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Background Review

Background Review. Elementary functions Complex numbers Common test input signals Differential equations Laplace transform Examples properties Inverse transform Partial fraction expantion Matlab. Elementary functions. The most beautiful equation.

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Background Review

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  1. Background Review • Elementary functions • Complex numbers • Common test input signals • Differential equations • Laplace transform • Examples • properties • Inverse transform • Partial fraction expantion • Matlab

  2. Elementary functions

  3. The most beautiful equation • It contains the 5 most important numbers: 0, 1, i, p, e. • It contains the 3 most important operations: +, *, and exponential. • It contains equal sign for equations

  4. Elementary functions

  5. Elementary functions

  6. Elementary functions

  7. Elementary functions

  8. Elementary functions • F(t)=3sin 3t +4cos 3t • F(t)=Asin(3t-d)=Acosd sin3t –Asin d cos3t • Acos d =3 • Asin d =-4 • A2=25, A=5 • tan d =-4/3, d=-53.13o • F(t)=5sin(3t+53.13o)

  9. Complex Numbers • X2+1=0  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) • z1 z2=(x1+iy1)(x2+iy2)=(x1x2 -y1y2) +i(x1y2 +x2y1)

  10. Polar form of Complex Numbers • z=x+iy, let’s put x=rcosq, y= rsinq • Then z = r(cosq+i sinq) = r cisq = rq • Absolute value (modulus) r2=x2+y2 • Argument q= tan-1(y/x) • Example z=1+i

  11. Euler 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(cos2 x+ sin2 x) = 1 • z=r(cosq+i sinq)=r eiq • Find e1+i • Find e-3i

  12. In Matlab >> z1=1+2*i z1 = 1.0000 + 2.0000i >> z2=3+i*5 z2 = 3.0000 + 5.0000i >> z3=z1+z2 z3 = 4.0000 + 7.0000i >> z4=z1*z2 z4 = -7.0000 +11.0000i >> z5=z1/z2 z5 = 0.3824 + 0.0294i >> r1=abs(z1) r1 = 2.2361 >> theta1=angle(z1) theta1 = 1.1071 >> theta1=angle(z1)*180/pi theta1 = 63.4349 >> real(z1) ans = 1 >> imag(z1) ans = 2

  13. Poles and zeros • Pole of G(s) is a value of s near which the value of G goes to infinity • Zero of G(s) is a value of s near which the value of G goes to zero.

  14. Poles and zeros in Matlab >> s=tf(‘s’) Transfer function: s >> G=exp(-2*s)/s/(s+1) Transfer function: 1 exp(-2*s) * ----------- s^2 + s >> pole(G) ans = 0, -1 >> zero(G) ans = Empty matrix: 0-by-1

  15. Test waveforms used in control systems

  16. 1st order differential equations • y’ + a y = 0; y(0)=C, and zero input • Solution: y(t) = Ce-at • y’ + a y = d(t); y(0)=0, input = unit impulse • Unit impulse response: h(t) = e-at • y’ + a y = f(t); y(0)=C, non zeroinput • Total response: y(t) = zero input response + zero state response = Ce-at + h(t) * f(t) • Higher order LODE: use Laplace

  17. Laplace Transform • Definition and examples Unit Step Function u(t)

  18. Laplace Transform

  19. Name:____________ The single most important thing to remember is that whenever there is feedback, one should worry about __________

  20. Laplace Transform

  21. Laplace Transform

  22. Laplace Transform

  23. Laplace Transform

  24. Laplace transform table

  25. Laplace transform theorems

  26. Laplace Transform

  27. Laplace Transform

  28. Laplace Transform

  29. 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 3t

  30. Matlab 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)

  31. 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)

  32. 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)

  33. Laplace 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)= 1

  34. 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

  35. Partial Fraction

  36. Partial Fraction

  37. Partial fraction; repeated factor

  38. Partial fraction; repeated factor But No FUN

  39. Partial fraction; exercise

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

  41. Matlab >> [r p k]=residue(n,d) r = 1.5000 -1.5000 1.0000 p = 3 -3 0 k = [] >> n=[1 9 -9] n = 1 9 -9 >> d=[1 0 -9 0] d = 1 0 -9 0 1.5/(s-3)-1.5/(s+3)+1/s

  42. Matlab >> [r p k]=residue(n,d) r = 2.0000 -3.0000 1.0000 p = 2.0000 -2.0000 1.0000 k = [] >> n=[11 -14] n = 11 -14 >> d=[1 -1 -4 4] d = 1 -1 -4 4 2/(s-2)-3/(s+2)+1/(s-1)

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

  44. >> 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^3 + 13 s^2 + 4 s - 12 ------------------------------------ s^5 - 6 s^4 + 11 s^3 - 6 s^2 >> [n,d]=tfdata(Y,'v') n = 0 1 -7 13 4 -12 d = 1 -6 11 -6 0 0 >> [r,p,k]=residue(n,d) r = 0.5000 -2.0000 -0.5000 3.0000 2.0000 p = 3.0000 2.0000 1.0000 0 0 k = [ ]

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