−∞−∞−Laplace Transform2202022022202020000sinsin)1(sin sin)1(coscos cos)1(sin1sin)(sinsin)(ωωωωωωωωωωωωωωωωωωωωωω+==+−=−−−−==−−−===∫∫∫∫∫∫∫∞−∞−∞−∞−∞−∞−∞−∞−∞−stdtestdtestdtesstdtesstestdtestdtestestdtetLttfstststststststststLaplace Transform)(sin)(cos)sin(cos)(1111 )(cos)( ,sin)(2222220)(0)(0tiLtLtitLeLsisssisisisisisiseisdtedteeeLttfttftitistististtiωωωωωωωωωωωωωωωωωωωωωω+=+=+++=++=++−=−−=−−=====∞−−∞−−∞−∫∫Laplace transform tableLaplace transform theoremsLaplace Transform22022020220)(22202)(0)(20)(0)(0)(0)(cos)(sin)(sin))(1(sin)()( sin)1(cos)( cos cos)1(sin1 sin)sin(ωωωωωωωωωωωωωωωωωωωωωωω+−−=+−=−=−+−−−=−−−−−−−=−=−−−−−==∫∫∫∫∫∫∫∫∞−∞−∞−∞−−∞−−∞−−∞−−∞−−∞−−∞−asastdteastdteastdteastdteasastdteasasteastdteastdteasteastdteeteLstststtastastastastastasatstatLaplace Transform ?)5223(sin2cos3)1213()123()3sin313(cos3sin313cos )3)2(13)2(2()3)2(3(3sin31)3331()31()91(21222212212222222122122122121=++−−=+−++=+−+=+=++++++=+++=+=+=+−−−−−−−−−−−sssLFindttsssLssLttetetesssLssLtsLsLsLtttLaplace Transform)0(')0()( )]0()([)0(')(')0(' )(')()(')(")}("{)0()()'()0()()'()()0()()0( )()()()(')}('{200000000fsfsFsfssFsfdttfesfdttfestfedttfetfLyssYyLxssXxLssFfdttfesfdttfestfedttfetfLstststststststst−−=−+−=+−=−−==−=−=+−=+−=−−==∫∫∫∫∫∫∞−∞−∞−∞−∞−∞−∞−∞−Laplace Transform• y”+9y=0, y(0)=0, y’(0)=2• L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2• L(y)=Y(s)•(s2+9)Y(s)=2• Y(s)=2/ (s2+9)• y(t)=(2/3) sin 3tMatlabF=2/(s^2+9)F =2/(s^2+9)>> f=ilaplace(F)f =2/9*9^(1/2)*sin(9^(1/2)*t)>> simplify(f)ans =2/3*sin(3*t)Laplace Transform• y”+2y’+5y=0, y(0)=2, y’(0)=-4• L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2s+4• L(y’)=sY(s)-y(0)=sY(s)-2• L(y)=Y(s)•(s2+2s+5)Y(s)=2s• Y(s)=2s/ (s2+2s+5)=2(s+1)/[(s+1)2+22]-2/[(s+1)2+22]• y(t)= e-t(2cos 2t –sin 2t)Matlab>> F=2*s/(s^2+2*s+5)F =2*s/(s^2+2*s+5)>> f=ilaplace(F)f =2*exp(-t)*cos(2*t)-exp(-t)*sin(2*t)Lapalce transform• Y”-2 y’-3 y=0, y(0)= 1, y’(0)= 7• Y”+2 y’-8 y=0, y(0)= 1, y’(0)= 8• Y”+2 y’-3 y=0, y(0)= 0, y’(0)= 4• 4Y”+4 y’-3 y=0, y(0)= 8, y’(0)= 0• Y”+2 y’+ y=0, y(0)= 1, y’(0)= -2• Y”+4 y=0, y(0)= 1, y’(0)= 1Partial Fraction103 3)2()2()3(12sput and 2,-sby multiply 61 32)3)(2(10sput and s,by multiply 32)3)(2(161)(y(t)? is What , 61)(2220002323=⇒+−+−+=++=−=⇒++−+=+−+=++−+=+−+=−++=−++=======BssCssABsssAsCssBsAssssCsBsAsssssssssYsssssYssssssPartial FractionttssseetyssssssssYCssAssBCsss322333315210361)(311522110316161)(152)3()2()3()2(13sput and 3,sby multiply −−=−=−=−+−=+−−+−=−++=−=++−++=−+−=+Partial fraction; repeated factor 2)23)(3(124137)(A?Abut before, as obtained becan D C, B, ,A123)()()()23)(3(124137)(3142)1(31-1(0)y' 1,y(0)