DERIVATIVES AND DIFFERENTIAL EQUATIONS

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DERIVATIVESANDDIFFERENTIAL EQUATIONS

• June 23,99

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WHAT IS A DERIVATIVE?

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

yf(x)
f(ah)
f(a)
a
ah
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SIGNIFICANCE OF DERIVATIVE

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

yf(x)
f(a)
a
ah
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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) doesnt exist

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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

f(c)0
xc
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DERIVATIVES IN MATLABdiff

• MATLAB computes two differentials, dy and dx,
  using diff, to arrive at the derivative
• If x and y are input array of numbers
• dydiff( y)
• dxdiff(x)
• Then,
• yprimediff(y)./diff(x)

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How diff works

• Take the array y3 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 to
• diff(y)2 2 -2 4
• Naturally, there is one fewer term in diff(y)
  than it is in y

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Using diff to find derivatives

• Let y10 25 30 50 10 for x1 2 3 4 5
• Then
• diff(y)15 5 20 -40
• diff(x)1 1 1 1 1
• Dividing term-by-term
• yprime15 5 20 -40

One fewer component
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WORKING WITH humps

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

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HOW IT LOOKS
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Try it!derivative of humps

• First, we must know how humps was created in
  order to know dx
• x00.011dx in this case is 0.01
• yhumps(x)
• Now use
• dydiff(y)
• dxdiff(x)

yprimedy./dx
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PLOT OF dy/dx
Function peaks
Deriv. goes to 0
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Finding critical pointstheory vs. application

• To find max. or min.we should look for instances
  where y0.
• This is where the difference between textbook
  methods and the real world shows up.
• To see for yourself look for instances where
  yprime0.

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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

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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
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FINDING ZERO CROSSINGS

• There are two ways to find where a function
  crosses zero
• 1. Numerical
• 2. 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

Zero crossing
X
X
X
X
X
X
X
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MATLABs WAY

• Let x be an n element arrayx1,x2,,xn. To
  generate the zero crossing array, do
• yx(1n-1).x(2n)
• Algebraically, this is what is happening
• yx1x2 x2x3 ,..., xn-1xn

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Try it!

• Evaluate humps in the range x0 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

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LOCATING SIGN CHANGES

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

1st0.29 2nd0.63 3rd0.88
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CLOSER LOOK
Sign change
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SECOND ZC
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THIRD ZC
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Differential Equations
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Simplest Diff.Eq

• A first order differential equation is of the
  form
• y dy/dxg(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 that
• y 3x2
• y y
• y (2x)cos2y
• Answers
• yx3
• yexp(x)
• y???

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SOLVING ODEs USING MATLAB

• MATLAB solves ordinary DE in two ways
  1)numerical and 2)symbolic
• For numerical solution use
• x,yode23(function,a,b,initial)
• See next slide for usage

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EXPLAINING ode23

• Here are the elements of
• x,yode23(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 interval
• b - right point of the interval
• initial - initial condition, i.e. y(a)

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USING ode23

• Lets solve y 3x2
• First write a function as follows
• function yprimeanything(x,y)
• yprime3x.2
• Then call
• x,yode23(anything,2,4,0.5)
• y is the solution. Plot x vs.y

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RC CIRCUIT

• Equation describing a source-free RC circuit is

vc
C
R
ic
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SOLVING FOR VOLTAGE

• The analytical solution is
• where Vo is the initial voltage across the
  capacitor

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MATLABs 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 RC10-3, write a function like
• function vcprimerc(t,vc)
• vcprime-(103)vc

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SOLVING FOR vc

• Use ode23 as follows
• t,vcode23(rc,0,1,2)

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CIRCUIT WITH FORCED RESPONSE

• Take the following RL circuit(p.497)

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WHAT IS i ?

• From circuits analysis we know

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SOME NUMBERS

• Let R2 ohm, L1 H and vs10cos3t
• Here is what we have
• Zsqrt(94)sqrt(13)
• Vm10
• ?56.3o

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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