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**Courses » Computational Electromagnetics & Applications**

**Unit 4 - Week**

**1: Finite Difference Method (FDM) - I**

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**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) *2)*

**1 point**

**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*)−2

*φ*(

*i*,

*j*)+

*φ*(

*i*−1,

*j*) Δ

*x*

*φ*(

*i*,

*j*+1)−2

*φ*Δ(

*iy*,

*j*)+

*φ*(

*i*,

*j*−1)

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**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**
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**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*)−2

*φ*(

*i*,

*j*)+

*φ*(

*i*−1,

*j*) Δ

*x*2

*φ*(

*i*,

*j*+1)−2

*φ*(

*i*,

*j*)+

*φ*(

*i*,

*j*−1) Δ

*y*2

### +

### = 0

*φ*(

*i*+1,

*j*)+

*φ*(

*i*−1,

*j*) Δ

*x*2

*φ*(

*i*,

*j*+1)+

*φ*(

*i*,

*j*−1) Δ

*y*2

### +

### = 0

*φ*(

*i*+1,

*j*)−

*φ*(

*i*,

*j*)+

*φ*(

*i*−1,

*j*) Δ

*y*2

*φ*(

*i*,

*j*+1)−

*φ*(

*i*,

*j*)+

*φ*(

*i*,

*j*−1) Δ

*x*2

### +

### = 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*

### /Δ

*x*

### )

2*k*

### Δ

*t*

### Δ

*x*

*x*

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*
*d*2
*dx*2

*x*

### Δ

*x*

### = 4

*y*(8)+

*y*(0) 4

*y*(8)−2

*y*(4)+

*y*(0) 4

*y*(8)−2

*y*(4)+

*y*(0) 16

*y*(8)+

*y*(0) 16

*y(8)−2y(4)+y(0)*16

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*

### ) =

Δ*x*

*u*

### (

*i*

### ,

*n*

### ) +

### (

### 1 −

### )

*u*

### (

*i*

### ,

*n*

### − 1)

Δ*t*2 Δ

_{Δ}

*x*

_{t}*u*

### (

*i*

### + 1,

*n*

### ) =

Δ*x*

*u*

### (

*i*

### ,

*n*

### ) +

### (

### 1 +

### )

*u*

### (

*i*

### ,

*n*

### − 1)

Δ*t*2 Δ

_{Δ}

*x*

_{t}*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

*u*

### (

*i*

### + 1,

*n*

### ) +

### (

### 1 −

### )

*u*

### (

*i*

### ,

*n*

### )

Δ*x*ΔΔ

*xt*

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

### (

*u*

### (

*i*

### + 1,

*n*

### ) +

*u*

### (

*i*

### ,

*n*

### − 1)

### )

### −

*u*

### (

*i*

### − 1,

*n*

### )

Δ*x*ΔΔ

*xt*

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

*u*

### (

*i*

### + 1,

*n*

### ) +

### (

### 1 −

### )

*u*

### (

*i*

### ,

*n*

### )

Δ*x*ΔΔ

*xt*

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

*u*

### (

*i*

### + 1,

*n*

### ) −

*u*

### (

*i*

### − 1,

*n*

### ) +

*u*

### (

*i*

### ,

*n*

### − 1)

Δ*x*ΔΔ

*xt*

*u*

### (

*i*

### + 1,

*n*

### ) =

Δ*x*

### (

*u*

### (

*i*

### + 1,

*n*

### ) +

*u*

### (

*i*

### ,

*n*

### )

### )

Δ*t*

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*x*

*u*

### (

*i*

### + 1,

*n*

### ) −

*u*

### (

*i*

### − 1,

*n*

### ) +

### (

### 1 −

### )

*u*

### (

*i*

### ,

*n*

### − 1)

Δ*t*ΔΔ

*xt*ΔΔ

*xt*

*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
8) * 1 point*
9)

*= = = = = None of these*

**1 point****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*

### 2Δ

*t*

### /Δ

*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*

### /Δ

*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### /Δ

*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*)

*d*3

*dx*3

*f*(

*i*+3)−

*f*(

*i*+2)+

*f*(

*i*+1)−

*f*(

*i*) (Δ

*x*)2

*f*(

*i*+3)+3

*f*(

*i*+2)+3

*f*(

*i*+1)+

*f*(

*i*) (Δ

*x*)3

*f*(

*i*+3)−3

*f*(

*i*+2)+3

*f*(

*i*+1)−

*f*(

*i*) (Δ

*x*)3

*f*(

*i*+3)+3

*f*(

*i*+2)−

*f*(

*i*) Δ

*x*2

*f(i+3)−3f(i+2)+3f(i+1)−f(i)*(Δx)3

### +

*y*

### (

*x*

### ) = 1

*y*(

*x*)

*d*2

*dx*2

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}

_{y}

_{(2) = 3 −}

_{h}

2 _{h}

_{h}

4
_{h}

*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) =

2Δ*t*

### [

*u*

### (

*i*

### + 1,

*n*

### ) +

*u*

### (

*i*

### ,

*n*

### − 1)] +

*u*

### (

*i*

### ,

*n*

### − 1)

Δ*x*

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

### [

*u*

### (

*i*

### + 1,

*n*

### ) − 2

*u*

### (

*i*

### ,

*n*

### ) +

*u*

### (

*i*

### ,

*n*

### − 1)]

(Δ*x*)2

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

### [

*u*

### (

*i*

### + 1,

*n*

### ) − 2

*u*

### (

*i*

### ,

*n*

### ) +

*u*

### (

*i*

### ,

*n*

### − 1)] +

*u*

### (

*i*

### − 1,

*n*

### )

Δ*x*

*u*

### (

*i*

### ,

*n*

### + 1) =

Δ*t*

### [

*u*

### (

*i*

### + 1,

*n*

### ) − 2

*u*

### (

*i*

### ,

*n*

### ) +

*u*

### (

*i*

### − 1,

*n*

### )] +

*u*

### (

*i*

### ,

*n*

### )

(Δ*x*)2

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*