7270675 Fast Fourier Transform Made Easy

Easy Fourier Analysis Part 1 Complextoreal.com

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SIGNAL PROCESSING & SIMULATION NEWSLETTER Fourier analysis made Easy Part 1 Jean Baptiste Joseph, Baron de Fourier, 1768 - 1830 While studying heat conduction in materials, Baron Fourier (a title given to him by

Napoleon) developed his now famous Fourier series, approximately 120 years after Newton published the first book on calculus. It took Fourier another twenty years to develop the Fourier transform which made the theory applicable to a variety of disciplines such as signal processing where Fourier analysis is now an essential tool. Fourier did little to develop the concept further and most of that work was done by Euler, LaGrange, Laplace and others. Fourier analysis is now also used in thermal analysis, image processing, quantum mechanics and physics.

Why do we need to do Fourier analysis � In communications, we can state the

problem at hand this way; we send an information-laced signal over a medium. The medium and the hardware corrupt this signal. The receiver has to figure out from the received signal which part of the corrupted received signal is the information signal and which part the extraneous noise and distortion. The transmitted signals have well defined spectral content, so if the receiver can do a spectral analysis of the received signal then it can extract the information. This is what Fourier analysis allows us to do. Fourier analysis can look at an unknown signal and do an equivalent of a chemical analysis, identifying the various frequencies and their relative �quantities� in the signal.

Fourier noticed that you can create some really complicated looking waves by just

summing up simple sine and cosine waves. For example, the wave in Figure 1a is sum of the just three sine waves shown in Figures 1b, 1c and 1d of assorted frequencies and amplitudes.


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(a) - A complicated looking wave

(b) - Sine wave 1 (c)- Sine wave 2 (d) - Sine wave 3 Figure 1 - Sine waves Let�s look at signal 1a in three dimensions. With time progressing to the right we

see the amplitude going up and down erratically, we are looking at the signal in Time domain. From this angle, we see the sum of the three sine waves as shown in Fig (1b,c,d).

When we look at the same signal from the side along the z-axis, what we see are the

three sine waves of different frequencies. We also see the amplitude but only as a line with its maximum excursion. This view of the signal from this point of view is called the Frequency Domain. Another name for it is the Signal Spectrum.

Figure 3 - Looking at signals from two different points of view

The concept of spectrum came about from the realization that any arbitrary wave is

really a summation of many different frequencies. The spectrum of the composite wave f(t) of Fig (1) is composed of just three frequencies and can be drawn as in Fig (3.1).


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This is called a one-sided magnitude spectrum. One-sided not because anything has been left out of it, but because only positive frequencies are represented. (So what is a negative frequency? Is there such a thing? We will discuss this in more detail in later. For now, suffice it to say that a negative frequency is simply a frequency which is lagging in phase.)

Figure 3.1 - The Frequency Domain spectrum of wave in Figure 1 Now let�s look at the signal in frequency domain. Think of it as a recipe, with x-

axis showing the ingredient and the y-axis, how much of that ingredients. The x-axis for a signal would show the different frequencies in the signal and y-axis the amplitude of each of those frequencies.

Let�s expand on this concept. V-8 juice for example has many different ingredients

such as celery juice, salt, water, spices, etc.. We can remove most of these ingredients one by one and the remaining liquid would still taste essentially like V-8. What we can not remove and have the item still retain its primary character is called the fundamental component. In V-8, that is tomato juice.

Signals carrying information, similarly, have a fundamental frequency along with

other lesser important frequencies. A noisy signal on the other has no single fundamental frequency. It has a flat spectrum. All frequencies are present in the signal in the same quantities. So a spectrum does not necessarily have a fundamental component. The spectrum of such a signal would be flat.

Let�s take the following complicated looking wave.


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Figure 4 - Another really complicated looking wave

The first thing we notice is that the wave is periodic. Fourier analysis tells us that

any arbitrary wave such as the above that is periodic, can be represented by a sum of other simpler waves.

Let�s try summing a bunch of sine waves to see what they look like.

Figure 5a - This is a wave of frequency 1 Hz , amplitude = 1

Figure 5b - This is a wave of frequency 2 Hz , amplitude = 1

Figure 5c - This is a wave of frequency 3 Hz , amplitude = 1

This wave is periodic with a period = 1 sec.


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Figure 5d - This is a wave of frequency 4 Hz , amplitude = 1 Each of the waves here have frequencies that are integer multiples. In more

scientific words, we say that they are harmonic to each other, similar to musical notes which are also called harmonic.

• What is a harmonic � It is a frequency that is integer multiple of the other

frequency. Waves of frequency 2 and 4 Hz are harmonics to a wave of frequency 1 Hz since they are both its integer multiples. Frequencies 2.4 and 3.6 Hz are harmonics to a wave of frequency 1.2 Hz since they are both integer multiples of 1.2 Hz.

• When the multiple factor is even, the harmonic is called an even harmonic and when the factor is odd, it is called the odd harmonic.

• Frequencies 66, 110, 154 Hz are odd harmonics of frequency 22 Hz, whereas 44, 88 and 132 Hz are even harmonics.

We write the sum of N such harmonics as

1( ) sin( )

N

nnf t tω

=

=∑ (1)

Each wave has a frequency that is integer multiple of the starting frequency ω,

which is equal to 2 (1)π in this case since f = 1 Hz. Here is what a sum of four sine waves of equal amplitude, each starting with a phase of 0 degrees at time 0 looks like.

( ) sin( ) s1 2in( ) sin( ) sin3 4( )f t t t t tω ω ω ω= + + +

Figure 6 - This is the sum of all four of the above sine waves.


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If we keep going and add a large number of sine waves of equal amplitude, the summation approaches an impulse function as shown below for N = 25. Since we added together 30 sine waves of amplitude 1, the maximum amplitude is 25.

Figure 7 - This is the sum of 25 sine waves. In the graph above, we allowed the amplitude of each harmonic to be one. Going to

the next level of abstraction, it is obvious that to represent an arbitrary wave, we need to allow the amplitude of each component to vary. Otherwise, all we will get is the scaled version of the signal in Fig (7). So we modify equation (1) by introducing a coefficient an to represent the amplitude of the nth sine wave as follows:

1( ) sin( )

N

nnf t n ta ω

=

=∑ (2)

The coefficient an allows us to vary the amplitude of each harmonic fn(t) = sin(nωt)

to create a variety of waves. Here is what one particular wave which is the sum of four sine waves of unequal amplitude looks like.

Figure 8 - Sum of four sine waves of unequal amplitude But looking at the original wave, f(t) in Fig (4), we see that it starts at a non-zero

value. No matter how many sine waves we add together, we can not replicate this wave because sine waves are always zero at time zero. But if we add some cosine waves to the sum in equation (2) which do not start at zero, we may be able to create the wave of Figure 2.


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So let�s add a bunch of cosine waves of varying amplitudes to our f(t) equation.

Figure 9a - A cosine wave of frequency 1 Hz, amplitude = 1

Figure 9b - A cosine wave of frequency 2 Hz, amplitude = 1

Figure 9c - A cosine wave of frequency 3 Hz, amplitude = 1

Figure 9d - A cosine wave of frequency 4 Hz, amplitude = 1 Once again the sum of the cosine waves of equal amplitude looks like this.


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Figure 10 - Sum of four cosine waves of equal amplitude

Figure 11 - Sum of 30 cosine waves of equal amplitude

A sum of 30 cosine waves looks like as in Fig (11). It approaches an impulse

function just as the sum of sine waves did but this one is an even function. • Even function � The function that is symmetrical about the y-axis. Cosine wave

is an even function. • Odd function � The function that is not symmetrical about the y-axis. Sine wave

is an odd function. The sum of the cosines is an even function. Contrast this with Fig (7), the sum of

sines, which is an odd function. These characteristics, odd and even, are useful when looking at real and imaginary components of signals.

Now let�s allow the amplitude of each cosine wave to vary. Here is what one

particular sum of four cosines of unequal amplitudes looks like.

Figure 12 - Sum of four cosine waves of unequal amplitude. Now let�s modify equation (2) to add the cosine waves.


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1 1( ) sin( ) cos( )

N N

n nn n

f t a n t b n tω ω= =

= +∑ ∑ (3)

The coefficients bn allow us to vary the amplitude of each cosine wave. Putting this

equation to work, we see in the following figure the sum of four sine and four cosine waves.

Figure 13 - Sum of five sine and cosine waves of unequal amplitudes We are very close to completing our equation for arbitrary periodic waves. There is

only one remaining issue. Sums of sine and cosines are always symmetrical about the x-axis so there is no possibility of representing a wave with a dc offset. To do that we add a constant, a0 to the equation. This constant moves the whole wave up (or down) along the y-axis offset.

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1( ) sin( ) cos( )

N N

n nn n

f t a n t b ta nω ω= =

= + +∑ ∑ (4)

The coefficient a0 provides us with the needed dc offset from zero. Now with this

equation we can fully describe any periodic wave, no matter how complicated looking it is. All arbitrary but periodic waves are composed of just plain and ordinary sines and cosines and can de composed in its constituent frequencies..

Equation (4) is called the Fourier Series equation. The coefficients a0, an, and

bn are called the Fourier Series Coefficients. An equation with many faces There are several different ways to write the Fourier series. One common

representation is by linear frequency instead of the radial frequency. Replace ω by 2πf and then write the equation as

01 1

( ) sin(2 ) cos(2 )N N

n nn n

f t a a nft b nftπ π= =

= + +∑ ∑ (5)


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The Fourier series equation allows us to represent any wave, low or high frequency, baseband or passband, large bandwidth or very small. N is the number of harmonics used in the summation. This is a variable and we can choose it to be anything, but for complete representation, N is set to infinity. This makes the equation completely general and we can represent even noise signals this way. The harmonics themselves do not have to be of integer frequencies such 1, 2, 3 etc.. The starting frequency can be any real or imaginary number. However, the harmonics of the starting frequency ARE its integer multiples.

1( ) ( )nf t n f t nT

= =

f(t) the smallest frequency is called the resolution frequency, determines how

finely we decompose the signal. It can be any arbitrary number, say for example 2.35. From that point on, the next harmonic is 2 times this, next one 3 times and so on. T, is the period of the first wave we pick, and each fn is an integer multiple of the inverse of that period. We can also start anywhere. We can pick a small resolution frequency and then start the analysis with the 100th harmonic for example.

Replace fn by n/T, where T is the period and replace N by ∞ to write equation (5)

in a different from.

( ) ( )01

( ) sin 2 cos 2/ /n nn

n Tf t a a t b tn Tπ π∞

=

= + +∑ (6)

We can also convert all sine waves and make them cosine waves by adding a half-

period phase shift. The cosine representation, used often in signal processing is written by adding a phase term to the equation.

sin(2 ) cos(2 2)/ft ftπ π π= + To create the f(t) we would add two cosine waves of the same frequency, except the

one of them would have a / 2π phase shift (that�s a sine wave, really.) Now we have only cosines. The name of the coefficient has been changed to cn, to reduce confusion between this term and the terms an and bn. a0 and C0 would be exactly the same as a0.

01

01

01

( ) cos( )

( ) cos( )

( ) c

2

2os( )

n

n

n nn

n nn

n nn

f t C C t

f t C C t

f

f

w

nt C C tT

π

π

φ

φ

φ

∞

=

∞

=

∞

=

= + +

= + +

= + +

∑

∑

∑

(6a)


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In complex representation, the Fourier equation is written as

/( ) nn

jn t Tf t C e π∞

=−∞= ∑ (7)

Complex notation, first given by Euler, is most useful-albeit scary-looking form. In

next part, we look at how it is derived and used for signal processing. All these different representations of the Fourier Series (4), (5), (6), (6a) and (7) are

identical and mean exactly the same thing. How to compute the Fourier Coefficients of an arbitrary wave In signal processing, we are interested in spectral components of a signal. We want

to know how many sines and cosines make up our signal and what their amplitudes are. Alternatively, what we really want are the Fourier coefficients of our signal. Once we know the Fourier coefficients, we know which frequencies are present in the signal and in what quantities. This is similar to doing chemical analysis on a compound, figuring out what elements are there and what relative quantity.

How do we compute the Fourier coefficients? Computing a0

01

( ) ( sin cos )n n n nn

f t a a t b tω ω∞

=

= + +∑

The constant a0 in the Fourier equation above represents the dc offset. But before

we compute it, let�s take a look at one particular property of the sine and cosine waves. Both sine and cosine wave are symmetrical about the x-axis. When you integrate a

sine or a cosine wave over one period, you will always get zero. The areas above the x-axis cancels out the areas below it. This is always true over one period as we can see in the figure below.

Figure 14 - The area under a sine or a cosine wave over one period is always zero.

Positive and negative area cancel.

Positive and negative area cancel. +Area

+Area +Area

-Area-Area


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1

1

sin 0

sin cos 0

T

n non

T

n n n non

a w t dt

a w t b w t dt

∞

=

∞

=

= + =

∑∫

∑∫

The same is also true of the sum of sine and cosines. Any wave made up of sum of

the sine and cosine waves also has zero area over one period. So we see that if we were to integrate our signal over one period the area obtained will have to come from coefficient a0 only. The harmonics can make no contribution and they fall out.

0

0

10

sin cos( )T

n n n no on

TT

f t dt a dt a w t b w t dt∞

=

+=

+∫ ∑∫∫644444474444448

(8)

The second term is zero in (8), since it is just the integral of a wave made up of sine

and cosines. Now we can compute a0 by taking the integral of our complicated looking wave over one period.

Figure 16 - Signal to be analyzed, looks like it has a dc offset since there is more area

above the x-axis than below.

Figure 16a - The dc component

All area comes from the a0 coefficient.

The wave has non-zero area in one period, which means it has a DC offset.


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Figure 16b - Signal without the dc component The area under one period of this wave is equal to

0 0

( )T T

of t dt a dt=∫ ∫ (9)

Integrating this very simple equation we get,

f t dt a TT

( )0

0∫ = (10)

We can now write a very easy equation for computing a0

aT

f t dtT

00

1= ∫ ( ) (11)

Since no harmonics contribute to area, we see that a0 is equal to simply the area

under our complicated wave for one period divided by T, the integral period. We can compute this area in software and if it is zero, then there is no dc offset. This is also the mean value of the signal. A signal with zero mean value has no dc offset.

Computing an

Now we employ a slightly different trick from basic trigonometry to compute the

coefficients of the sine waves. Here is a sine wave of an arbitrary frequency nω that has been multiplied by itself.

( ) sin *sinf t t tn nω ω=

Area under the wave when shifted down is zero.


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Figure 17 - The area under a sine wave multiplied by itself is always non-zero. We notice that the resulting wave lies entirely above the x-axis and has a net

positive area. From integral tables we can compute the area as equal to

( ) ( )0

sin sin / 2T

n na n t m t dt a T for n mω ω = =∫ (12)

Where T is the period of the fundamental harmonic. But now let�s multiply the sine

wave by an arbitrary harmonic of itself to see what happens to the area.

( ) sin *sinf t t tn mω ω=

Figure 18 - The area under a sine wave multiplied by its own harmonic is always zero. The area in one period of a sine wave multiplied by its own harmonic is zero. We

conclude that when we multiply a signal by a particular harmonic, the only contribution comes from that particular harmonic. All others harmonics contribute nothing and fall out.

Multiplying one sine wave by any other causes the area under the new wave to become zero.

Sine wave multiplied by another of a different harmonic


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( ) ( )

( ) ( )

0

0

sin sin 0

sin sin / 2

T

n

T

n n

a n t m t dt for n m

a n t m t dt a T for n m

ω ω

ω ω

= ≠

= =

∫

∫

(12)

Now let�s multiply a sine wave by a cosine wave to see what happens.

( ) sin *cosf t t tn mω ω=

Figure 19 - The area under a cosine wave multiplied by a sine wave is always zero. It seems that the area under the wave which is multiplication of a sine and cosine

wave is always zero whether the harmonics are the same or not. Summarizing, by setting n nω ω=

( ) ( )

( ) ( )

( ) ( )

0

0

0

sin sin 0

sin sin / 2

cos sin 0

T

n n m

T

n n m n

T

n n m

a t t dt for n m

a t t dt a T for n m

a t t dt for all n and m

ω ω

ω ω

ω ω

= ≠

= =

=

∫

∫

∫

(13)

Rules: 1. The area under one period of a sine or a cosine is zero. 2. The area under one period of a wave that is a product of two sine or cosine waves of non-harmonic frequencies is zero.

Sine wave multiplied by a cosine wave for any n and m


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3. The area under one period of a wave that is a product of two sine or cosine waves of same harmonic frequency is non-zero and not equal to / 2na T , where T is the period of the resolution frequency we have chosen. 4. The area under one period of a wave that is a product of a sine wave and a cosine wave of any frequencies (different or equal) is equal to zero. Recall that in vector representation, sine and cosines are orthogonal to each other.

So all harmonics are by definition orthogonal to each other. A very satisfying interpretation of the above rules is that sine and cosine waves can

act as filtering signals. In essence they act as narrow-band filters and take out all frequencies except the one of interest. This forms the basic concept of a filter.

Now let�s use this information. Successively multiply the Fourier equation by a sine

wave of a particular harmonic and integrate over one period as in equation below.

( ) ( ) ( ) ( ) ( ) ( )

0

0

0

00

sin cos sin( )0

sin sin sinT T

no

T T

nf t nwt dt a n ta wt dt b n t n t dn t tt dω ω ω ω= + +∫ ∫∫ ∫6447448 64444744448

We know that the integral of the first and the third term is zero since the first term

is the integral of a sine wave multiplied by a constant (Rule 1) and the third is a sine wave multiplied by a cosine wave (Rule 3). This simplifies our equation considerably. The integral of the second term is

( ) ( )0

sin sin2

Tn

na Ta n t n t dtω ω =∫ (13)

From this we write the equation to obtain an, which are the coefficients of each of

the sine waves as follows

( )0

2 ( ) sinT

na f t n t dtT

ω= ∫ (14)

The an is then computed by taking the signal over one period, successively

multiplying it with a sine wave of n times the starting fundamental frequency and then integrating. This gives the coefficient for that particular harmonic.

Imagine we have a signal that consists of just one frequency, we think it is around 5

Hz (and is a sine wave from). We begin by multiplying this signal by a sine wave of frequency .2 and each of its harmonics which are .4, .6, .8 ,�..10 and so on. Actually


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since we know it is in the range of 5 Hz, we can dispense with the lower harmonics say up to 4 and start with 4.2 and go to 5.8 Hz.

Here is all the math we do. 1. Multiple the wave with a sine wave of frequencies 4.2 and integrate the result.

Most likely the result will be zero. 2. Go to next harmonic, which 4.4. This is 22nd harmonic of the resolution

frequency .2 Hz. 3. Repeat step 1 and 2 and continue until harmonic frequency is equal to 5.8 Hz. The results will show that the integrals of all harmonics frequencies are zero, except

for the 25th harmonic, the integral of which will be equal to

25252.5

2a T a= =

25One period integral

2.5a =

Where T = 1/f = 1/.2 = 5 sec. The coefficient can now be calculated which gives the

amplitude of the wave. (We already know its frequency, which is 5 Hz, since the integral is non-zero for that component.).

Computing coefficient of cosines, bn Now instead of multiplying by a sine wave we multiply by a cosine wave. The

process is exactly the same as above.

( ) ( ) ( ) ( ) ( ) ( )

0 0

00 0

cos( ) cos cos cossin cos0

T T

n

T T

no

a n t dt a n tf t n t d n t dt b n t n t dttω ω ωω ω ω += +∫ ∫ ∫∫6447448 6444447444448

Now terms 1 and 2 become zero. (First term is zero from rule 1, the second term

due to rule 3.) The third terms is equal to

( ) ( )0

cos cos2

Tn

nb Tb n t n t dtω ω =∫ (14)

and the equation can be written as


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( )0

2 ( ) cosT

nb f t n t dtT

ω= ∫ (15)

So the process of finding the coefficients is multiplying our signal with

successively larger frequencies of a fundamental wave and integrating the results. This is easy to do in software. The results obtained successively are the coefficient for each frequency of the harmonic wave. We do the same thing for sines and cosine coefficients.

Following this process, we compute the coefficients of the following wave

Figure 20 - The signal to be analyzed Without going through the math, we will give the answers in two vectors, first is the

coefficients of the sine and second the cosine waves and the dcoffset. an = [.4 .3 .7 .3 .3 .3 .2 .3 .4] bn = [.05 .2 .7 .5 .2 .2 .1 .05 .02] a0 = .32 From this we can write the equation of the above wave as

4sin(2 1) .3sin(4 ) .7sin(6 ).3sin(8 ) .3sin(10 ) .3sin(12 ) .2sin(14 ) ..05cos(2 1) .2cos(4 ) .7cos(6 ) .5cos(8 ).2c

( ) .

os(10 ) .2cos(12 ) .1cos(14

.

) ..

3

.

.2

..f t

t t t tt t

t t tt t t t

tπ π π π

π π ππ π π π

π π π

+ ++ + + + ++ + + + ++ + + +

= +

The coefficients are the amplitudes of each of the harmonics. The resolution frequency is 1 Hz and the harmonics are integer multiples of this frequency. Now we know exactly what the components of the received wave are. If the transmitted wave consisted only of one of these frequencies, then, we can filter this wave and get back the transmitted signal.


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Summary The Fourier series is given by

01 1

( ) sin( ) cos( )n n n nn n

f t a a t b tω ω∞ ∞

= =

= + +∑ ∑

where

2n nfω π= The coefficients of the Fourier series are given by

aT

f t dtT

00

1= ∫ ( )

( )0

2 ( ) sinT

na f t n t dtT

ω= ∫

( )0

2 ( ) cosT

nb f t n t dtT

ω= ∫

where ω is the fundamental frequency and is related to T by

2 2n nnfT

ω π π= =

Coefficients become the spectrum Now that we have the coefficients, we can plot the magnitude spectrum of the

signal.


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00.10.20.30.40.50.60.70.8

Frequency

Mag

nitu

de

SineCosine

Figure 21 - The Fourier series coefficients for each harmonic You may now say that this spectrum is in terms of sines and cosines, and this is not

the way we see it in books. The spectrum ought to give just one number for each frequency.

We can compute that one number by knowing that most signal are represented in

complex notation where sine and cosine waves are related in quadrature. The total power shown on the y axis of the spectrum is the power in both the sine and cosine waves in the real and imaginary components of the same frequency. We can compute the magnitude by from the root sum square of the sine and cosine coefficients for each harmonic including the dc offset of the zero frequency value.

2 2

n nMagnitude a b= + Plot the modified spectrum

0.00

0.20

0.40

0.60

0.80

1.00

1.20

Frequency

Mag

nitu

de

Magnitude


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Figure 22 - A traditional looking spectrum created from the Fourier coefficients Voila! Although this is not a real signal, we see that it now looks like a traditional

spectrum. The largest component is at frequency = 3. The y-axis can easily be converted to dB. In complex representation, the phase of the signal is defined by

φn n nb a= −tan ( / )1 For every frequency, we can also compute and plot the phase. Phase plays a very

important role in signal processing and particularly in complex representation and shows useful information about the signal.

One thing you may not have noticed during this computation of the coefficients is

that they will be different depending on what you pick as the resolution frequency. We will get different answers depending on the choice we make for this number. In essence depending on the resolution, the signal energy leaks from one frequency to the next so we get different answers, but the overall picture remains the same. The issues of leakage will discussed later.

We also stated that the wave has to be periodic. But for real signals we can never

tell where the period is. Random signals do not have discernible periods. In fact, a real signal may not be periodic at all. In this case, the theory allows us to extend the �period� to infinity so we just pick any representative section of our signal or even the whole signal and call it �The Period�. Mathematically this assumption works out just fine for real signals.

Figure 23 - We call the signal periodic, even though we don�t know what lies at

each end.

Figure 24 - Our signal repeated to make it mathematically periodic, but ends

do not connect and have discontinuity


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The part of the signal that we pick as representing the real period is only a sample of the whole and not really the actual period. The end section of the chosen section will most likely not match as they would for a real periodic signal. The error introduced into our analysis due to this end mismatch is called aliasing. Windowing functions are used to artificially shape the ends so that they are zero at the ends and so the chosen signal portion is made artificially periodic. This introduces errors in the analysis which have to be dealt with by other techniques.

Next the complex representation. Copyright 1998, All rights reserved C. Langton Revised 2002 I can be reached at [email protected] Other tutorials at www.complextoreal.com

Tutorial 6 - Fourier Analysis Made Easy Part 2 Charan Langton, complextoreal.com

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SIGNAL PROCESSING & SIMULATION NEWSLETTER Tutorial 6 - Fourier Analysis Made Easy Part 2

Complex representation of Fourier series

e wt ijwt = +cos sin wt (1) Bertrand Russell called this equation “the most beautiful, profound and subtle expression in mathematics.”. Richard Feyman., the noble laureate said that it is “the most amazing equation in all of mathematics”. In electrical engineering, this enigmatic equation is equivalent in importance to F = ma. This perplexing looking equation was first developed by Euler (pronounced Oiler) in the early1800’s. A student of Johann Bernoulli, Euler was the foremost scientist of his day. Born in Switzerland, he spent his later years at the University of St. Petersburg in Russia. He perfected plane and solid geometry, created the first comprehensive approach to complex numbers and is the father of modern calculus. He was the first to introduce the concept of log x and ex as a function and it was his efforts that made the use of e, i and pi the common language of mathematics. He derived the equation ex + 1 = 0 and its more general form given above. Among his other contributions were the consistent use of the sin, cos functions and the use of symbols for summation. A father of 13, he was a prolific man in all aspects, in languages, medicine, botany, geography and all physical sciences. ejwt in Euler’s equation is a decidedly confusing concept. What exactly is the role of j in ejwt? We know that it stands for −1 but what is it doing here? Can we visualize this function?

Before we continue the discussion of Fourier Series and its complex representation, let’s first try to make sense of ejwt as it relates to signal processing. Take any real number, say 3, and plot it on a X-Y plot as in Fig 1a. Multiply this number by j, so it becomes 3j. Where do we plot it now? Herein lies our answer to what multiplication with j does.

3

3j

Y

Each time the number ismultiplied by i,it shifts by 90o.

X

-3j

-3

3

3i

Y

Phase shift due tomultiplication with j

X

Figure 1a - Relationship of real Figure 1b – Multiplication with j and imaginary numbers represents a phase shift

The number stays exactly the same, 3j is the same as 3, except that multiplication with j shifts the phase of this number by +90o. So instead of an X-axis number, it becomes a Y-axis number. Each subsequent multiplication rotates it further by 90o in the X-Y plane as shown in Figure 1b. 3 become 3j, then -3 and then -3j and back to 3 doing a complete 360 degree turn. Division by j means the opposite. It shifts the phase by -90o. (Question: What does division by -j mean?)

This is essentially the concept of complex numbers. Complex numbers often thought of as “complicated numbers” follow all of the common rules of mathematics. Whereas in calculus of real numbers, we deal with numbers along a line in one dimension, in complex math, we allow numbers to move in many dimensions and have an another property called phase associated with them. Perhaps a better name for complex numbers would have been 2D numbers.

To further complicate matters, the axes, which were called X and Y in our Cartesian mathematics are now called respectively Real and Imaginary. Why so? Is the quantity 3j any less real than 3?

This semantic confusion is the unfortunate result of the naming convention of

complex numbers and helps to make them confusing, complicated and of course complex


Page 2

3

3

Y

X

3+j3

cos wt

sin wt

X

ejwt

Figure 2 – a. Plotting complex numbers b. plotting a complex function

Now let’s plot a complex number, 3 + j3. In Cartesian math we would write this

number as (3,3) indicating 3 units on the X-axis and 3 units in the Y-axis. Similarly, the real quantity is plotted on the X-axis (real part) and the j coefficient (imaginary part) is plotted on the Y-axis. These are the X-Y projections of this number. The projection magnitudes are real and not encumbered by the vexing j.

A complex number can have for its coefficients, instead of numbers, equations

(cos x, sin x). We plot these in exactly the same way as shown in Figure 2b except that X and Y projections instead of being numbers, are functions, namely sine and cosine in this case. Now let’s take a look at the ejwt again. It is called a Cisoid {(cos x + j sin x)usoid} from contraction of the parts of the Euler’s equation.

Now forget about the ejwt part and concentrate only on the RHS containing sines and cosines.

e t jjwt = +cos sinω ωt We plot this function by setting the X-axis = cos wt and the Y-axis = sin wt. This

plot is shown in Figure 3.


Page 3

Figure 3 – ejwt plotted in three dimensions is a helix

In Figure 3 cos wt is plotted on the Real axis and sin wt is plotted on the Imaginary axis. The function looks like a helix moving forward in time to the right. The X-Z and the Y-Z projections, if plotted, would be the sine and cosine functions.

Had we plotted the function e-jwt, we would have seen that it moves to the left instead of to the right. This direction of rotation has important implications for the definition of frequency.

The quantity “ee-to-the-jay-omega-tee” is a mouthful and is commonly called a Phasor, particularly in electrical engineering. Phasors are plotted with time dimension suppressed, so they look like a vector frozen in time with its plane rotating with the angular frequency of the cisoid. Now let’s express sines and cosines in terms of our new quantity ejwt. So we have

Timecos wt

sin wt

Imaginary Axis

Real Axis

wte wt jjwt = +cos sin

and (2)

e wt jjwt− = −sin cos wt


Page 4

Manipulating these two equations, we get (3)

jeewt

jwtjwt

2sin

−−=

cos wt e ejwt jwt

=+ −

2

Now let’s just substitute Q+, for ejwt and Q- for e-jwt , we get (4) 2

cos −+ +=

QQwt

sin wt Q Qj

=−+ −

2

The use of Q is just to make it easier to see what is happening. We have redefined

sine as a difference between two phasors Q+ and Q- and cosine as the sum of the same of the same two phasors. The presence of j in the definition of sine means that it is -90o to the other term and nothing more. So mentally erase the j in the denominator, if it bothers you. The phasor Q+ is arbitrarily defined to rotate in the counterclockwise direction and the Q- phasor in the clockwise direction. The vector sum of these two phasors is changing with time and represents the cosine and sine functions. In Figure 4 we show two phasors at a particular time. They always rotate in opposite directions and meet each other at 0 and 180 degrees. Their instantaneous vector sum equals the quantity (2 cos(wt)) and their vector difference equal (2 sin(wt).)

Y

X

ejwt

Phasor Q+ rotatescounterclockwisewith time

Phasor Q- rotatesclockwisewith time

e-jwt

Q- Q+

2sinwt = ejwt - e-jwt

2coswt = ejwt + e-jwt

Figure 4 – ejwt and e-jwt phasors


Page 5

In Figure 5 we plot the progression of these two phasors to see how their sum and differences would equal the cosine and sine function. Each picture depicts the phasors at a particular time. Time is increasing as one moves from left to right then to retrace as in reading a page.

Imaginary

ejwt

e-jwtReal

Imaginary

ejwt

e-jwt

2 coswt = 2/sqrt(2)Real

1

1

22 sinwt = 01

1

2 sinwt = 2/sqrt(2)

2 cos wt = 2

Figure 5a - Phasor representation of sine and cosine, 1. Wt = 0, 2. Wt = pi/4 At t = 0, both phasors are horizontal. Their vector sum is twice the length of each.

So cos wt = 1 and since the difference is zero, sin wt = 0

At t = pi/4, the Q- phasor has rotated up to pi/4 and the Q- phasor has rotated to -pi/4. Now their vector sums, give us cos wt = 1/sqrt 2 and their difference gives also 1/sqrt2.

Imaginary3. wt = π/2

ejwt

e-jwt

Real

Imaginary

At wt = -3pi/4

ejwt

e-jwt

Real

2 cos wt = 0

2 sinwt = 2

2 coswt = -2/sqrt(2)

2 sinwt = 2/sqrt(2)

4. wt = 3π/4

Figure 5b - Phasor representation of sine and cosine, 1. Wt = pi/2, 2. Wt = 3pi/4 At t = pi/2, Q+ phasor has rotated upright and the Q- has rotated down to the

opposite side. Now the vector sum gives us the cos wt = 0 and sin wt = 1.


Page 6

At t = 3pi/4, we get the same situation as at t = 0, but the cosine term is negative

as it should be.

Imaginary

ejwt

e-jwt-2 cos wt = -2

Real

2 sin wt = 0

5. wt = π

Figure 5c- Two phasors at wt = pi/2

At wt = pi/2, the phasors meet again. The sine term which is the difference is once

again zero and the cosine term is the sum of the two magnitudes and as such cos wt = 1 and sin wt =0.

By following each phasor we see that at every t, we get the conventional and correct values of sine and cosines. Now we make the following important points that will help us in dealing with concepts of negative frequency and signals in quadrature. 1. Cosine wave is sum of two phasors rotating in opposite directions divided by 2. 2. Sine wave is difference of the same two phasors divided by 2. 3. Since any real periodic signal can be represented as a sum of sines and cosines, then

it also be represented as a sum of positive and negative phasors (also called exponential).

4. Just as we could create a spectrum out of the coefficients of the sinusoids, we can do the same thing out of the coefficients of the phasors.

If we think about sine and cosines strictly in terms of phasors and forget about the

old trigonometric definition of sine in terms of frequency and amplitude, we can talk about (but using old terminology) the concept of negative frequency.


Page 7

We can say that both sine and cosine waves are made up of two quantities called phasors, a phasor of positive frequency, ejwt and a phasor of negative frequency, e-jwt. So both sine and cosine contain negative frequency content. The idea is similar to talking about negative colors or negative people. These are perceived as physical properties and it is hard for us visualize them as negative. But when seen from a mathematical perspective, there is such a thing as a negative color; white can bee seen as negative of black and according to my esteemed colleague Dr. Dave Watson, there is definitely such a thing as “negative person.” but of course none in Advanced Systems!

This terminology is confusing because in complex domain we are not talking about frequency at all but the exponent of the exponential, ejwt. The Q+ phasor represents the positive frequency content and the Q- phasor the negative frequency because of the sign of the exponent. Each phasor then represents only the positive or the negative frequency.

Here is a hardware oriented view of negative frequency. A two-pole permanent

magnet AC generator connected to same shaft with their field windings in space quadrature will produce a positive frequency output by driving the shaft in one direction of rotation. And a negative frequency output when driven in the opposite direction. So it is direction of the motion that determines the sign of the frequency.

The difficulty is that frequency is really a two dimensional concept but is often

seen only as one. Two dimensions are needed to describe a frequency, its cycles per second and its direction of rotation. Historically we have always talked of frequency as a physical quality of a wave. Spectrum analyzers and other electrical measuring devices are one dimensional as well which limits our understanding of the general concept of frequency.

The general concept of frequency can be written as follows

f ddt

=φ

We can define frequency as the rate of change of phase over time. So a + 2π rotation over half second means the frequency is 2. And here we see that if phase rotates around counter-clockwise, then we have the definition of positive frequency and when it goes the other way then it is negative. A - 2π rotation over half second means the frequency is -2. Velocity or speed which we also tend to think of as a scalar has a similar confusing aspect. We can talk about 60 miles per hour and this makes perfect sense. But what does –60 miles per hour mean? Mathematically it is a perfectly OK construct. It just means same speed but going backwards. The concept of negative frequency is just as simple as that.

What use is ejwt? Why bother with it at all?


Page 8

Recall that we can use a sinusoid as a filter. When we multiply a signal by a sinusoid of a particular frequency, the product when integrated reveals the frequency of the sinusoid hidden in the signal. To compute trigonometric coefficients, this is essentially what we do, we multiply a random signal by sinusoids of different frequencies to yield all its frequency components. Multiplying by ejwt does exactly same thing. Except that now instead of doing sines and cosines one at a time we can do them both together. The function allows us to deal with two dimensional signals together.

We can also interpret the multiplication as a form of frequency shifting. When we

multiply a signal by ejw0t, then we are essentially isolating and shifting that signal to the w0 frequency to the right. When we multiply it by a e-jw0t, then we are shifting it leftwards to - w0.

Figure below shows the effect of this multiplication. Figure 8a shows the

Amplitude spectrum centered about frequency = 2. Multiplying this signal by e where f = 2 causes the spectrum of the new signal to shift to 4 for a total shift of f = 2.

j f t( )2π

When we multiply this signal by where f = -2 causes the spectrum of the new signal to shift to 0 for a total shift of f = -2 as in Figure 8c.

e j f t( )2π

This important property of Cisoid allows us to shift signals from baseband to

carriers and vice versa. It is a fundamental equation whenever we talk about modulation.

1 2 3 Frequency0

Figure 6a - Amplitude Spectrum of an arbitrary signal f(t)


Page 9

SIGNAL PROCESSING & SIMULATION NEWSLETTER Discrete Fourier Transform (DFT) and the FFT Let’s take this continuos signal composed of three sinusoids.

Assume that fa = 1, fb = 3, and fc = 5. The waveform is plotted below.

Figure 1 - A signal representing a square wave We can draw the Fourier transform of this signal easily by examining the amplitudes of each of these frequencies and then putting one-half on each side of the y-axis as shown in Fig. 2 for a two sided spectrum. Real signals such as this one produce only one sided spectrums also shown below.

Figure 2 - The two-sided and single-sided Fourier Transform of g(t)

This is the theoretical Fourier Transform of the continuos waveform g(t). The Fourier Transform tells us that there are just three frequencies in the signal and no others. There is no ambiguity in the results. This ideal Fourier Transform is what we want to see when do the Fourier Transform on a analyzer or on a computer but in reality this is nearly impossible to obtain. All implementations of the Fourier Transform are attempts to achieve the theoretical results, however, digital signal processing introduces approximations and truncation effects which keep us from realizing the ideal.

Figure 3 shows the outline of the same signal along with dots that represents what we actually see of the signal on a oscilloscope. This is because most signals we capture are sampled versions of the real analog signal. We pulse the analog signal every so often and then plot these sampled values. We connect the samples and get a proxy to the signal.

g t f t f t f ta b c( ) cos( ) cos( ) cos( )= − +2 2 3 2 513

15π π π

f

G(f)

1/21/6

1/10

f

G(f)

1

1/31/5

Page 1 of 20Here is a wave

5/4/2006http://complextoreal.com/fft3.htm

Figure 3 - The discrete samples of a real signal as shown by dots. We really do not know the underlying shape of the

signal. Mathematically, the sampled signal is obtained by multiplying the target signal with an impulse train of

period τ. Since we usually collect only a limited number of samples we limit the length of the impulse train to a certain time window. The discrete signal is expressed as

So before we can even look at a signal, two things have happened. 1. we have multiplied the target signal

by an impulse train and made the continuos signal a discrete signal and 2. we have chosen to collect only a limited number of samples, in effect windowing the sequence with a rectangular window function.

Fig 4a shows the original signal and its Fourier Transform. In (b) we have the Fourier Transform of a

pulse train which is used for sampling the original signal. The Fourier transform of the impulse train consists of just one frequency, the sampling frequency.

Next step is the rectangular window that limits the infinite impulse train. Its Fourier Transform is shown

in c and is the well-known sinc function. Now we have a signal which is a product of three signals.

for t < T

1 2 3 The first is the original signal, the second is the impulse train of period τ and the third is a step function

lasting for time T. What about the Fourier Transform of the product of these three signals? Mathematically we know that multiplication in time domain of two signals results in convolution of their spectrums in the frequency domain. So we have an inkling that the convolution of all three of the Fourier Transforms may not give us the spectrum in (a). But is it close enough to (a), and if not how different is it?

g k g t t k( ) ( ) ( )τ δ τ= −

g k g t t k u t( ) ( ) ( ) ( )τ δ τ= −



Figure 4 - The quandary of digital signal processing a. The continuos target signal, b. is multiplied by an impulse train of frequency fs, c. The limiting window of length T,

d. The sampled and limited signal, what do we get here? Let’s define some terms. Sampling frequency: fs

Sampling frequency is a measure of how often we pulse the continuos signal to obtain the samples. The

quantity sampled is the amplitude of the signal. Sample Time: τ

Τime between each sample. It is also the inverse of the sampling frequency. If sampling frequency, fs = 20 samples/sec, then

τ = 1/fs = .05 sec

(a)g(t)

fs(t)

g(kτ)

f

G(f)g(t)

t

f

F(f)

t fs-fs

g(t)

t f

?

(b)

(c)

(d)

w(t)

W(f)

G(f)

f

The target signal

Infinite impulse train

Limiting window

Sampled and limited signal of length N samples



Total Number of Samples: N

The total number of samples collected or available = N

Sequence Time Length: T This the time length of the collected sequence and is equal to the product of the sample time and the total number of samples. = N τ Sample index: k The sequence time is no longer continuos so instead of t, we use a discrete time measure called kth sample. This is an index of the samples. Its range extends from 0 to N-1, where N is the last sample. Each kth sample of total N samples, is located at time k times τ secs.

Kth

Figure 5 - Referring to individual samples

In a continuos signal we refer to a particular point at its instantaneous value of t. For discrete signals, We refer to any particular sample as g(kτ). So each sample differs in time from the previous one by τ secs. For example the 3rd sample similarly is located at (3 x .05) = .15 secs and the 10th sample is located at time (10 x .05) = .5 secs..

Define the g(t) in its discrete form as g(t) = g(k τ)

Harmonic index: n Frequencies that are integer multiple of a fundamental frequency are referred to as harmonics of that fundamental frequency. In computing DFT, we use the concepts of harmonics in a special way. From the sampling theorem, we know if we want to recreate a signal from its samples then we must collect at least twice its frequency number of samples per second. This also says that we can detect frequencies in a signal only up to one half of its sampling frequency. So the values of n ranges from

This says that we can only detect half as many harmonics as the total number of samples. However the index itself goes from -(N-1) to +(N-1) and spans both sides of the spectrum reflecting the positive and negative components of the frequency.

0 1 2 3 4 . k N-1

τ τ τ τ τ ττ

n fN

f

or n N

s s≤

≤

2

2



Relating the Discrete Fourier transform to the Continuous Fourier Transform Recall that the Fourier transform is given by

-1-

The above equation says that if we multiply the target signal by a complex sinusoid of harmonic frequencies 1, 2, 3, ..n one at a time and then integrate the results, the integration yields the amplitude of the nth harmonic. Why? Because multiplication by the sinusoid acts as filter for all other frequencies. (Refer to Fourier Transform Tutorials No. 1 and 2). The objective is to compute the complex coefficients cn, which when plotted give us the frequency content of the signal.

= amplitude of the nth harmonic sine wave

= amplitude of the nth harmonic cosine wave

the complex coefficients which produce the spectrum

The integration limits in the Fourier transform formula of Eq. 1 go from - in to + in. What does that mean for our sampled sequence of N samples? Fourier Transform also requires that the signal be periodic. But looking at only N samples we can not tell if the samples cover one exact period, more than one or less than one period. In order to do the Fourier Transform, we need at least one whole period or the result is suspect. We already see some problems as we go from continuous to discrete processing, The problems are

1. We do not have an infinitely long series and 2. We do not know if the N samples we have observed cover a single period, less than a single period or more than one.

Let’s continue despite the fact that we don’t know if what we are about to do is right. We are going to assume that these two things will not cause us much trouble and the results will be acceptable. Now let’s change time from continuos to discrete by making the following substitutions for time and frequency.

g(t) = g(k τ)

Equation 1 becomes

-2-

G f g t e dtj t( ) ( )= −

−∞

+∞

∫ ω

a g t n t dtn = −∞

+∞

∫ ( )sin( )ω

b g t n t dtn = −∞

+∞

∫ ( )cos( )ω

c a bn n n= +2 2

∞ ∞

ω πn nf= 2

G f g k enj f k

k

Nn( ) ( ) ( )( )= −

=

−

∑ τ π τ2

0

1



We have also changed the integral to summation to note the change from continuous to discrete as two processes are equivalent in the two domains. Fundamental frequency of the signal By applying the Fourier Transform algorithm on these N samples, we have made an implicit assumption. We have assumed that the signal is periodic over N samples, so we have assumed that the fundamental frequency of our signal is equal to the inverse of time T of the N samples. We express the fundamental frequency as

We can rewrite T as a function of sample time, τ and total number of samples chosen, N to alternately express the fundamental frequency in terms of fs and N.

This is a very important concept to understand. It says that you have artificially set the fundamental frequency to the sampling frequency divided by the total number of samples observed. It is a strange idea seemingly having nothing to do with the target signal and in fact this is true. This frequency referred to as the fundamental frequency of the signal really is kind of a resolution frequency and has nothing to do with the target signal. It just means that we resolve the target signal components in integer multiples of this resolution frequency. Let’s say that we sampled the above signal at sample time of 1/20th sec and observed 60 samples. Then the fundamental frequency is

Now when we compute the Fourier Transform we will be stepping this fundamental frequency by integer multiples. With f0 = .333, the next harmonic would be f1 = .666 and so on. The harmonics used in the analysis are not integers but integer multiple of the fundamental frequency of the signal as determined by the sampling frequency and the N samples observed. An alternate way to see these harmonics is see them as bins which collect energy. In DFT they are also called cells. Now the nth harmonic can be expressed as n times f0.

this is also equal to

fT01

=

fN s

samplex Total no of sample

01 1

= =τ sec .

f fN

s0 =

f fN

Hzs0

2060

333= = = .

f n fn = 0



Now we rewrite the Fourier Transform substituting above expression for fn in Eq. 2.

τ’s cancel and we get,

-3-

The above form of the Fourier Transform is called the Discrete Fourier Transform (DFT). The Division by N is used to normalize the values. DFT is a special case of the Fourier Transform and is actually an approximation of the real thing. The validity of the approximation is effected by the type of waveform we are dealing with as well as the parameters fs and N. Computing the DFT The process of computing the DFT is identical to computing the Fourier coefficients we did in Tutorial 1. Here you need to know 1. What is the sampling frequency of the target signal? Is the sampling frequency large enough so that it covers all significant frequencies in the signal? 2. How many samples do we need? First compute the fundamental frequency, and starting with the fundamental frequency we multiply the discrete signal by a complex exponential and perform summation on the result. Do you recall what it means to multiply by a complex exponential? How do you interpret the following equation?

The figure below shows what is happening in real-life.

f nN

n fNn

s= =τ

G f g k en

j nN

k

k

N( ) ( )

( )( )=

−

=

−

∑ τπτ

τ2

0

1

G nN

g k ej n

N k

k

N

ττ

π⎛⎝⎜

⎞⎠⎟ =

−

=

−

∑ ( )2

0

1

G nN N

g k ej n

N k

k

N

ττ

π⎛⎝⎜

⎞⎠⎟ =

−

=

−

∑1 2

0

1( )

f t e j ft( ) − 2π



Figure 6 - What we are really doing when we multiply by a complex exponential

The signal in fact is being split into two parts, 1. multiplied by a sine wave and the other by a cosine of the same frequency. The resulting two signals are orthogonal and are the result of multiplication with the complex exponential or phasor. DFT Step by step Now we compute the DFT of the signal in Fig 1. Step 1 - Multiply the target signal in 7a by a cosine wave in 7b of frequency f0. For this demonstration, we assume that f0 = 1. (Although only cosines are shown, we do this for both sines and cosines and keep track of the results separately.)

Figure 7a - The sampled target signal

Figure 7b - First Harmonic f1 = 1 * f0

f(t)

cos(2π ft)

sin(2π ft)

X

X f(t) sin(2π ft)

f(t) cos(2π ft)



Figure 7c - result of multiplying the first harmonic with the target signal. The waveform has positive area.

The multiplication gives us the waveform in 7cNow integrate this waveform over the N samples. In a discrete case, we integrate by multiplying the sample amplitude by the width of the base which is equal to τ, the sample time, using the trapezoidal rule. We are in effect adding up the areas of all the small gray rectangles in Figure 7c. Each sample value is multiplied by τ and these areas are summed. The result of the multiplication tells us something interesting. We see that the resulting waveform is not even, so it has net area under it. This means that there is a signal hiding in this frequency. What is the amplitude of this frequency? That we know only when we complete the summation. The result of the summation gives us the amplitude of this harmonic in the target signal. Step 2: Now multiply the Signal in 7a with the second harmonic as shown in Figure 7d. The multiplication gives the waveform in Figure 7e.

Figure 7d - 2nd Harmonic f2 = 2 * f0

Figure 7e - Result of multiplying the 2nd harmonic with the target signal. The waveform has no area.

The waveform of 7e is even, which means that the summation of the little gray rectangles will give zero area. Since it has no net area means there is nothing of interest here. Let’s go to the next harmonic. Now multiply the target signal with the 3rd harmonic as in Fig 7f. The resulting waveform is shown in Fig 7g.



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