© 1999BG Mobasseri 1 5/30/99

DERIVATIVESANDDIFFERENTIAL EQUATIONS

June 23,’99

i

i =Cdvdt

v+

-

© 1999BG Mobasseri 2 5/30/99

WHAT IS A DERIVATIVE?

• Derivative of f(x) at x=a is given by f’(a)

a a+h

y=f(x)

f(a)

f(a+h)

′ f a( )=limh→0f a+h( )−f a( )

h

© 1999BG Mobasseri 3 5/30/99

SIGNIFICANCE OF DERIVATIVE

• In the limit, derivative at x=a is equal to the slope m of the tangent line at a.

a a+h

y=f(x)

f(a)

© 1999BG Mobasseri 4 5/30/99

APPLICATIONS OF DERIVATIVE

• The most widely used application of derivative is in finding the extremum (max or min) points of a function.

• If a function has a local extremum at a number c then either f’(c)=0 or f’(c) doesn’t exist

© 1999BG Mobasseri 5 5/30/99

CRITICAL NUMBERS

• A number c in the domain of a function f is a critical number of f if either f’(c)=0 or f’(c) does not exist

x=c

f’(c)=0

© 1999BG Mobasseri 6 5/30/99

DERIVATIVES IN MATLAB:diff

• MATLAB computes two differentials, dy and dx, using diff, to arrive at the derivative

• If x and y are input array of numbersdy=diff( y)dx=diff(x)

• Then, yprime=diff(y)./diff(x)

© 1999BG Mobasseri 7 5/30/99

How diff works

• Take the array y=[3 5 7 5 9];• diff(y) simply is the pairwise

differences, i.e. diff(y)=[5-3 7-5 5-7 9-5]

which is equal todiff(y)=[2 2 -2 4]

• Naturally, there is one fewer term in diff(y) than it is in y

© 1999BG Mobasseri 8 5/30/99

Using diff to find derivatives

• Let y=[10 25 30 50 10] for x=[1 2 3 4 5]

• Thendiff(y)=[15 5 20 -40]diff(x)=[1 1 1 1 1]• Dividing term-by-termyprime=[15 5 20 -40]

One fewer component

© 1999BG Mobasseri 9 5/30/99

WORKING WITH humps

• humps is a built-in MATLAB function, like peaks and is given by

y = 1(x−.3)2 +0.01[ ]

+ 1x−.9( )

2 +.04[ ]−6

© 1999BG Mobasseri 10 5/30/99

HOW IT LOOKS

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

© 1999BG Mobasseri 11 5/30/99

Try it!:derivative of humps

• First, we must know how humps was created in order to know dx

x=0:0.01:1;%dx in this case is 0.01y=humps(x);• Now usedy=diff(y);dx=diff(x);

yprime=dy./dx;

© 1999BG Mobasseri 12 5/30/99

PLOT OF dy/dx

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

20

40

60

80

100HUMPS

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1000

-500

0

500

1000FIRST DERIVATIVE OF HUMPS

Function peaks

Deriv. goes to 0

© 1999BG Mobasseri 13 5/30/99

Finding critical points:theory vs. application

• To find max. or min.we should look for instances where y’=0.

• This is where the difference between textbook methods and the real world shows up.

• To see for yourself look for instances where yprime=0.

© 1999BG Mobasseri 14 5/30/99

Where are the zeros?

• If you did the search right, you will realize that nowhere in yprime you can actually see a zero

• Since there are no points where y’ is exactly zero, we should look for points of transition from positive to negative, i.e. sign changes

+ +

© 1999BG Mobasseri 15 5/30/99

Zero crossing and sign transition

• In Matlab, all derivatives are discrete. Some are positive, some negative but no zeros

• Since derivative goes from + to -, we can infer that it went to zero somewhere in between

Zero crossing

© 1999BG Mobasseri 16 5/30/99

FINDING ZERO CROSSINGS

• There are two ways to find where a function crosses zero;

1. Numerical2. Algebraic(polynomial fitting)

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ZC USING ARRAY OPERATION

• To find where a sign transition takes place, multiply two consecutive numbers in the derivative array. If there is a sign change, the product is negative

[10 15 16 9 -4 -3 -4 -9]X X X X X X X

+ + + - + + -

Zero crossing

© 1999BG Mobasseri 18 5/30/99

MATLAB’s WAY

• Let x be an n element array[x1,x2,…,xn]. To generate the zero crossing array, do

y=x(1:n-1).*x(2:n)• Algebraically, this is what is

happeningy=[x1x2 x2x3 ,..., xn-1xn]

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Try it!

• Evaluate humps in the range x=0 to 1 in increments of 0.01

• This gives you a 1x101 array. Find yprime. This is now a 1x100 array

• Find where in this 101 positions sign changes occur

© 1999BG Mobasseri 20 5/30/99

LOCATING SIGN CHANGES

• Just look for instances of negative sign in the zero crossing array

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-800

-600

-400

-200

0

200

400

600

800

1st=0.292nd=0.633rd=0.88

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CLOSER LOOK

0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4-800

-600

-400

-200

0

200

400

600

800

Sign change

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SECOND ZC

0.55 0.6 0.65 0.7 0.75-50

-40

-30

-20

-10

0

10

20

30

40

50

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THIRD ZC

0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95-50

-40

-30

-20

-10

0

10

20

30

40

50

© 1999BG Mobasseri 24 5/30/99

Differential Equations

i

i =Cdvdt

v+

-

© 1999BG Mobasseri 25 5/30/99

Simplest Diff.Eq

• A first order differential equation is of the form

y ’=dy/dx=g(x,y)• We are looking for a function y

such that its derivative equals g(x,y)

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FEW EXAMPLES

• Find y that such thaty ’=3x2

y ‘=yy ‘=(2x)cos2y

• Answers:y=x3

y=exp(x)y=???

© 1999BG Mobasseri 27 5/30/99

SOLVING ODE’s USING MATLAB

• MATLAB solves ordinary DE in two ways: 1):numerical and 2):symbolic

• For numerical solution use[x,y]=ode23(‘function’,a,b,initial

)• See next slide for usage

© 1999BG Mobasseri 28 5/30/99

EXPLAINING ode23

• Here are the elements of [x,y]=ode23(‘function’,a,b,initial

) function - this is g(x,y) in y

’=g(x,y). Must be written as a separate MATLAB function

a - left point of the intervalb - right point of the intervalinitial - initial condition, i.e. y(a)

© 1999BG Mobasseri 29 5/30/99

USING ode23

• Let’s solve y ‘ =3x2

• First write a function as followsfunction yprime=anything(x,y)

yprime=3*x.^2;• Then call

[x,y]=ode23(‘anything’,2,4,0.5)• y is the solution. Plot x vs.y

© 1999BG Mobasseri 30 5/30/99

RC CIRCUIT

• Equation describing a source-free RC circuit is

dvc

dt+ 1

RCvc =0 RC vc

ic

© 1999BG Mobasseri 31 5/30/99

SOLVING FOR VOLTAGE

• The analytical solution is

• where Vo is the initial voltage across the capacitor

vc t( ) =Voe−t RC

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MATLAB’s WAY

• To use ode23, we need to cast the problem in the form of y ‘=g(x,y)

• Here, y is vc. . Therefore,

vc’=-(1/RC)vc

• With RC=10^-3, write a function likefunction vcprime=rc(t,vc)

vcprime=-(10^3)*vc

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SOLVING FOR vc

• Use ode23 as follows[t,vc]=ode23(’rc’,0,1,2)

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CIRCUIT WITH FORCED RESPONSE

• Take the following RL circuit(p.497)

Ldidt+Ri=Vmcosωt

L

Rvs=Vmcosωt~

i

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WHAT IS i ?

• From circuits analysis we know

i=VmZcosωt−β( )

whereZ= R2+ω2L2

β=tan−1ωLR

L

Rvs=Vmcosωt~

© 1999BG Mobasseri 36 5/30/99

SOME NUMBERS

• Let R=2 ohm, L=1 H and vs=10cos3t

• Here is what we haveZ=sqrt(9+4)=sqrt(13)Vm=10

β=56.3o

i = 1013

cos3t−56.3( )

© 1999BG Mobasseri 37 5/30/99

EXERCISE- VERIFY WITH MATLAB

• Use ode23 to solve for current. Match the two results, graphically, to see if the amplitude, phase and frequencies match the theoretical result