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Mathematical Details

Task 1 a:

1. Convert 58.40625₁₀ to Binary number.

Hence the converted decimal number to binary is

(111010.01101)_­2

2. Convert 101.0101₂ to Decimal number.

 2⁰+2² = 5

2^-2+2^-4 = 0.375

Hence the converted binary number to decimal number is

(5.375)_­10

3. Convert 1A4E₁₆ to Decimal number.

16³*1+16²*10+16¹*4+16⁰*15 = 4096+2560+64+15

= (7199)₁₀

Task 1.b

1. Perform the binary addition

                               

Task 1.c

1. Convert 5613.90625₁₀ to octal number.

Converted octal number is

(12755.72)₈

2. Convert 11010110₂ to Binary number.

This number is already in binary form.

Task 2:

Find the magnitude of impedance (modulus |z|) and the phase difference between the current and voltage (Argument) when Z = 5 – 35i Ω

∣Z∣= √ R2+(XL​−XC​)2​

= √ 25 + 1225

= √ 1250

= 35.3553

Phase difference = tan^-1 ( 35/5) = tan^-1(7) = -10.9018

Task 3:

Compute (2 + i) (3 – 4i) then convert the outcome to polar and exponential forms   

= (2 + i)(3 – 4i)

= 6 + 3i – 8i + 4

= 10 – 8i

Polar form = A ∠±θ

= √ 100+64

= √164 = 12.8

Θ = tan^-1(8/10) = 38.66

So polar form is 12.8∠38.66

Exponential form

= 12.81e^i0.67

Task 4:

Find the identity for (cosθ +i sinθ ) ² Using De Moivre’s theorem.

De Moivre’s theorem states that

For

(cosθ +i sinθ ) ⁿ = cosn θ +i sinn θ where n is positive or negative integer or n is positive or negative fraction

Now proving for

(cosθ +i sinθ ) ² = cos ² θ +i sin ²θ  + 2isin θcos θ

= cos ² θ – sin ²θ  + i2sin θcos θ

= cos 2 θ + iSin 2 θ

Hence proved

Task 5:

Using De Moivre’s theorem simplify: 3(cos 4θ + i sin 4θ)⁷

= 3(cos 28 θ + iSin 28 θ)

= 3 cos 28θ + i3Sin 28θ

 Task 6:

Find the determinant for matrix A

= 25 (8*1-12*1) -5(64*1-144*1) +1(64*12-8*144)

= 25*(-4) -5*(-80) – 384

= -100 +400-384

= -84

Task 7:

Using Graph express the following complex numbers in polar form

• 5+j4
• -5+j4
• -5-j4
• 5-j4

Task 8:

Solve the given set of linear equation using the method of matrix inverse

M2

2x – y = 7

5x – 3y = 18

Now for finding solution we need to multiply both side by inverse

Hence x = -11 and y = -29

Task 9:

Evaluate and validate solution for the given set of linear equation by finding the value of x, y and z

      x+2y+z=4

    3x-4y-2z=2

   5x+3y+5z=-1

Solving

Solving

-7 y = -19

Y = 19/7

-10y -5z = -10

= – 10*(19/7) -5z = -10

Z = -24/7

X+2y+z = 4

X = 4-2y-z

= 4 – 38/7 +24/7

X =2

Therefor x =2, y =19/7 and z = -24/7

Task 10:

Calculate the roots of f(x) = x² – 9 = 0 using three different iterative techniques

Applying Newton method for iteration

Let’s say our root is X₁ =4

Solving by putting

f(x) = 4² – 9 = 7

f `(x) = 2x

X₂ = 4 – 7/8

= 3.125

X₃ = 3.125 – 0.7656/6.25

= 3.125 – 0.123

= 3.002

X4 = 3.002 – 0.012/6.004

     = 3

Hence the solution of the equation is 3

Task 11:

Solve the following differential equation.

(a) With initial condition given by

Solving the equation

Integrating both side with initial condition

Y = – 1/ (x³-2)

(b)  Show the value of y when x =1

Integrating both side with initial condition x=1

 Y = – 2/(2x³-1)

Task 12:

Solve the following second order differential equation.

Converting in the form of quadratic

r²+11r+24 =0

(r+3)(r+8)

R = -3, -8

Y(t) = C₁e^-3t + C₂e^-8t

Now using initial conditions y(0) = 0

Y(0) = C₁+C₂ = 0 —————————-eq. 1

Y’(t) = -3 C₁e^-3t – 8 C₂e^-8t

Now using initial conditions (0) = – 7

(0) = -3C₁-8C₂ = -7 ———————–eq. 2

Putting value of C₁ in terms of C₂ in equation 2

(0) = -3C₁-8C₂ = -7

         = 3C₂ -8C₂ = -7

C₂ = 7/5

C₁ = – 7/5

So, our solution will be

Y(t) = -7/5e^-3t + 7/5e^-8t

Task 13  

Find the Laplace transforms of the given function:

• f(t) = 6e^-5t + e^3t + 5^t3 – 9

f(s) = 6/s+5 + 1/s-3 + 30/s^4 -9/s

• h(t) = 3sinh (2t) + 3sin (2t)

H(s) = 6/s²-4 + 6/s²+4

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