## Description

General Instructions

Question 1

Given {a,b,c} denotes a set of linearly independent vectors and {b1,b2,b3} denotes the orthogonal basis obtained by

applying the Grahm-Schmidt algorithm on {a,b,c}.

⟹ b1 = a,

b2 = b − Projb1b, b3 = c − Projb1c − Projb2c

a) b3 ⊥ b1 ⟺ ⟨b1,b3⟩ = 0

First, we will prove that b1 ⊥ b2 ⟺ ⟨b1,b2⟩ = 0

⟨b1,b2⟩ = ⟨a,b − Projb1b⟩

= ⟨a,b⟩−⟨a,⟨eb1,b⟩eb1⟩

= ⟨a,b⟩ − ⟨a,b⟩ = 0

Therefore, b2 ⊥ b1. Now,

⟨b1,b3⟩ = ⟨a,c − Projb1c − Projb2c⟩

= ⟨a,c⟩ − ⟨a,⟨eb1,c⟩eb1⟩ − ⟨a,⟨c,eb2⟩eb2⟩

= ⟨a,c⟩ − ⟨eb1,c⟩⟨a,eb1⟩ − ⟨c,eb2⟩⟨a,eb2⟩

= ⟨a,c⟩ − ∥a∥⟨ea,c⟩ − 0 = ⟨a,c⟩ − ⟨a,c⟩ = 0

Thus, we have shown that b3 ⊥ b1. Now b2 ⊥ b3 ⟺ ⟨b2,b3⟩ = 0

⟨b2,b3⟩ = ⟨b2,c − Projb1c − Projb2c⟩

= ⟨b2,c − Projb2c⟩ − ⟨b2,Projb1c⟩

= (⟨b2,c⟩−⟨b2,⟨eb2,c⟩eb2⟩) − ⟨b2,Projb1c⟩

= 0 − ⟨b2,Projb1c⟩

= −⟨b − ⟨eb1,b⟩eb1,(⟨c,eb1⟩)eb1⟩ = −(⟨c,eb1⟩)(⟨b,eb1⟩ − ⟨b,eb1⟩) = 0

Thus, we have shown that b2 ⊥ b3.

b) Given W = Span{b2,b3}.

⟹ ProjWb1 = Projb2b1 + Projb4b1

= ⟨b1,b2⟩ eb2 + ⟨b1,b3⟩ eb3 = 0

∥b2∥2 ∥b3∥2

Therefore, ProjWb1 = 0.

Question 2

(⟨a,eb2⟩ = ∥b∥2⟨b1,b2⟩ = 0)

Given finite length continuous time signals: s1(t) = ⎧0 if t0≤≤0t ≤ 2

0 if t ≥ 2

⎧0 s2(t) = ⎩

⎧

2(t), s3(t) and s4(t) can be found using Grahm

s3(t) = ⎩

s

Schmidt Algorithm.

⟨s1(t)

but s1(t), s2(t), s3(t) and s4(t) are real

⟹ ⟨s1(t),s1(t)⟩ = ∫ (s1(t))2

Let the basis be ϕ1(t), ϕ2(t), ϕ3(t) and ϕ4(t).

∥s

⟹ ϕ′ (t) s (t) ⎧0 ≤ 0 ≤ 2

Calculating ϕ2(t). Defining ϕ′ (t) as follows

ϕ′ (t) = s2(t) −)⟩ϕ1(t)

then, ϕ′ (t)

ϕ′ (t) = s2(t) − (∫)ϕ1(t)

−

⟩ dt 0

t ≤ 1 t ≤ 2

2

Calculating ϕ3(t)

ϕ3(t) = s3(t) − ⟨s3( ),ϕ1(t)⟩ϕ1(t) − ⟨s3( ( )

∞

= s3(t) −3(t)ϕ1( )dt)ϕ1(t) − (∫ s3( )ϕ2(t)dt)ϕ2(t)

−

2

= s3(t) −3(t)dt)s

0 )

ϕ3(t) = −1 2 ≤ t ≤ 3

0 t ≥ 3

Calculating ϕ4(t)

ϕ′4(t) = s4(t) − ⟨s4(t),ϕ1(t)⟩ϕ1(t) − ⟨s4(t),ϕ2(t)⟩ϕ2(t) − ⟨s4(t),ϕ3(t)⟩ϕ3(t)

ϕ4(t) = 0

⟹ ϕ4(t) doesnt exist, i.e, the signals ϕ1(t) ϕ2(t) ϕ3(t) are the orthogonal(orthonormal to be specific) basis of the signal space spanned by 1(t),s2(t),s3(t) 4(t).

b) A brute force method to check whether the ordering of signals changes the basis would be to perform the Grahm Schmidt algorithm on the new ordering and check if the basis is the same as the previous one. Let the orthogonal basis of the signal space spanned by the signals 2(t), s3(t) 1(t) 4( ) be 1( ), ϕ2(t), ϕ3(t) and ϕ4(t).

ϕ1(t) = 2(t)

2

⟹

⎧

t ≤ 1

ϕ1(t) = ⎨

t ≤ 2 ⎩

Calculating ϕ2(t). Defining ϕ′2(t) as follows

ϕ′2(t) = s3(t) − ⟨s3(t),ϕ1(t)⟩ϕ1(t)

dt)ϕ1(t)

1

= s3(t) −s

0

dt)

s

⟹ ϕ (t) 3(t)

⎧

ϕ

≤ ≤ 3 ⎩

Calculating ϕ3(t). Defining ϕ′3(t) as follows

ϕ′3(t) = s1(t) − ⟨s1(t),ϕ1(t)⟩ϕ1(t) − ⟨s1(t),ϕ2(t)⟩ϕ2(t)

1

= s1(t) − (∫ ss

= s1(t) − 2 3(t)

⎧

ϕ′3(t) = ⎨

⎩

0 ≤ 0

⟹

t ≥ 3

Calculating ϕ4(t). Defining ϕ′4(t) as follows

′

ϕ′4(t) = s4(t) − ⟨s4(t),ϕ1(t)⟩ϕ1(t) − ⟨s4(t),ϕ2(t)⟩ϕ2(t) − ⟨s4(t),ϕ3(t)⟩ϕ3(t)

ϕ′4(t) = 0

Note that these signals obtained are different from the ones obtained in part (a). Hence, the ordering of the signals does affect the basis. This also goes to show that the computational effort varies based upon the order in which the signals are taken.

Question3

a) The discrete time signals given can be represented in a vectorial format as follows:

x1 = [1 x2 = [1 x3 = [0 1 1

1 1

0.25 1

1

0.5 1

−1

1 1 1

−1

1 1]T

−1 −1]T

0.5 0.25 0]T

x4 = [−1 −1 −1 −1 1 1 1 1]T

b) Let the orthogonal basis be B = {u1,u2,u3,u4}. In accordance with the Grahm Schmidt Algortihm, the basis generated would be as follows: (Note how normalization helps with projection calculations)

u x1 x2 − ⟨x2,u1⟩ x3 − ⟨x3,u1⟩ − ⟨x3,u2⟩ x4 − ⟨x4,u1⟩ − ⟨x4,u2⟩ − ⟨x4,u3⟩

Solving, we get the basis as:

T

u

T

u

T

u

u4 = [0 0 0 0 0 0 0 0]T

Thus, u4 isn’t a basis vector, since it is redundant. B = {u1,u2,u3}

c) The signal representation of the basis vectors calculated above is:

T

u

T

u

u3 = [0 0 0 0 0 0 0 0]T

T

u

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