Experiment 1: Fourier Analysis And Synthesis Of Waveforms

Experiment 1: Fourier Analysis and Synthesis of Waveforms

Leslie Boachie Yiadom Student number: 071512103 EE number : ee07u085 Tutor: Akram Alomainy

Introduction In this experiment Fourier Analysis is used to gain a better understanding of signals in both the time and frequency domains, it also can be used to see what affect certain processes and techniques have on different types and forms of signals and understand the reasons for this, it should also help gain a understanding of different types of signals by looking at similarity and differences, in order to do this and gain this better understanding of signals, a java applet is to be used to create different types of signals and waveforms to be analysed, the experiment consisted of 4 main parts A to D each having sub parts to be completed by performing different processes on the signals using the applet. In part A, a square waveform and also a saw tooth waveform are used to develop and basic understanding of how Fourier series can be used to define types of signals and also what the affects of phase shifting are on wave forms, in part B a saw tooth and cosine waveform are used to understands what quantisation can do to signals and the affects it has, part C was where a triangle waveform is used and clipped to see what affect this will have on the reliable of a signal compared to it’s original form and finally in part D is where a noise waveform was used to understand what they really consist of and why it exists.

Background Theory Signals are a form of sending information from one system to another, signals can be represented in many forms such as square, triangle or sinusoid and they can be show in different domains such as time or frequency, in a time domain where it’s possible to define a signal at every point of the wave form it’s know as continuous time signals because they can be defined continuously over a period time, these continuous signals produce over certain amount of time can be classed as two main types one being deterministic and the other being random, deterministic type of signals can be determined at all point and the future positions and values of the signal can be estimated, but for a signal which is said to be random it can not be determined as there is no patterns or repetition in this signal it is said that a random signal contains information as we do not know what we will get from the signal. Signals can also be describe as discrete where sampled points from a continuous signal are taken to define the signal by certain points.

Continuous signals can be even or odd functioned and these can easily be recognised by looking at the waveforms they produce and they can be defined in both continuous and discrete forms where t = 0 in continuous form and n = 0 in discrete form, If you look over the y axis an even functions signal in the –y axis is identical to the signal in the +y axis, another an even function can be defined is by x (t) = x (-t), in an odd function the signal in –y axis is the inverse to the signal in the +y axis, another an even function can be defined is by x (t) = -x (-t) [4] ∞

So x (t) = ½ ao + ∑

∞

an . cos (n.ω.t) + ∑ bn . sin (n.ω.t), this equation we can find the value of the

n=1

n=1

sine and cosine components in a signal and this is know as Fourier’s series T/2

where ao = 2/T∫

x (t) dt

-T/2 T/2

and an = 2/T∫

x (t) . cos (n.ω.t) dt

-T/2 T/2

and

bn = 2/T∫ x (t) . sin (n.ω.t) dt -T/2

If a signal is said to be odd then in most cases an will be equal to zero also it can be said if a signal

is even then in most cases bn will be equal to zero. [4] Fourier’s series is a form of Fourier’s Analysis know which relates to Fourier transforms, which where created by French mathematician Jean Baptiste Joseph Fourier, the main idea of Fourier’s series is that almost all period signal in the time domain can be broken down to sums of sine and cosine signals.[2]

In general the idea of Fourier’s transformer can be used as a method of changing a signal in the time domain to a signal in the frequency domain or the from the frequency domain to the time domain.

A signal in the frequency domain can be represented by pulse of amplitude representing the value of the frequency at different time periods.

In a continuous signal if we have to function x (t) and y (t) over a interval of [ - T , T ] T

then 2/T∫ x(t) y(t) dt = 0 , this is said to be orthogonally -T

In discrete signal, orthogonality can be defined when the product of two signals over the interval [ - T , T ] tends to 0 [4]

In time domain representation the signal is represented over a period of time

Sine waveform in the time domain

Cosine waveform in the time domain

In frequency domain representation the signal is represented at a frequency pulse Sine wave in frequency domain

Cosine wave in frequency domain

Experimental Processes In order to complete the experiment the use of a java applet was needed using many function of this applet many a wide way of processes can be completed.

A) Use Fourier Analysis to observe the effect of limiting the bandwidth of a real signal.

1) The first step for this experiment was to use the Square wave function and reduce the number of terms to 0, doing this should produce a flat red line, this red line is know as the synthesised signal and it is this signal which will be used to carry out the analyse of different types of signals. Also produced is a single white dot, these white dots can be used to define spectrum components in the sine and cosine frequency domain. The next step was to increase the numbers by two, this should cause the synthesised signal, to change and also produce two new components, where the sine one components should have a non-zero value and the cosine component should have a zero value, increasing the number of terms further should change the synthesised signal and increase the number of sine and cosine components at this point the use Fourier Series analysis can be used to check that these values are correct.

2) The second experiment in Part A used the square wave function along with the rectifying function in order to produce a square wave similar to the first but halved in size, using this signal the processes done in the first of Part A should be repeated.

3) Using the square wave function and increasing the number of terms to a level where the synthesised signal represent a square wave and using the phase shift function to synthesised signal should start to move this function should be used 10 times to move the synthesised signal to a specific distant to be analysed by looking at what happens to the synthesised signal and the frequency components.

4) The last experiment in Part A used the saw tooth wave function, reduce the number of terms to 0, once again the synthesised signal is should be flat. Increasing the number of terms should change the synthesised signal and new components should appear, increase the number of terms until the synthesised signal looks very much like the saw tooth waveform, at this point the number of terms should be looked at and the frequency of the highest non zero component. Increasing the number of terms to about half way and use the phase shift function 10 times the affects occurring to the synthesised signal and frequency components should be analysed. Reduced the numbers of terms to just one terms and repeat the processes of using phase shift function analyse the affects.

B) Use Fourier analysis to decide on the bandwidth needed to support a binary representation of an analogue signal.

1) The first experiment for Part B, the saw tooth waveform was used increasing the number of terms so the synthesised signal looks very much like the saw tooth waveform, using the quantizing function 3 should cause the synthesised signal to look like a staircase, this should be used to define the levels need for quantizing and periods for sampling. Repeat the first process but this time use a cosine waveform function and analyse the results. Once again using the a saw tooth waveform repeat the first process a carry out more analysis such as calculating number of bits needed to represent a certain numbers of levels, this third process was then repeated but this time the quantizing function was only used twice.

C) Use Fourier Analysis to look at an “unusual” signal. 1) For Part C the triangle wave form function was used, the number of terms had to be increased to a level were the synthesised signal had a very good representation of the original signal, using the clip function such cause the top of the waveform to flatten and start to look like a square waveform signal, while this is happening analysis of the affects of the original white signal and frequency components should be carried out. The step was to reduce the number of terms, until there were a few frequency components left, when this was done it should show the synthesised signal is a poor representation of the original white signal.

D) Use Fourier Analysis to examine noise.

1) The last part of the whole experiment used the noise waveform function, by increasing the number of terms the affects on the original white signal, the synthesised signal and frequency components should be analysed.

Results and Discussion Part A Using square waveform and reducing its number of terms to zero, the red synthesised signal became flat, this was because at that point there were no sine components to create the square waveform, when there were no sine components there was still one cosine component with zero value, this point is known as the zero frequency term. Increasing the number of terms produced more sine and cosine components and checking the sine and cosine dots value of the components could be seen with the frequency and a visual representation of the signal, by calculating Fourier’s series a comparison was made to check the

similarities between the java applets results and the calculations, to do these calculations the formulas for an and bn were used with the overall limits to being T/2 to -T/2 then by breaking the integral down into two parts with T/2 to 0 and 0 to – T/2, the value of x (t) was taken as 1 for the integral T/2 to 0 and for the integral 0 to – T/2 it was -1, where ω is equal to 2 π / T, all these value allowed the first case where n = 1 to be calculated. 0

For a1 = 2/T∫

T/2

(-1) . cos (1.ω.t) dt + 2/T∫ (1) . cos (1.ω.t) dt

-T/2

0 0

T/2

a1 = 2/T [(- sin (1.ω.t) ) + (sin (1.ω.t) )] -T/2

0

a1 = 2/T [ 0 – (- sin(-π) T/ 2π) + (sin(π) T/ 2π) – 0] = 0 + 0 + 0 – 0 = 0 0

For b1 = 2/T∫

T/2

(-1) . sin (1.ω.t) dt + 2/T∫ (1) . sin (1.ω.t) dt

-T/2

0 0

T/2

b1 = 2/T [( cos (1.ω.t) ) + (-cos (1.ω.t) )] -T/2

0

b1 = 2/T [(cos(0) T/ 2π) – (cos(-π) T/ 2π) + (- cos(π) T/ 2π) – (- cos (0) T/ 2π)] [(cos(0) / π) – (cos(-π) / π) + (- cos(π) / π) – (- cos (0) / π)]=1/π +1/π +1/π +1/π = 4/π After completing the first term, the second to the fifth terms were also calculated, then compared with the values from the java applet. 0

For a2 = 2/T∫

T/2

(-1) . cos (2.ω.t) dt + 2/T∫ (1) . cos (2.ω.t) dt

-T/2

0 0

T/2

a2 = 2/T [(- sin (2.ω.t) ) + (sin (2.ω.t) )] -T/2

0

a2 = 2/T [ 0 – (- sin(-2π) T/ 4π) + (sin(2π) T/ 4π) – 0] = 0 + 0 + 0 – 0 = 0 0

For b2 = 2/T∫ -T/2

T/2

(-1) . sin (2.ω.t) dt + 2/T∫ (1) . sin (2.ω.t) dt 0

0

T/2

b2 = 2/T [( cos (2.ω.t) ) + (-cos (2.ω.t) )] -T/2

0

b2 = 2/T [(cos(0) T/ 4π) – (cos(-2π) T/ 4π) + (- cos(2π) T/ 4π) – (- cos (0) T/ 4π)] [(cos(0) /2π) – (cos(-2π) /2π) + (- cos(2π) /2π) – (- cos (0) /2π)]=1/2π -1/2π -1/2π +1/2π = 0

0

For a3 = 2/T∫

T/2

(-1) . cos (3.ω.t) dt + 2/T∫ (1) . cos (3.ω.t) dt

-T/2

0 0

T/2

a3 = 2/T [(- sin (3.ω.t) ) + (sin (3.ω.t) )] -T/2

0

a3 = 2/T [ 0 – (- sin(-3π) T/ 6π) + (sin(3π) T/ 6π) – 0] = 0 + 0 + 0 – 0 = 0 0

For b3 = 2/T∫

T/2

(-1) . sin (3.ω.t) dt + 2/T∫ (1) . sin (3.ω.t) dt

-T/2

0 0

T/2

b3 = 2/T [( cos (3.ω.t) ) + (-cos (3.ω.t) )] -T/2

0

b3 = 2/T [(cos(0) T/ 6π) – (cos(-3π) T/ 6π) + (- cos(3π) T/ 6π) – (- cos (0) T/ 6π)] [(cos(0) /3π) – (cos(-3π) /3π) + (- cos(3π) /3π) – (- cos (0) /3π)]=1/3π +1/3π +1/3π +1/3π = 4/3π

0

For a4 = 2/T∫

T/2

(-1) . cos (4.ω.t) dt + 2/T∫ (1) . cos (4.ω.t) dt

-T/2

0 0

T/2

a4 = 2/T [(- sin (4.ω.t) ) + (sin (4.ω.t) )] -T/2

0

a4 = 2/T [ 0 – (- sin(-4π) T/ 8π) + (sin(4π) T/ 8π) – 0] = 0 + 0 + 0 – 0 = 0

0

For b4 = 2/T∫

T/2

(-1) . sin (4.ω.t) dt + 2/T∫ (1) . sin (4.ω.t) dt

-T/2

0 0

T/2

b4 = 2/T [( cos (4.ω.t) ) + (-cos (4.ω.t) )] -T/2

0

b4 = 2/T [(cos(0) T/ 8π) – (cos(-4π) T/ 8π) + (- cos(4π) T/ 8π) – (- cos (0) T/ 8π)] [(cos(0) /4π) – (cos(-4π) /4π) + (- cos(4π) /4π) – (- cos (0) /4π)]=1/4π -1/4π -1/4π +1/4π = 0

0

For a5 = 2/T∫

T/2

(-1) . cos (5.ω.t) dt + 2/T∫ (1) . cos (5.ω.t) dt

-T/2

0 0

T/2

a5 = 2/T [(- sin (5.ω.t) ) + (sin (5.ω.t) )] -T/2

0

a5 = 2/T [ 0 – (- sin(-5π) T/ 10π) + (sin(5π) T/ 10π) – 0] = 0 + 0 + 0 – 0 = 0 0

For b5 = 2/T∫

T/2

(-1) . sin (5.ω.t) dt + 2/T∫ (1) . sin (5.ω.t) dt

-T/2

0 0

T/2

b5 = 2/T [( cos (5.ω.t) ) + (-cos (5.ω.t) )] -T/2

0

b5 = 2/T [(cos(0) T/ 10π) – (cos(-5π) T/ 10π) + (- cos(5π) T/ 10π) – (- cos (0) T/ 10π)] [(cos(0) /5π) – (cos(-5π) /5π) + (- cos(5π) /5π) – (- cos (0) /5π)]=1/5π +1/5π +1/5π +1/5π = 4/5π

When comparing the calculated with those produced from the applet waveforms it showed the calculation were correct

After calculating 5 terms and looking at the 7th term is showed many combination of sine wave signals know as harmonics.

The point where the synthesised signal matched the original white waveform, contained a lot of sine and cosine components, where the cosine components had no value but the sine components increased in frequency, this high frequency would be a issues to look at from an engineering point of views as it mean this signal would have a large bandwidth so high cost and also there would was quite a bit of noise in the signal.

The second experiment in part A used the rectifying function on the java applet for the square wave causing the signal to be rectified leaving only a positive signal, when the number of terms were zero once again there were no sine components to create the square waveform, as in the first part of A when there was no sine components there and still one cosine component it’s value was now 0.5, but like the first case this point is known as the zero frequency term, but because the signal had been rectified it’s caused this zero frequency term to change.

Once again by increasing the number of terms produced more sine and cosine components and checking the sine and cosine dots gave a value for the components, the value for these signals were half of those produced in the first experiment, once again the uses of Fourier’s series could be used to calculate the values of an and bn but as the signal was rectified the limits were no longer T/2 to -T/2 and were now T/2 to 0 making x (t) now equal to 1 for the integral T/2 to 0, by calculating the first five terms in the Fourier’s series, but this time it was a little different in all cases, when n =1 the calculation was, T/2

For a1 = 2/T∫

(1) . cos (1.ω.t) dt

0 T/2

a1 = 2/T [(sin (1.ω.t) )] 0

a1 = 2/T [(sin(π) T/ 2π) – 0] = 0 – 0 = 0 T/2

For b1 = 2/T∫

(1) . sin (1.ω.t) dt

0 T/2

b1 = 2/T [(-cos (1.ω.t) )] 0

b1 = 2/T [(- cos(π) T/ 2π) – (- cos (0) T/ 2π)] [(- cos(π) / π) – (- cos (0) / π)]=1/π +1/π = 2/π

T/2

For a2 = 2/T∫

(1) . cos (2.ω.t) dt

0 T/2

a2 = 2/T [(sin (2.ω.t) )] 0

a2 = 2/T [(sin(2π) T/ 4π) – 0] = 0 – 0 = 0 T/2

For b2 = 2/T∫

(1) . sin (2.ω.t) dt

0 T/2

b2 = 2/T [(-cos (2.ω.t) )] 0

b2 = 2/T [(- cos(2π) T/ 4π) – (- cos (0) T/ 4π)] [(- cos(2π) / 2π) – (- cos (0) / 2π)]= -1/2π +1/2π = 0 T/2

For a3 = 2/T∫

(1) . cos (3.ω.t) dt

0 T/2

a3 = 2/T [(sin (3.ω.t) )] 0

a3 = 2/T [(sin(3π) T/ 6π) – 0] = 0 – 0 = 0

T/2

For b3 = 2/T∫

(1) . sin (3.ω.t) dt

0 T/2

b3 = 2/T [(-cos (3.ω.t) )] 0

b3 = 2/T [(- cos(3π) T/ 6π) – (- cos (0) T/ 6π)] [(- cos(3π) / 3π) – (- cos (0) / 3π)]=1/3π +1/3π = 2/3π

T/2

For a4 = 2/T∫

(1) . cos (4.ω.t) dt

0 T/2

a4 = 2/T [(sin (4.ω.t) )] 0

a4 = 2/T [(sin(4π) T/ 8π) – 0] = 0 – 0 = 0 T/2

For b4 = 2/T∫

(1) . sin (4.ω.t) dt

0 T/2

b4 = 2/T [(-cos (4.ω.t) )] 0

b4 = 2/T [(- cos(4π) T/ 8π) – (- cos (0) T/ 8π)] [(- cos(4π) / 4π) – (- cos (0) / 4π)]= -1/4π +1/4π = 0 T/2

For a5 = 2/T∫

(1) . cos (5.ω.t) dt

0 T/2

a5 = 2/T [(sin (5.ω.t) )] 0

a5 = 2/T [(sin(5π) T/ 10π) – 0] = 0 – 0 = 0 T/2

For b5 = 2/T∫

(1) . sin (5.ω.t) dt

0 T/2

b5 = 2/T [(-cos (5.ω.t) )] 0

b5 = 2/T [(- cos(5π) T/ 10π) – (- cos (0) T/ 10π)] [(- cos(5π) / 5π) – (- cos (0) / 5π)]=1/5π +1/5π = 2/5π These calculation and values produced by the java applet showed that Fourier’s series could be used to calculate the values

Using the phase shift function of the applet on the square waveform synthesised signal, the signal shifted and by using this functions 10 times the red line waveform had moved a whole half cycle forwards, this movement was because the synthesis red line signal was being advanced in time, using the principles of Fourier transform where a signal in the time domain would be transformed and represented in the frequency domain, it is possible to see that by an advancing phase shift in the time domain causes a reduction in frequency in the frequency domain.

The signal in the time domain after 10 clicks of the phase shift button showed that signal had been advanced by half a time period, the line spectrums also showed that pattern was the same as the original pattern but this time it was in a negative direction, the reason the frequency spectrum patterns similar to the original pattern was because the signal had not actually changed it has just been advanced in time, phase shift can be measured in degrees but it is normally found in radians using the equation x(t) = A sin (ω.t + φ) [5], where φ represents phase shifts, by taking ω is equal to 2π / T were 2π/T is also equal 2π x f , it is possible to take ω to be 2π x f to give the equation x(t) = A sin (2π x f . t + φ) [5]by rearranging this equation it is possible to find the value for phase φ = -2π x f .t [5] taking in the case where the phase shift function was used once the phase shift could be calculated to be -2π x 220 x -0.05 = 69.115 rad

Using the phase shift function, the frequency spectrums for the sine components to change to have a negative values, but this process would not affect the bandwidth needed to transmit the signal as the frequency for the signal would say the same.

The relationship between phase shift (φ) and time shift (∆t) can be found using the equation -ω x ∆t = φ, by rearranging this equation it is possible to find ∆t = φ/ω which is also equal to ∆t = -φ/2π x f, so this would mean that f = -φ/2π x ∆t [5]

Using a saw tooth shaped waveform and reducing the number of terms allowed a waveform with no sine components and a single cosine component to be produced, the analysis carried out on the

square wave form was similar to the analysis carried out here, because when there were no sine components the line was flat, because like the square wave the saw tooth signal created was an odd functioned signal, so it was made from sine components and as there were no sine terms there was nothing for the signal to be created from, similar to the square wave form the cosine term had a zero value and this was also because of the same reason which was because it was the zero frequency term, a increase in the number of terms created more sine components which had non zero values whereas the cosine components did not but this should have been expected after noticing that like the square wave form first used the saw tooth wave form was a odd signal and by using Fourier’s series on the square wave form, it was found cosine components mainly equalled zero and the reason being that an is the even function of a signal and bn being the odd.

To produce a better looking saw tooth wave form, the number of terms were increased to 21 terms and the this point the frequency was 4635 Hz, at this point the bandwidth 4415 Hz.

Increasing the number of terms to about half way produced a well represented saw tooth waveform from the red synthesised signal, using phase shift function shifted the signal in the time domain and after 10 clicks the signal moves half a cycle, but in the frequency domain representation of the signal the spectrum were different to the original signal as it by using phase shift with just the first number of terms it showed that the line spectrum for the sine components decreased and after 5 clicks it was actually zero while the cosine function has taken the value of the original sine component at this point the time domain representation of the saw tooth signal was a cosine wave form but by pressing the phase shift function a further 5 times both the cosine and sine started to go down with the cosine component returning to zero and the sine continuing to decrease down to a negative value of it’s starting value.

Part B Focusing on the process of quantizing a signal, the first step was to select a saw tooth wave form and increasing the number of terms to a stage where it represent the saw tooth waveform quite well using the quantize function of the java applet three times a staircase representation of the saw tooth signal was produced which had 5 quantization levels and this allowed number of samples per saw tooth period to be calculated as 5.

With one time period being defined by T and with frequency being found by 1/T = f, using 5 samples per one saw tooth period the sampling frequency could be calculated by multiplying the fundamental frequency of 1Hz with the numbers of 5 per period to give the sampling frequency as 5Hz.

Using a cosine wave form signal also produced 5 quantization levels but had 10 samples per cosine period and found that the sampling frequency was 10Hz

After complete this small analysis of the cosine waveform, the saw tooth signal was used again and quantised to have 5 quantization levels and 5 samples per period , if the quantized saw tooth signal was to be sent using a digital communication system it could be sent as a stream of bits, if the signal had 4 quantization levels it could easily be represented using by 2 bits with 4 levels, so for 5 levels the and this can be equation L = 2 to the power of n, were L is equal to the quantizing levels and n is the number of bits needed so 2 to the power of 3 equals and this would be enough to represent 5 levels, so by using 3 bits bit to represent these levels a simple scheme can be developed, where the first level L0 can be represented by 000, L1 by 001, L2 by 010, L3 by 011 and L4 by 100.

If the saw tooth to be sent used binary where the frequency was 1 Hz then the bit rate needed could be calculated by multiplying the fundamental frequency by the number of bits per sample period [6] so taking the fundamental frequency as 1Hz and having 15 bits per sampling period to represent the levels the bit rate can be calculated to be 15 bits per second.

The fundamental frequency could be found using the applet by doing this the frequency was take to be 22 kHz, the reason it was best to used a square wave form for this was because the voltage value of 0 and 1 could easily be represented in the form of a square wave form.

By comparing the fundamental frequency here with that gain from part A, it showed that to send the sampled saw tooth in a digital form it would have a higher bandwidth which would also mean that this process would be costly to produce, by repeating this process again but only pressing the quantize button twice I saw that the number of levels had increased around 9 which would mean that 4 bits would be needed to represent each quantizing levels but this would also improve the quality when reproducing the signal, but this property of better quality would cause the problem of more cost.

Part C Using a triangle waveform and increasing the number of terms to make the cosine waveform look very much like a triangle wave form, then by clipping the signals, the shape started to change a looked slightly like a square wave waveform and the line spectrums also change with the first term increasing highly and others move in different directions but when the number of terms were decreased the red signal did not look to similar to the original white signal by looking at the signal where a few number of terms were used it could be seen that the signal started to look like a cosine wave form again, this meant in order to produce a triangle wave a cosine wave would not need a lot of components because the triangle wave form had wide sloping side but when using the cosine signal to represent a square waveform a lot more components were needed this was because the square wave form had very sharp edges.

Part D Noise signals are said to be random as that can not be determined at any point in time by selecting a noise type signal and then starting with no terms and increasing them to the highest number possible it could be seen that cosine and sine components both had near zero values different values and were positioned randomly, which means noise is made up of many frequencies where components values are very difficult to define at any point in time, so from this knowledge I would expect noise power spectral density to have a random appearance.

Conclusion After completing the whole experiment using Fourier‘s Analysis, it was more understandable that changes in time domain would also cause changes to occur in the frequency domain, also notice from this experiment was that even though many of these signals were different they could be analysis effective using very similar methods, so the use Fourier’s Analysis is not only way for signals to be analysed to find what components, because these process indiscreetly can help engineer’s determine other properties and characteristic in order to improve quality or reduce costs, and from this it shows that Fourier’s Analysis is a simple but very important tool to engineer’s.

Resources and References [1] http://www.falstad.com/fourier/ [2] http://en.wikipedia.org/ [3] Signals and Systems by M.L Meade and C.R Dillon [4] Signals and Systems by Alan V Oppenheim, Alan S. Willsky and Ian T Young [5] Signal Processing First by James H. McClellan, Ronald W. Schafer and Mark A. Yoder [6] Communication Systems by Simon Haykin