1 / 44

# 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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Background Review' - nat

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

• Elementary functions

• Complex numbers

• Common test input signals

• Differential equations

• Laplace transform

• Examples

• properties

• Inverse transform

• Partial fraction expantion

• Matlab

• 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

• 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)

• 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)

• 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

• 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

>> 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

• 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.

>> 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

• Definition and examples

Unit Step Function u(t)

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

Laplace transform theorems

• 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

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)

• 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)

>> 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)

• 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

>> 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

>> [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

>> [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

>> [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)

>> [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

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 = [ ]