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SIGNALS ANALYSISSIGNALS ANALYSIS

Analogy Between Vectors and Signals

There is a perfect analogy between vectors and signals.

Vector

A vector contains magnitude and direction. The name of the vector is denoted by bold face typeand their magnitude is denoted by light face type.

Example:V is a vector with magnitude V. Consider two vectors V1and V2as shown in the

following diagram. Let the component of V1along with V2is given by C12V2. The component of a

vector V1along with the vector V2can obtained by taking a perpendicular from the end of V1to

the vector V2as shown in diagram:

The vector V1can be expressed in terms of vector V2

V1= C12V2+ Ve

Where Ve is the error vector.

But this is not the only way of expressing vector V1in terms of V2. The alternate possibilities are:

V1=C1V2+Ve1

V2=C2V2+Ve2

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The error signal is minimum for large component value. If C12=0, then two signals are said to be

orthogonal.

Dot Product of Two Vectors

V1. V2= V1.V2cos

= Angle between V1 and V2

V1. V2=V2.V1

The components of V1alognV2= V1Cos =

From the diagram, components of V1alognV2= C 12V2

Signal

The concept of orthogonality can be applied to signals. Let us consider two signals f1 and f2 .

Similar to vectors, you can approximate f1 in terms of f2 as

f1 = C12f2 + fe for (t1< t < t2)

fe = f1 C12f2

One possible way of minimizing the error is integrating over the interval t1to t2.

However, this step also does not reduce the error to appreciable extent. This can be corrected bytaking the square of error function.

Where is the mean square value of error signal. The value of C12which minimizes the error, you

need to calculate

Derivative of the terms which do not have C12 term are zero.

V1.V2V2

.V1 V2= 2V2 C1 V2

=C12.V1 V2

V2

t tt t

t t t

t t t

[ (t)]dt1t2 t1

t2t1

fe

[ (t) (t)]dt1

t2 t1 t2

t1

f1 C12f2

= [ (t) dt1t2 t1

t2t1

fe ]2

[ (t) dt1t2 t1

t2t1

fe C12f2]2

= 0ddC12

[ [ (t) (t) dt] = 0ddC12

1t2 t1

t2t1

f1 C12f2 ]2

[ (t) 2 (t) (t) + (t) ]dt= 01t2 t1

t2t1

ddC12

f21d

dC12f1 C12f2

ddC12

f22 C212

2 (t) (t)dt + 2 [ (t)]dt= 0t2t1

f1 f2 C12t2t1

f22

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If component is zero, then two signals are said to be orthogonal.

Put C12= 0 to get condition for orthogonality.

0 =

Orthogonal Vector Space

A complete set of orthogonal vectors is referred to as orthogonal vector space. Consider a threedimensional vector space as shown below:

Consider a vector A at a point (X1, Y1, Z1). Consider three unit vectors (VX, VY, VZ) in the direction

of X, Y, Z axis respectively. Since these unit vectors are mutually orthogonal, it satisfies that

You can write above conditions as

The vector A can be represented in terms of its components and unit vectors as

Any vectors in this three dimensional space can be represented in terms of these three unitvectors only.

If you consider n dimensional space, then any vector A in that space can be represented as

As the magnitude of unit vectors is unity for any vector A

The component of A along x axis = A.VX

The component of A along Y axis = A.VY

=C12(t) (t)dtt2

t1f1 f2

(t)dtt2t1

f22

(t) (t)dtt2t1

f1 f2

(t)dtt2t1

f22

(t) (t)dt= 0 t2t1

f1 f2

. = . = . = 1VX VX VY VY VZ VZ

. = . = . = 0VX VY VY VZ VZ VX

. =

{Va Vb

1

0

a= b

a b

A= + + . . . . . . . . . . . . . . . . (1)X1VX Y1VY Z1VZ

A= + + +. . . + . . . . . (2)X1VX Y1VY Z1VZ N1VN

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The component of A along Z axis = A.VZ

Similarly, for n dimensional space, the component of A along some G axis

Substitute equation 2 in equation 3.

Orthogonal Signal Space

Let us consider a set of n mutually orthogonal functions x1 , x2 ... xn over the interval t1to t2. As

these functions are orthogonal to each other, any two signals xj , xk have to satisfy the

orthogonality condition. i.e.

Let a function f , it can be approximated with this orthogonal signal space by adding thecomponents along mutually orthogonal signals i.e.

Mean sqaure error

The component which minimizes the mean square error can be found by

Let us consider

All terms that do not contain Ckis zero. i.e. in summation, r=k term remains and all other terms

are zero.

=A. V G. . . . . . . . . . . . . . . ( 3 )

CG= ( + + +. . . + . . . + )X1VX Y1VY Z1VZ G1VG N1VN VG

= + + +. . . + . . . +X1VXVG Y1VYVG Z1VZVG G1VG VG N1VNVG

= since = 1G1 VG VG

If 1 i.e. =kVG VG VG VG

A = = KVG G1VG VG G1

=G1(A )VG

K

t t tt t

(t) (t)dt= 0 wherejk t2t1

xj xk

Let (t)dt= t2t1

x2k

kk

t

f(t) = (t) + (t)+. . . + (t) + (t)C1x1 C2x2 Cnxn fe

= (t)nr=1Crxr

f(t) = f(t) (t)nr=1Crxr

= [ (t) dt1t2 t2

t2t1

fe ]2

= [f[t] (t) dt1t2 t2

t2t1

r=1

nCrxr ]

2

= =. . . = = 0d

dC1

d

dC2

d

dCk

= 0ddCk

[ [f(t) (t) dt] = 0ddCk

1t2 t1

t2t1

nr=1Crxr ]

2

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Mean Square Error

The average of square of error function fe is called as mean square error. It is denoted by

.

.

You know that

The above equation is used to evaluate the mean square error.

Closed and Complete Set of Orthogonal Functions

Let us consider a set of n mutually orthogonal functions x1 , x2 ...xn over the interval t1to t2. This

is called as closed and complete set when there exist no function f satisfying the condition

If this function is satisfying the equation then f is said to be

orthogonal to each and every function of orthogonal set. This set is incomplete without f . Itbecomes closed and complete set when f is included.

f can be approximated with this orthogonal set by adding the components along mutuallyorthogonal signals i.e.

If the infinite series converges to f then mean square error is

zero.

Orthogonality in Complex Functions

If f1 and f2 are two complex functions, then f1 can be expressed in terms of f2 as

..with negligible error

2f(t) (t)dt + 2 [ (t)]dt= 0 t2

t1

xk Ck t2

t1

x2k

=Ckf(t) (t)dtt2

t1xk

in (t)dttt2t1

x2k

f(t) (t)dt= t2

t1

xk CkKk

tepsilon

= [ (t) dt1t2 t1

t2t1

fe ]2

= [ (t) (t) dt1t2 t1

t2t1

fe nr=1Crxr ]

2

= [ [ (t)]dt + (t)dt 2 (t)f(t)dt1t2 t1

t2t1

f2e nr=1C2r

t2t1

x2r nr=1Crt2t1

xr

(t)dt= (t)f(d)dt=C2r t2t1

x2r Crt2t1

xr C2rKr

= [ [ (t)]dt + 2 ]1t2 t1

t2t1

f2 nr=1C

2rKr nr=1C

2rKr

= [ [ (t)]dt ]1t

2 t

1

t2t1

f2 nr=1C

2rKr

= [ [ (t)]dt + ( + +. . . + )]1t2 t1

t2t1

f2 C21 K1 C22 K2 C

2n Kn

t t tt

f(t) (t)dt= 0t2t1

xk

f(t) (t)dt= 0 for k= 1, 2, . .t2t1

xk t

tt

t

f(t) = (t) + (t)+. . . + (t) + (t)C1x1 C2x2 Cnxn fe

(t) + (t)+. . . + (t)C1x1 C2x2 Cnxn t

t t t t

(t) = (t)f1 C12f2

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Where

Where = complex conjugate of f2 .

If f1 and f2 are orthogonal then C12= 0

The above equation represents orthogonality condition in complex functions.

=C12(t) (t)dtt2

t1f1 f

2

| (t) dtt2t1

f2 |2

(t)f2 t

t t

= 0(t) (

t)

dt

t2

t1f

1 f

2

| (t) dtt2t1

f2 |2

(t) (dt) = 0 t2t1

f1 f2