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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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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

  • 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


Elementary functions


Elementary functions


Elementary functions


Elementary functions


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)


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)


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


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


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


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.


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


Test waveforms used in control systems


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


Laplace Transform

  • Definition and examples

Unit Step Function u(t)


Laplace Transform


Name:____________

The single most important thing to remember is that whenever there is feedback, one should worry about __________


Laplace Transform


Laplace Transform


Laplace Transform


Laplace Transform


Laplace transform table


Laplace transform theorems


Laplace Transform


Laplace Transform


Laplace Transform


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


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)


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)


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


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


Partial Fraction


Partial Fraction


Partial fraction; repeated factor


Partial fraction; repeated factor

But No FUN


Partial fraction; exercise


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


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


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)


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


>> 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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