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# Background Review - PowerPoint PPT Presentation

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

### 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

• 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

### 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

• 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

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 = [ ]