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##### 1. The Discrete Fourier Transform Given a function s(t) defined-(Answered)

Description

Question

wave physics I have to submit it tomorrow I want correct answer and I give high price and I attach file you need it ?and we have to use?ft-eqs-2.pdf I already attach .

1.

The Discrete Fourier Transform

Given a function s(t) defined over the interval

?T

2

? t ?

T

2

sampled at discrete intervals ?t = ?, we measure sample values sr

where ?n ? r ? n ? 1 and n =

T

2?

(total number N = 2n).

We can represent sr by a set of N coefficients, S m where :

sr =

n?1

X

mr

S m e2?j( N

)

m=?n

and we can find the coefficients using sr in :

Sm =

X

mr

1 n?1

sr e?2?j( N )

N r=?n

This works because of the orthogonality relations :

n?1

X

r=?n

sin(

2?k r

2?k r

) cos(

) = 0 for all k, r integers

N

N

and

n?1

X

2?k r

2?k r

) sin(

)=

sin(

N

N

r=?n

0

k 6= r

N

2

k = r 6= 0

0

k=r=0

and

n?1

X

2?k r

2?k r

) cos(

)=

cos(

N

N

r=?n

2.

0

N

2

k 6= r

k = r 6= 0

N k=r=0

The Fourier Series

Now we extend the Discrete Fourier Transform to continuous

sampling. Rewrite equation 2 as :

Sm =

1

N?

P

mr?

sn ? e?2?j N?

Take the limit as ? ? 0, N ? ? such that N? = T . Then r? ? t,

S r ? = S(t)dt and

1

? Sm =

T

Sm

Z

T

2

?T

2

s(t) e?2?j

mt

T

dt

but still

?

X

s(t) =

Sm e2?j

mt

T

m=??

3.

The Continuous, Infinite Fourier Transform

Finally we can take the limit as T ? ?, but first rewrite 4 as

s(t) =

?

X

(T Sm ) e2?j

mt

T

?

m=??

then as T ? ? let

s(t) =

S(f ) =

?

Z

??

Z

m

T

? f,

1

T

1

T

? df and T S? S(f ) to get :

S (f ) e2?jf t df

?

s(t) e?2?jf t dt

??

4.

Convolution

Imagine a process which returns y(t) given input x(t), where :

y(t) =

Z

?

0

h(u) x(t ? u) du

or the causal, physically realizable version :

y(t) =

Z

t

??

x(u) h(t ? u) du

then the Fourier Transform of y(t) is :

Y(f ) =

Z

=

Z

?

hZ

e?2?jf t dt

??

??

?

??

?

h(u)x(t ? u)du

h(u) e?2?jf u du ?

Z

?

??

i

x(t ? u)e?2?jf (t?u) d(t ? u)

= H(f )?X (f )

which is just the multiplication of the transforms, so that :

Z

y(t) =

?

??

X (f ) H(f ) e2?jf t df

5.

The Autocorrelation Function

Given the series of sample values sr , we can compute the

Autocorrelation Function (ACF, not normalized here) as :

cs (u) =

1

N

n?1

X

(sr )(sr+u )

r=?n

where s =

1

N

n?1

X

sr and note that when r + u &gt; n ? 1 or r + u &lt; ?n

r=?n

we can either assume that the function sr repeats with period N so

that sr+u = sr+u?N or sr+u = sr+u+N or we can just take sr+u = 0

when (r + u) &gt; n ? 1 or (r + u) &lt; n.

Compare the autocorrelation function (acf) with the variance :

cs (0) =

1

N

P

(sr ? s)2

Divide cs (u) by cs (0) to get the normalized autocorrelation

function :

rs (u) =

cs (u)

cs (0)

Autocorrelation is sometimes called ?self convolution?. In the

continuous case the ACF is given by :

cs (u) =

1

T

Z

T

2

[s(t)][s(t + u)] dt

? T2

The normalization is sometimes taken as

1

|T ?u|

since in fact the limits

on the integral are max(? T2 , ? T2 + u) to min( T2 ,

T

2

+ u). This applies

to the case where we assume that the functions do not repeat, if we

assume that they repeat for values of t &gt;

normalization stays simply

6.

T

2

or t &lt; ? T2 then the

1

.

T

The Autocorrelation Function is the Fourier Transform

of the Power Spectrum

6.1.

The Discrete Case

Consider the squares of the Fourier Series Coefficients :

?t

N

C s (m) =

n?1

X

?2?j mr

N

sr e

r=?n

=

?t

N

XX

r

sr sr? e?2?j

2

m(r?r ? )

N

r?

Here neither of the sums is over m, so there is no orthogonality, no

delta function, no collapse of the double sum as in the simple Fourier

Series. To simplify it we define u = (r ? r ? ) and v = r ? so that :

C s (m) = ?t

X

e?2?j

mu

N

u

(

X

1 n?1

sv sv+u )

N v=?n

the term in parentheses on the right is the ACF, and the formula is

simply the Fourier transform of it ! Thus :

C s (m) = ?t

X

u

e?2?j

mu

N

cs (u)

6.2.

C s (f ) =

=

1

T

1

T

Z

R

T

2

The Continuous Case

2

s(t) e?2?jf t dt

?T

2

Z

dt

T

2

?T

2

?

dt? s(t) s(t? ) e?2?jf (t?t )

Again, set u = (t ? t? ), v = t? to get :

C s (f ) =

1

T

Z

T

du e?2?jf u

?T

Z

min( T2 , T2 +u)

max(? T2 ,? T2 ?u)

dv s(v) s(u ? v)

= F.T.[cs (f )]

So finally

C s (f ) =

1

T

Z T

2

?T

2

and in the discrete case

C s (f ) =

?t

N

?2?jf t

s(t) e

dt

2

|Sm |2

where again ?t is the sampling interval, and the frequency variable f

ranges from :

1

? 2?t

? f ?

1

2?t

Paper#9210535 | Written in 27-Jul-2016

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