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Courses » Computational Electromagnetics & Applications

Unit 4 - Week

1: Finite Difference Method (FDM) - I

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How to access the portal? Pre-requisite Assignment MATLAB Week 1: Finite Difference Method (FDM) - I Lecture 1 Lecture 2 Lecture 3 Lecture slides week 1 Exercise 1 Exercise 2 Exercise 3 Matlab program for exercise1 Matlab program for exercise2 First Matlab program for exercise3 Second Matlab program for exercise3 Lab tour 1 Summary week 1 Quiz : Week 1 Assignment 1 Week 1 -Feedback:

Due on 2018-08-15, 23:59 IST.

1) 1 point 2) 1 point

Week 1 Assignment 1

The due date for submitting this assignment has passed. As per our records you have not submitted this

assignment.

A finite-differencing scheme is stable in the following cases

Statement 1: The numerical domain of influence is smaller than the physical domain of influence. Statement 2: The numerical domain of influence for the problem contains the physical domain of influence.

Statement 3: The numerical velocity must be more than the physical velocity. Statement 4: The numerical velocity must be less than the physical velocity.

Only statement 1 is correct Only statement 2 is correct Only statement 3 is correct Only statement 4 is correct Only statements 1 and 3 are correct Only statements 1 and 4 are correct Only statements 2 and 3 are correct Only statements 2 and 4 are correct None of the statements are correct No, the answer is incorrect.

Score: 0

Accepted Answers:

Only statements 2 and 3 are correct

Assume and to be the step-sizes with indices and along - and -axes respectively.

The finite-difference representation for two-dimensional Laplace equation using central differencing scheme is

Δ

x

Δ

y

i

j

x

y

φ

(

x

,

y

) = 0

2

+

= 0

φ(i+1,j)−2φ(i,j)+φ(i−1,j) Δx φ(i,j+1)−2φΔ(iy,j)+φ(i,j−1)

A project of In association with

Funded by

(2)

Powered by Week 2: Finite Difference Method (FDM) - II Week 3: Finite Difference Method (FDM) -III Week 4: Boundary Conditions Week 5: Variational Methods Week 6: Finite Element Method - I Week 7: Finite Element Method - II Week 8: Method of Moment Week 9: Finite Volume Method -I & -I-I Week 10: Finite Volume Method -III Week 11: Algebraic Topological Method - I Week 12: Algebraic Topological Method - II and Mimetic Method Video Download TEXT TRANSCRIPTS Computational Electromagnetics & Applications 3) 1 point None of these No, the answer is incorrect. Score: 0

Accepted Answers:

Which of the following is(are) the correct computational stencil(s) for a one-dimensional

wave equation with aspect ratio where, is a constant, and are step-sizes in time and -axis respectively? (filled-squares represent known points and empty-circles represent unknown point in the figure)

Statement 1: Explicit scheme with aspect ratio <1

Statement 2: Implicit Scheme with aspect ratio = 1

Statement 3: Explicit scheme with aspect ratio = 1

Statement 4: Implicit Scheme with aspect ratio < 1 Only statement 1 is correct.

+

= 0

φ(i+1,j)−2φ(i,j)+φ(i−1,j) Δx2 φ(i,j+1)−2φ(i,j)+φ(i,j−1) Δy2

+

= 0

φ(i+1,j)+φ(i−1,j) Δx2 φ(i,j+1)+φ(i,j−1) Δy2

+

= 0

φ(i+1,j)−φ(i,j)+φ(i−1,j) Δy2 φ(i,j+1)−φ(i,j)+φ(i,j−1) Δx2

+

= 0

φ(i+1,j)−2φ(i,j)+φ(i−1,j) Δx2 φ(i,j+1)−2φ(i,j)+φ(i,j−1) Δy2

r

= (

k

Δ

t

x

)

2

k

Δ

t

Δ

x

x

(3)

4) 1 point

5) 1 point

Only statement 2 is correct Only statement 3 is correct Only statement 4 is correct. Only statements 1 and 2 are correct. Only statements 1 and 3 are correct. Only statements 2 and 4 are correct. Only statements 3 and 4 are correct. None of the statements are correct. No, the answer is incorrect.

Score: 0

Accepted Answers:

Only statements 1 and 3 are correct.

The central-difference formulation of the differential at point = 4 with step-size is

None of these. No, the answer is incorrect. Score: 0

Accepted Answers:

Which of the following statement(s) is(are) correct for a first-order, one-dimensional partial differential equation (PDE) with constant spatial step-size?

Statement 1: The order of truncation error in forward-differencing is less than that of central-differencing.

Statement 2: The order of truncation error in backward-differencing is more than that of central-differencing.

Statement 3: The order of truncation error in forward-differencing is same as that of central-differencing.

Statement 4: The order of truncation error in forward-differencing is same as that of backward-differencing.

Only statement 1 is correct Only statement 2 is correct Only statement 3 is correct Only statement 4 is correct. Only statements 1 and 2 are correct Only statements 1 and 3 are correct. Only statements 1 and 4 are correct

y d2 dx2

x

Δ

x

= 4

y(8)+y(0) 4 y(8)−2y(4)+y(0) 4 y(8)−2y(4)+y(0) 16 y(8)+y(0) 16 y(8)−2y(4)+y(0) 16

(4)

6) 1 point

7) 1 point

Only statements 2 and 3 are correct Only statements 2 and 4 are correct. Only statements 3 and 4 are correct Only statements 1, 2, and 3 are correct Only statements 1, 2, and 4 are correct Only statements 1, 3, and 4 are correct Only statements 2, 3, and 4 are correct None of the statements are correct No, the answer is incorrect.

Score: 0

Accepted Answers:

Only statements 1 and 4 are correct

Assume is the step-size with index along the -axis and is the time-step with index . What is the forward-difference (FD) and central-difference (CD) formulation for the given equation? FD: CD: FD: CD: FD: CD: FD: CD: None of these. No, the answer is incorrect. Score: 0

Accepted Answers:

FD: CD:

Assume is the step-size with index along the -axis and is the time-step with index .

The central-difference explicit-formulation and aspect-ratio for the given equation are

=

Δ

x

i

x

Δ

t

n

u

(

x

,

t

) =

u

(

x

,

t

)

t

x

u

(

i

+ 1,

n

) =

Δx

u

(

i

,

n

) +

(

1 −

)

u

(

i

,

n

− 1)

Δt2 ΔΔxt

u

(

i

+ 1,

n

) =

Δx

u

(

i

,

n

) +

(

1 +

)

u

(

i

,

n

− 1)

Δt2 ΔΔxt

u

(

i

,

n

+ 1) =

Δt

u

(

i

+ 1,

n

) +

(

1 −

)

u

(

i

,

n

)

Δx ΔΔxt

u

(

i

,

n

+ 1) =

Δt

(

u

(

i

+ 1,

n

) +

u

(

i

,

n

− 1)

)

u

(

i

− 1,

n

)

Δx ΔΔxt

u

(

i

,

n

+ 1) =

Δt

u

(

i

+ 1,

n

) +

(

1 −

)

u

(

i

,

n

)

Δx ΔΔxt

u

(

i

,

n

+ 1) =

Δt

u

(

i

+ 1,

n

) −

u

(

i

− 1,

n

) +

u

(

i

,

n

− 1)

Δx ΔΔxt

u

(

i

+ 1,

n

) =

Δx

(

u

(

i

+ 1,

n

) +

u

(

i

,

n

)

)

Δt

u

(

i

,

n

+ 1) =

Δx

u

(

i

+ 1,

n

) −

u

(

i

− 1,

n

) +

(

1 −

)

u

(

i

,

n

− 1)

Δt ΔΔxt ΔΔxt

u

(

i

,

n

+ 1) =

Δt

u

(

i

+ 1,

n

) +

(

1 −

)

u

(

i

,

n

)

Δx ΔxΔt

u

(

i

,

n

+ 1) =

Δt

u

(

i

+ 1,

n

) −

u

(

i

− 1,

n

) +

u

(

i

,

n

− 1)

Δx ΔxΔt

Δ

x

i

x

Δ

t

n

u

(

x

,

t

) =

u

(

x

,

t

)

tt

xx

r

u

(

i

,

n

+ 1) =

r

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

− 1,

n

)] + 2

u

(

i

,

n

) −

u

(

i

,

n

− 1)

r

Δ

t

2

/

Δ

x

2

(5)

8) 1 point 9) 1 point = = = = = None of these No, the answer is incorrect. Score: 0

Accepted Answers:

=

If is the index in -direction then, the forward-difference approximation of a third-order differential is

None of these. No, the answer is incorrect. Score: 0

Accepted Answers:

Common data for questions 9 and 10

For an ordinary differential equation (ODE), in 0 < x < L with boundary conditions y(0) = y(L) = 0. The domain is divided as shown in the figure with step-size h and index i in x-direction.

The finite-difference formulation using central-differencing scheme is

u

(

i

,

n

+ 1)

r

[

u

(

i

+ 1,

n

) +

u

(

i

− 1,

n

)] +

u

(

i

,

n

− 1)

r

t

x

u

(

i

,

n

+ 1) =

r

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

− 1,

n

)] + 2

u

(

i

,

n

) −

u

(

i

,

n

− 1)

r

Δ

x

t

2

u

(

i

,

n

+ 1)

r

[

u

(

i

+ 1,

n

) −

u

(

i

,

n

) +

u

(

i

− 1,

n

)] +

u

(

i

,

n

− 1)

r

Δ

x

2

t

u

(

i

,

n

+ 1) =

r

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

− 1,

n

)] + 2

u

(

i

,

n

) −

u

(

i

,

n

− 1)

r

Δ

t

2

/

Δ

x

2

i

x

f(x) d3 dx3 f(i+3)−f(i+2)+f(i+1)−f(i) (Δx)2 f(i+3)+3f(i+2)+3f(i+1)+f(i) (Δx)3 f(i+3)−3f(i+2)+3f(i+1)−f(i) (Δx)3 f(i+3)+3f(i+2)−f(i) Δx2 f(i+3)−3f(i+2)+3f(i+1)−f(i) (Δx)3

+

y

(

x

) = 1

y(x) d2 dx2

(6)

10) 1 point

11) 1 point

None of these No, the answer is incorrect. Score: 0

Accepted Answers:

The value of and are

and

and

and

and None of these No, the answer is incorrect. Score: 0

Accepted Answers:

and

Common data for questions 11 and 12

Assume is the step-size with index in -direction and is the time-step with index . For the one-dimensional heat-equation,

The finite-difference formulation of the above equation, using forward-differencing in time and central-differencing in space, is

None of these No, the answer is incorrect. Score: 0

y

(

i

+ 1) = ( − 2)

h

2

y

(

i

) +

y

(

i

− 1)

y

(

i

+ 1) =

h

2

− [

y

(

i

) +

y

(

i

− 1)]

y

(

i

+ 1) =

h

− [(

h

− 2)

y

(

i

) +

y

(

i

− 1)]

y

(

i

+ 1) =

h

2

− [( − 2)

h

2

y

(

i

) +

y

(

i

− 1)]

y

(

i

+ 1) =

h

2

− [( − 2)

h

2

y

(

i

) +

y

(

i

− 1)]

y

(1)

y

(2)

y

(1) =

h

2

y

(2) = 0

y

(1) =

h

2

y

(2) = 3 −

h

2

h

4

y

(1) = 0

y

(2) =

h

2

y

(1) =

h

y

(2) = 3

h

h

2

y

(1) =

h

2

y

(2) = 3 −

h

2

h

4

Δ

x

i x

Δ

t

n

u

(

x

,

t

) −

u

(

x

,

t

) = 0

t

xx

u

(

i

,

n

+ 1) =

t

[

u

(

i

+ 1,

n

) +

u

(

i

,

n

− 1)] +

u

(

i

,

n

− 1)

Δx

u

(

i

,

n

+ 1) =

Δt

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

,

n

− 1)]

x)2

u

(

i

,

n

+ 1) =

Δt

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

,

n

− 1)] +

u

(

i

− 1,

n

)

Δx

u

(

i

,

n

+ 1) =

Δt

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

− 1,

n

)] +

u

(

i

,

n

)

x)2

(7)

12) 1 point Accepted Answers:

For = 1 and = 1 with the initial-conditions given as

, , calculate and using

the result from the above problem.

= 1 and = 3

= 0 and = -2

= 7 and = -1

= 2 and = 1 None of these. No, the answer is incorrect. Score: 0 Accepted Answers: = 1 and = 3

u

(

i

,

n

+ 1) =

Δt

[

u

(

i

+ 1,

n

) − 2

u

(

i

,

n

) +

u

(

i

− 1,

n

)] +

u

(

i

,

n

)

(Δx)2

Δ

x

Δ

t

u

(0, 0) =

u

(2, 0) =

u

(3, 0) =

u

(4, 0) = 0

u

(1, 0) = 1

u

(2, 1)

u

(1, 2)

u

(2, 1)

u

(1, 2)

u

(2, 1)

u

(1, 2)

u

(2, 1)

u

(1, 2)

u

(2, 1)

u

(1, 2)

u

(2, 1)

u

(1, 2)

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