HI GUYS,THIS IS THE PRESENTATION FOR COLLEGE STUDENTS FOR A SMALL LEVEL PRESENTATION

1. PRESENTED BY
M.PRAVEEN
M.YOGESH
K.VISHAL
S.SURYA
R.VENGATESH
GUIDED BY
ASHOK
(AP/S&H)

2. SUMMARY
NUMERICAL METHODS IN ENGINEERING
NUMERICAL METHODS IN CRIME DETECTION
NUMERICAL METHODS IN SCIENTIFIC
COMPUTING
NUMERICAL METHODS IN FINDING ROOTS
NUMERICAL METHODS IN HEAT EQUQTION

3. Introduction
Why are Numerical Methods so widely used in
Engineering?
 Engineers use mathematical modeling (equations and
data) to describe and predict the behavior of systems.
 Closed-form (analytical) solutions are only possible
and complete for simple problems (geometry,
properties, etc.).
 Computers are widely available, powerful, and
(relatively) cheap.
 Powerful software packages are available (special or
general purpose).

4. A few applications of Numerical Methods in
Engineering:
• Structural/mechanical analysis, design, and
behavior.
• Communication/power
Network simulation
Train and traffic networks
• Computational Fluid Dynamics (CFD):
Weather prediction
Groundwater & pollutant movement

5. Numerical Methods in Engineering

6. Topic 1
Overview of the Modelling Process

7. Why numerical modelling?
To reduce the number of prototypes
constructed.
To use as the forward solver in an
optimization routine to gain the best
possible design.

8. What are the alternatives?
Design and build lots of physical models
Time consuming
Expensive
Measurement difficulties (cables, calibration,
mechanical precision, interference, mutual coupling,
field confinement, etc)
Analytical solutions
Limited to simple models because of the computational
complexity

9. Numerical Modelling Procedure
Select/write your numerical modelling
code.
Find a simple analytical model and divide
into segments/pixels/voxels.
These MUST have every dimension less than
l/10 where l is the wavelength in the
material being modelled. (Note that this is
NOT the free space wavelength).

10. Let’s do the calculation of l
The complex propagation coefficient g is given by the
equation
where
s is the conductivity,
e is the absolute permittivity,
m is the absolute magnetic permeability,
w is the angular radiation frequency, and j is the imaginary number,
a is the attenuation constant
b is the phase constant
)( weswmbag jjj 

11. The wavelength in the material
b

l
2

At this frequency we have
The wavelength depths on the electromagnetic
properties of the material.
Remember NO segment/pixel/voxel can be larger
than in any dimension l/10 .
This can make life very difficult when you have
electrically large conductive materials.

12. Solution
MoM and FDTD both use perfectly
conducting materials.
To introduce finite conductivity, we
can use lumped impedance elements.

13. Maths and crime: Deblurring a
number plate
A short crime story
• Burglar robs a bank
• Escapes in a getaway car
• Pursued by police

14. GOOD NEWS
Police take a photo
BAD NEWS
Photo is blurred

15. SOLUTION
Find a model of the blurring process
Blurring function g
Original image
f
• Blurring formula
• Inverting the formula we can get rid the blur
• BUT need to know the blurring function g
ydygyxfxh   2
)()()(

16. 
2/)/( 222
dxgdexhdeef xixiyi



Inversion formula
h(x) f(x)
An example of Image
Processing

17. Scientific computing
Design and analysis of algorithms for
numerically solving mathematical problems
in science and engineering
Considers the effect of approximations and
performs error analysis
modern simulations of engineering
applications

18. Computational problems:
attack strategy
Develop mathematical model (usually requires a combination of math
skills and some a priori knowledge of the system)
Come up with numerical algorithm (numerical analysis skills)
Implement the algorithm (software skills)
Run, debug, test the software
Visualize the results
Interpret and validate the results
Mathematical modeling

19. Computational problems:
well-posedness
The problem is well-posed, if
(a) solution exists
(b) it is unique
(c) it depends continuously on problem data
Simplification strategies:
Infinite finite
Nonlinear linear
High-order low-order

20. What computers can’t do
Solve (by reasoning) general mathematical problems
 they can only repetitively apply arithmetic
primitives to input.
Solve problems exactly.
Represent all numbers. Only a finite subset of the
numbers between 0 and 1 can be represented.

21. Finding roots / solving equations
General solution exists for equations such as
ax2 + bx + c = 0
The quadratic formula provides a quick answer to all
quadratic equations.
However, no exact general solution (formula) exists
for equations with exponents greater than 4.

22. Finding roots…
Even if “exact” procedures existed, we are stuck with
the problem that a computer can only represent a
finite number of values… thus, we cannot “validate”
our answer because it will not come out exactly
However we can say how accurate our solution is as
compared to the “exact” solution

23. Finding roots, continued
Transcendental equations: involving geometric
functions (sin, cos), log, exp. These equations cannot
be reduced to solution of a polynomial.
Convergence: we might imagine a “reasonable”
procedure for finding solutions, but can we guarantee
it terminates?

24. Square root program
If the input c < 1, the root lies between c and 1.
Else, the root lies between 1 and c.
The (positive) square root function is continuous
and has a single solution.
c = x2
F(x) = x2 - c
Example:
F(x) = x2 - 4
-6
-4
-2
0
2
4
6
0 0.5 1 1.5 2 2.5 3

25. double Sqrt(double c, double tol)
{
double a,b, mid, f;
// set initial boundaries of interval
if (c < 1) { a = c; b = 1}
else { a = 1; b = c}
do {
mid = ( a + b ) / 2.0;
f = mid * mid - c;
if ( f < 0 )
a = mid;
else
b = mid;
} while( fabs( f ) > tol );
return mid;
}
-6
-4
-2
0
2
4
6
0 0.5 1 1.5 2 2.5 3

26. Heat equations
The heat capacity c of a material gives the amount of
energy needed to raise the temperature T one
kilogram of the material by one degree K (Kelvin).
The thermal conductivity k of a material indicates the
amount
of energy transmitted trough a plate with thickness 1
m and 1sq.m area if the temperatures at the two sides
differ by 1 K.
A(x) ρ c (∂ T (t, x)/∂t)=∂/∂x(k A(x)(∂T (t, x)/∂x))+A(x)f(t,x)

28. The flux of thermal energy is a vector indicates the
direction of the flow and the amount of thermal
energy flowing per second and square meter
Fourier’s law of heat conduction can be stated as
= −k ∇T

29. One Dimensional Heat Equation
If a temperature T over a solid (with constant cross
section A) is known to depend on one coordinate x
only, then the change of temperature ∆T measured
over a distance ∆x will lead to a flow of thermal energy
∆Q.
If the time difference is ∆t then
∆Q/∆t = −k A (∆T/∆x)

30. Two Dimensional Heat Equation
If the domain G ⊂ R2 with boundary curve C describes
a thin plate with constant thickness h then we may
assume that the temperature will depend on t, x and y
only and not on z.
E (t) =