Inverse Laplace Transform

Laplace Transform of derivatives

Laplace Transform applications y”+9y=0, y(0)=0, y’(0)=2 L(y”)=s 2 Y(s)-sy(0)-y’(0)= s 2 Y(s)-2 L(y)=Y(s) (s 2 +9)Y(s)=2 Y(s)=2/ (s 2 +9) y(t)=(2/3) sin 3t

Matlab s=sym('s'); (F=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”)=s 2 Y(s)-sy(0)-y’(0)= s 2 Y(s)-2s+4 L(y’)=sY(s)-y(0)=sY(s)-2 L(y)=Y(s) (s 2 +2s+5)Y(s)=2s Y(s)=2s/ (s 2 +2s+5)=2(s+1)/[(s+1) ]- 2/[(s+1) ] 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)

ODE to State space

Y”+2 y’+ y=0, y(0)= 1, y’(0)= -2 >> A=[0 1;-1 -2]; B=[0;1]; C=[1 0]; D=0; >> x0=[1;-2]; >> t=sym('t'); >> y=C*expm(A*t)*x0 y = exp(-t)-t*exp(-t) Y”+2 y’+ y=f(t)=u(t), y(0)= 2, y’(0)= 3 >>s=sym('s'); Fs = 1/s; Gs = 1/(s^2+2*s+1) >>x0=[2;3]; >> y=C*expm(A*t)*x0 + ilaplace(Gs*Fs) y = 2*exp(-t)+5*t*exp(-t)+1-exp(-t)*(t+1)

Partial Fraction

Partial fraction; repeated factor

But No FUN

Partial fraction; exercise

>> [r p k]=residue(n,d) r = 1 2 p = 1 0 k = [] >> d=[1 -1 0] d = >> n=[3 -2] n = /(s-1) + 2/s Matlab

>> n=[1 9 -9] n = >> d=[ ] d = >> [r p k]=residue(n,d) r = p = k = [] 1.5/(s-3)- 1.5/(s+3)+1/s Matlab

>> n=[11 -14] n = >> d=[ ] d = >> [r p k]=residue(n,d) r = p = k = [] 2/(s-2)- 3/(s+2)+1/(s-1) Matlab

>> b=[1 2 1] b = >> a=[1 0] a = 1 0 >> [r p k]=residue(a,b) r = 1 p = k = [] 1/(s+1)- 1/(s+1) 2 Matlab

>> Y=(s^4-7*s^3+13*s^2+4*s-12)/s^2/(s-3)/(s^2-3*s+2) Transfer function: s^4 - 7 s^ s^2 + 4 s s^5 - 6 s^ s^3 - 6 s^2 >> [n,d]=tfdata(Y,'v') n = d = >> [r,p,k]=residue(n,d) r = p = k = [ ]

Initial value theorem Conditions: f(t) is continuous except finite jump at t=0 or F(s) is proper: num deg <= den deg

Final value theorem Conditions: f(t) is finite and converges or every pole of sF(s) has real part < 0

Examples

>> s=sym('s') s = s >> F=(s+1)/(s+2)/(s+3) F = (s+1)/(s+2)/(s+3) >> f=ilaplace(F) f = -exp(-2*t)+2*exp(-3*t) f(0)=1, f(inf)=0

Examples

Transfer Function

Example

A, B, C, D are matrices

Input Output System Input Output System: –If the input x(t) is unit impulse, x(t)=δ (t), the output is called the impulse response. – If the input x(t) is unit step, x(t)=u(t), the output is called the step response. Input  (t),u(t) Output y(t) System

X(s) H(s) Y(s)

Frequency Response Frequency Response: Given T.F. H(s) H(s)

Example Works only if H(s) is stable

Poles and zeros Poles: values of s at which TF  infinity –Most time, poles = roots of denominator –When there are common factors in numerator and denominator, cancel them first Zeros: values of s at which TF = 0 –Finite zeros: roots of numerator –Number of zeros at infinity: n-m, n= den deg and m = num deg Totally n poles and n zeros n is called the order of the system n – m the relative order

Example Order: n = 4 = den deg m= num deg = 1 Relative order = n-m = 3 4 poles at: 0, 0, -4, -6 One finite zero at -1 3 zeros at infinity

Stability System is BIBO stable if any bounded input generates bounded output Simple criteria: –After common factor cancellation –All poles have strictly negative real parts

Example G1 is not BIBO stable G2 and G3 are

Example

example

>> s=sym('s'); >> Y=1/(s^2+3*s+2)/s*exp(-2*s) Y = 1/(s^2+3*s+2)/s*exp(-2*s) >> ilaplace(Y) ans = 1/2*heaviside(t-2)*(1+exp(-2*t+4)-2*exp(2-t))

2.7

2.8c