use of frequency domain

LECTURE 4 4-1Copyright 1998, Texas Instruments Incorporated All Rights Reserved

Use of Frequency Domain

Telecommunication Channel |A|

ffc

Frequency Response

Signal frequencies > fc are attenuated and distorted|A|

f

JOHN’S “A”

IAN’S “A”

SPEECH ANALYSIS RECOGNITION IDENTIFICATION

Spectrum analyzers convert a time-domain signal into frequency domain


Fourier Series

Tp

Original Signal xp(t) First 4 Terms of Fourier SeriesSum of First 4 Terms of Fourier Series and xp(t)

Original xp(t)

First 4 Terms of Fourier Series

Periodic signal expressed as infinite sum of sinusoids.

Ck’s are frequency domain amplitude and phase representationFor the given value xp(t) (a square value), the sum of the first four terms of

trigonometric Fourier series are:

xp(t) 1.0 + sin(t) + sin(3t) + sin(5t)

dte)t(xT

1c

where,ec)t(x

p

0

0

T

tjkp

pk

k

tjkkp


Fourier Transform

Ck =1

Tp

Tp/2

-Tp/2

x(t) e-j(kt) dtWHERE

C() = d2 x(t) e-j t dt

Increase TP = Period Increases No Repetition1

Tp

=2

d2

k 0

Discrete coefficients Ck become continuous C()

Discrete frequency variable becomes continuous

x(t) e -jt dt

C()

d2= X() =

X() e jt d

x(t) =1

2

normalize

INVERSE

FT PAIR

x(t) = Ckej(k0 t)

k=-

TP


Discrete Time Fourier Transform

Replace t with Tsn Continuous x(t) becomes discrete x(n) Sum rather than integrate all discrete samples

Fourier Transform

Limits of integration need not go beyond ± because the spectrum repeats itself outside ± (every 2)

Keep integration because X() is continuous

Discrete Time Fourier Transform

dte)t(x)(xtj

n

j n

e)n(x)(x

de)(x2

1)t(x

tj

de)(x2

1)n(x

)n(j

Inverse Fourier Transform

Inverse Discrete Time Fourier Transform


Discrete Fourier Transformation

Recall DTFT Pair:

de)(X2

1)n(x

2

jn

n

jne)n(x)(X

To Make DTFT Practical:

There are an infinite number of time domain samples

is continuous

Take only N time domain samples

Sample the frequency domain, i.e. only evaluate x() at N discrete points. The equal spacing between points is = 2/N

The result is the Discrete Fourier Transform (DFT) pair:

FactorTwiddleeWwhere

W)k(XN

1)n(xandW)n(x)k(X

N

2j

N

1N

0k

kNN

1N

0n

kNN

nn


DFT Relationships

Time Domain Frequency Domain

N Samples

0

0

Ts

1

2Ts

2

3Ts

3 N-1

(N-1)Ts

|x(k)|

0

0

1 2 N/2 N-2 N-1

N

Fs

N

F2 s

2

Fs

N

F2 sN

Fs

N Samples

k

f

X(n)

t

n


Practical Considerations

8-point DFT requires 8 2 = 64 multiplications

1000-point DFT requres 10002 = 1 million multiplications And all of these need to be summed

Standard DFT

An example of an 8 point DFT

Writing this out for each value of n

0k7W)0(x Each term such as requires 8 multiplications

Total number of multiplications required: 8 * 8 = 64Do not forget that each multiplication is complex

1Nk0,W)k(X)k(X1N

0n

knNnN

7,...,1,0k,W)7(x.......W)1(xW)0(x)k(X 7k7

1k7

0k7n

7,...,2,1,0k,W)k(X)k(X7

0n

kn7nn


Fast Fourier TransformationSymmetry Property

Periodicity Property

THE FAST FOURIER TRANSFORM

Splitting the DFT in two

or

Manipulating the twiddle factor

kN

2/NkN WW

kN

NkN WW

2/N

)2/N

2(j)2

N

2(j2

N WeeW

1

2

N

0r

rk2N

12

N

0r

kN

rk2NN )W).(1r2(xW)W).(r2(x)k(X

1

2

N

0r

k)1r2(N

12

N

0r

rk2NN W).1r2(xW).r2(x)k(X

1

2

N

0r

12

N

0r

rk2N

kN

rk2Nn W)1r2(xWW)r2(x)k(X


Time Savings

N/2 Multiplications

(N/2) 2 Multiplications (N/2) 2 Multiplications

For an 8-point FFT, 42 + 42 + 4 = 36 multiplications, saving 64 - 36 = 28 multiplications

For 1000 point FFT, 5002 + 5002 + 500 = 50,500 multiplications, saving 1,000,000 - 50,500 = 945,000 multiplications

Time savings assume 50ns cycle time 8-point FFT saves 1.2s

1000-point FFT saves 47.25ms

1

2

N

0r

rk2/N

kN

12

N

0r

rk2/NN W)1r2(xWW)r2(x)k(x


Decimation in Time

Decimate once Called Radix-2 since we divided by 2

Splitting the original series into two is called decimation in time

n = {0, 1, 2, 3, 4, 5, 6, 7} n = { 0, 2, 4, 6 } and { 1, 3, 5, 7 }

Let us take a short series where N = 8

Decimate again

n = { 0, 4 } { 2, 6 } { 1, 5 } and { 3, 7 }

The result is a savings of N2 – (N/2)log2N multiplications

1024 point DFT = 1,048,576 multiplications1024 point FFT = 5120 multiplication

Decimation simplifies mathematics but there are more twiddle factors to calculate

A practical FFT incorporates these extra factors into the algorithm


4-Point FFT

x(n)

0

X4(k) = W4

kn3

Let us consider an example where N=4:

Decimate in time into 2 series: n = {0 , 2} and {1, 3}

+x(2r)

r=0

X4(k) =1 rk

W2W4

kx(2r+1)

r=01

rkW2

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W2

kW4

kW2

k

We have two twiddle factors. Can we relate them?

WN = e

2N

-jk kREMEMBER:

2kW2

k= e

22

-j k*= W4

= e2

4-j 2k*

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W4

2kW4

kW4

2k Now our FFT becomes:


Flow Diagram

Represent with a flow diagram:

Write out values for k=0 only:

Two DFTs:

x(0)

x(1)

x(2)

x(3)

X4(0)0

0

0

0 W4

0=

= [ x(0) + x(2) ] + [ x(1) + x(3) ], k=0,1,2,3W4

2kW4

kW4

2k

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W4

0W4

0W4

0

This is only one quarter of the flow diagram

X4(0)

X4(k)

x02

x13


Full Flow Diagram

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W4

2W4

1W4

2

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W4

0W4

2W4

0

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W4

2W4

3W4

2

= [ x(0) + x(2) ] + [ x(1) + x(3) ]W4

0W4

0W4

0

Write out all values for k:

SPOT THE BUTTERFLY ?

x(0)

x(1)

x(2)

x(3)

X4(0)

X4(1)

X4(2)

X4(3)

0

2

0

2 3

2

1

0

W4

6W4

224

-j= e

*6= -1 =

NOTICE

W4

4W4

024

-j= e

*4= 1 =


The Butterfly

W43 = j

W42 = -1

W41 = -j

W40 = 1

X0

X1

X3

X2

x3

x2

x1

x0

W40

W41

4 Point FFT Butterfly

X1

X2x2

WNk

x1 x1X1 = + WNk x2

x1X2 = – WNk x2

Typical Butterfly Twiddle Conversions

X0 = (x0 + x2) + W40 (x1+x3)

X1 = (x0 – x2) + W41 (x1–x3)

X2 = (x0 + x2) – W40 (x1+x3)

X3 = (x0 – x2) – W41 (x1–x3)

4 Point FFT Equationsa

b


A Practical Example

AMPLITUDE OF X1= 12+j2 = 2

X0=x0+ x2 + x1+x3

X1=x0–x2 + -j(x1–x3) = 1–0 + -j(0–1) = 1 + j

X2=(x0+ x2) – (x1+ x3) = (1 + 0) - (0 + 1) = 0

X3=(x0–x2)– -j (x1–x3) = (1–0)– -j(0–1) = 1– j

= 1 + 0 + 0 + 1 = 2

FFT Conversion

SAMPLED AT 10kHzTs= 100 uS

1

NTs

=F =

xk = {1,0,0,1}

Amplitude

Time0

1

2

(nTs)0 1 2 3

Time DomainAmplitude

Frequency

1

2

kHz

Frequency Domain

03 1 220-2.5 2.5 5.0-5.0

Frequency Spacing


DSP and FFT

Fast Fourier Transform is a generic name for reducing DFT computations. We considered Radix-2 here, but many other algorithms exist.

The simplified butterflies can be implemented with a DSP very efficiently Special FFT chips implement it even faster

But DSPs are programmable

And they can perform other operations on the signal

FFT requires address shuffling for faster data table access Most DSPs can perform shuffling in the background

Modern DSPs can perform an FFT of 1024 samples in well under 5 ms


Summary Frequency domain information for a signal is important for

processing

Sinusoids can be represented by phasors

Fourier series can be used to represent any periodic signal

Fourier transforms are used to transform signals From time to frequency domain From frequency to time domain

DFT allows transform operations on sampled signals

DFT computations can be sped up by splitting the original series into two or more series

FFT offers considerable savings in computation time

DSPs can implement FFT efficiently


Sliding Windows

Window is time shiftedby one sample

T

T

T

TT= 0Add new value

Input signal samplebuffer

Originalwindowed signal

New-old samplewindow

Next windowedsignal

Subtractold

T=N

Figure 1

Figure 2

Figure 3

Figure 4

T= 0

T


Example Frequency Bin

N Delay

New Sample

K2 = K1N

-

+

Further Frequency

Bins

x[N-1] x K2

K1 x (X[k]*ej2n/N)

X[k]

K1 x ej2n/N

Old Sample

Note: K1=0.999

x[0]


The Phasor ModelIm = ImaginaryRe = Real

COMPLEX PLANE

PHASOR = VECTOR ROTATING

Amplitude = ASpeed = rad per second

Re

Im

A

a

b

x(t) = a + jb j = -1where

A = a2 + b2 = t = tan -1ba

and

1. Rectangular Form 2. Polar Form

e j(t) = cos(t) + j sin(t)

x(t) = Ae j(t) where

= 2f

= 180 degrees


Modeling Sinusoids

Im

A

a

R/2

R/2

(t + ) OR(nTs + )LET sin= e j()- e - j()

2jTHEN

x(t) = R cos(t + )

x(t) = j(t + )

(e- j(t + )

+ eR2

)

Or as a sum of two phasors:

cos= e j()+ e - j()

2AND

Drawing the phasors for cos

In general:

e j() = cos() + jsin() e - j() = cos() - jsin()

REWRITE: e jt

ANDAS

b

-b

x(t) = R cos(t + )

use of frequency domain

Documents