ENG317 Jan 2011 Lab Manual

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ENG317

Optical Communications Systems

LABORATORY MANUAL

January 2011 Presentation



ENG317 Optical Communication Systems - January 2011

2

_____________________________________________________________________

ENG317

Optical Communication Systems

________________________________________________________________

LABORATORY MANUAL

You are advised to work through this laboratory manual to familiarize yourself with

using MATLAB. You do not need to hand in your solutions to the exercises in this

laboratory manual.

Modeling Communication SystemsUsing MATLAB

Introduction: What is MATLAB?

This laboratory tutorial will show you how to use MATLAB features and operators tomodel communications systems. MATLAB is a programming language whose basic data

structure is the row vector (or a 1 x N matrix i.e. a matrix with one row and N columns).In most cases, MATLAB operators will perform their calculations directly on all

elements of a vector in parallel. So there is often no need to specify an “loop index” asdone in other programming languages like BASIC, or C.

It is recommended you avoid using loops (for loops, while loops etc) in MATLAB so thatyour computations can be performed faster.




3

The Colon Operator

Colon notation is a convenient part of MATLAB that lets you specify a portion of avector or matrix. In many cases, this notation will allow you to avoid for loops.

The colon notation works by specifying an index range with a start, a skip, and a final

value. A regularly spaced vector of integers (or reals) is obtained by writing:

>> my_signal = [start:skip:end]

If you omit the “skip” parameter as in:

>> my_signal = [start:end]

the increment will be 1.

Exercise 1:

Using the colon operator, generate the following row vectors:

(a) a = [0 1 2 3 4 5 6 7 8 9 10 11]

(b) b = [1 3 5 7 9 11 13 15 17 19]

(c) c = [0 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 0.020]

(d) d = [1 4 7 10 13 16 19 22 25 28]

The Semi-Colon Operator

The semi-colon placed at the end of a statement, separates one statement from the next

and also suppresses the immediate display of the computed result.

When creating matrices, the semi-colon is used as a separator between the rows of the

matrix.




4

Exercise 2:

Using the semi-colon operator, generate the following matrices:

(a) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

15

42 M

(b) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

715

83 N

(c)

⎟

⎟⎟

⎠

⎞

⎜

⎜⎜

⎝

⎛ =

365

490

11122

P

(d)⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ =

023

327

941

Q

(e)⎟⎟⎟

⎟⎟

⎠

⎞

⎜⎜⎜

⎜⎜

⎝

⎛

=58

3

7

5

1

4

2

R

(f) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

6405

81103S

How to get HELP

MATLAB has an in-built help function that lists the available operators and features. The

index of the help menu can be obtained by typing help at the command prompt:

>> help

Figure 1 shows a sample output listing available subtopics.




6

Creating a Sinusoidal Input Signal

A sine wave signal is mathematically represented by:

wave.theof amplitudeis

and Hertzinfrequencyis

frequencyradiantheisf 2 where

)sin(

A

f

t A y

⋅=

⋅⋅=

π ω

ω

The time variable, t has indefinite limit (i.e. from 0 to infinity). However, in simulation

work, we run the simulation for a fixed time limit from a start_time to an end_time. In

addition, we cannot test every point between the start and end times; so we must sample

the signal at specific times.

The Sampling Frequency: Nyquist Theorem:

The distance between the sampling points is called the sampling interval and is denoted

by T s. The reciprocal of the sampling interval is known as the sampling frequency, F s.To correctly represent the signal’s waveform, Fs must be chosen in accordance with

Nyquist Sampling Theorem which requires the sampling frequency to be more than

twice the frequency being sampled. For instance, to create a 100Hz sine waveform, F s must be greater than 200 Hz (i.e. Ts < 5 ms). A possible value would be 250 Hz.

MATLAB’s colon operator can be used to do the sampling of a signal as follows:

>> ω = 2*pi*f;>> Ts = 0.0001;

>> start_time = 0;

>> end_time = 0.02;

>> t = [start_time:Ts:end_time];

The time units are in seconds. The sinusoidal waveform values can then be computed in a

row vector as follows:

>> A = 2;

>> y = A*sin(ω*t);

Plotting the Signal Waveform:

MATLAB’s plot command can be used to visualize the sampled signal’s waveform:

>> plot(t,y)

This plots signal’s amplitude values, vector y, versus the sampling time vector t .




7

If you use the command:

>> plot(y)

the amplitude, y, will be plotted against the integer index of values starting from

start_time to end_time.

To see the horizontal and vertical grid lines, enter the following command after the plot

command:

>> grid

Exercise 4:

This exercise explores the effect of the choice of sampling interval on the signal’swaveform. Using amplitude of A = 1, create the following sine waveforms:

(a) Plot

(i) ONE cycle and(ii) TWO cycles of the wave f = 100 Hz with F s =5000 Hz.

(b) Now change the sampling frequency to Fs = 250 Hz and plot a cycle. Increase thesampling frequency, Fs in steps of say 50 Hz.

(i) What’s the lowest sampling frequency, Fs, at which the signal frequency is

correct?(ii) What’s the lowest Sampling frequency at which the signal waveform is seen to be

a sine wave?

(c) Generate one cycle of a 100 Hz wave using the colon operator: t = [0:0.001:0.01] andstore it in a row vector S1. This sets Ts = 0.001 or Fs = 1000 Hz. Next double the

sampling rate to Fs = 2000 Hz and store the new samples in a vector S2. Repeat up to

a sampling rate of 8000 Hz. Discuss what happens to the number of samples as thesampling rate is increased. Is there any difference in the waveform as the sample rateis increased?

(d) Alternative plotting method for discrete signals: Use the MATLAB operator, stem(t,s)to plot the sampled signal, s versus the discrete time variable, t. Use the vectors from

part(c) above.




8

Plotting the Signal Spectrum:

The spectrum of a signal can be found computing the Discrete Fourier Transform (DFT)

of the signal. MATLAB make use of an algorithm called the Fast Fourier Transform

(FFT) to compute the DFT. The DFT takes in time samples of the signal’s waveform;similar to the vectors S1, S2 or S3 that you computed in exercise 4(c) – see figure 2.

The FFT will compute the discrete samples of the frequency components from the 100

Hz time samples. You need to supply the vector of time samples, s1 and the number of samples in the vector s1 i.e. the vector’s length which can be found using the MATLAB’s

length() operator. This operator will return a value, N. Usually, the index used in the

derivation of the DFT starts from 0, and hence the computation must end at N-1 to

preserve the same number of sample values. So in the FFT equation, if there are Nsamples, the summation index is from 0 to N-1. Hence in figure 4, the number of samples

is set to length(s1) – 1.

» s1=sin(2*pi*100*[0:0.001:0.02])

s1 =

Columns 1 through 7

0 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878

Columns 8 through 14

-0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511


0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000

Figure 2: Discrete time samples of a 100 Hz signal




9

You should observe that if there were N time samples there would be N frequency

components output by the FFT. In addition observe that each of the N frequencycomponents is a complex number (having a real and imaginary part). Thus the magnitudeof a frequency will be the absolute value of the corresponding complex component while

the phase will be its argument.

The amplitude calculated by the FFT is a function of half the number of samples N/2

(only half the number of samples is needed to compute the DFT due to symmetry of its

calculations). Thus to get the true amplitude of each frequency component, its absolutevalue must be scaled by N/2. This is illustrated in figure 4.

» f1=fft(s1,length(s1)-1)

f1 =

Columns 1 through 4

0.0000 0.0000 - 0.0000i 0.0000 -10.0000i 0.0000 - 0.0000i

Columns 5 through 8

0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i


0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i


0.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i


0.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 +10.0000i 0.0000 - 0.0000I

Figure 3: The FFT Output




10

Observation of Figure 4 reveals that there are only two frequency components with non-

zero values; implying that the signal has only 2 frequency components and that the

amplitude is 1.

Exercise 5:

This exercise explores the FFT output. In Figure 2, change the amplitude of the sampled

sine waveform to A = 2, A = 10 etc. Re-compute the FFT as shown in figures 3 and 4.

Check that the magnitude response changes according.

The Frequency Resolution of the Spectrum:

We now know how to get the amplitude and phase of the signal spectrum but haven’t yet

considered the actual frequencies at which these amplitudes are computed! In the FFT,the sampling interval in the time domain signal, i.e. Ts, determines the frequency range

over which the FFT computes the frequencies.

If Ts = 0.001 seconds, then the FFT will return a frequency range of 0 Hz to 1000 Hz. In

general, a sampling interval of Ts seconds will correspond to a FFT frequency range of 0

to Fs = 1/Ts Hz. If N samples are used in the calculations, the spacing (also referred to asthe resolution) between the frequency components will be:

N

F f s=Δ Hz (1)

Consequently, if a specific frequency resolution is desired, the sample length N must beappropriately chosen. If the number of samples is not enough, the signal vector is zero-

» magf1=abs(f1)/((length(s1)-1)/2)

magf1 =

Columns 1 through 7

0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 0.0000


0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000


0.0000 0.0000 0.0000 0.0000 1.0000 0.0000

Figure 4: The Magnitude of the Frequency Response




11

padded (i.e. a string of zeros is added to the end of the vector to increase its length) so

that the desired resolution may be satisfied. In such cases N is chosen to be a power of 2to facilitate fast computation of the FFT.

To plot the example in figure 4, we observe that Ts = 0.001 seconds and N = 20 (seeFigure 2) so Fs = 1000 Hz. The FFT frequency resolution will be 50 Hz. Note that the

MATLAB FFT operator uses only the indices starting from 1 to N and ignores the 0.

Thus although Figure 2 shows N = 21, MATLAB will only use the 20 samples from 1 to21 giving a total of 20 sample used.

We can create a frequency scale starting from 0 Hz in steps of 50 Hz ending just before1000 Hz to get 20 points. This is easily done using the colon operator and then the

spectrum can be plotted:

>> f = 0:50:999>> plot(f,magf1)

>> grid

The spectrum output is as shown in figure 6.

» f=0:50:999

f =


0 50 100 150 200 250 300 350 400 450 500 550


600 650 700 750 800 850 900 950

» plot(f,magf1)

» grid

Figure 5: Frequency Spectrum Output from FFT




12

The FFT has a mirror symmetrical computation along the frequency axis. The point of

mirror symmetry is at f = Fs/2. This means that the value at frequency point k is identical

to that at point N - k. In our example, the value at k = 100 Hz will be equal to that at N –

k = 1000 - 100 = 900 Hz. You can check this out using the axis operator as shown infigures 7 and 8.

Figure 6: Spectrum from FFT

» axis([0 500 0 2])

Figure 7: Mirror Symmetry of FFT - positive frequency




13

Positive and Negative Frequencies:

In theory, the frequencies from 0 to Fs/2 represent positive frequencies while those fromFs/2 to Fs represent the negative frequencies. MATLAB has an operator called the fft-

shift that will swap the amplitude values from the range [Fs/2 to Fs] (i.e. the negative

frequencies) to be below the positive frequencies. The frequency values can then besimply adjusted by subtracting Fs/2 from each value to set the new plot range to be from [

-Fs/2 to +Fs/2] instead of from [0 to Fs]. This is shown in Figure 9.

» axis([500 1000 0 2])

Figure 8: Mirror Symmetry of the FFT – negative frequency




14

» magf1shift=fftshift(magf1)

magf1shift =

Columns 1 through 7

0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000


0.0000 1.0000 0.0000 0.0000 0.0000 1.0000 0.0000


0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

» fshift=f-500

fshift =


-500 -450 -400 -350 -300 -250 -200 -150 -100 -50 0 50


100 150 200 250 300 350 400 450

» plot(fshift,magf1shift)» grid

» axis([-300 300 0 1.5])

»

Figure 9: Creating the Positive and Negative Frequencies




15

Improving the Frequency Resolution:

To improve the frequency resolution, the number of samples can be increased in

accordance with equation (1). This can be done by zero-padding the signal sample usingthe MATLAB operator fftseq as shown in figures 11, 12 and 13.

Figure 10: The Positive & Negative Spectrum Plot




16

» [M m df]=fftseq(s1,0.001,25)

M =

Columns 1 through 4

0.0000 2.1828 + 1.4585i 2.9208 - 1.2098i 0.1512 - 0.7603i

Columns 5 through 8

2.5586 + 2.5586i 7.7556 - 1.5427i 3.8338 - 9.2556i -5.0387 - 7.5409i


-5.7677 + 0.0000i -1.0389 + 1.5549i -0.3396 - 0.8199i -1.6747 - 0.3331i


-0.6893 + 0.6893i -0.0431 - 0.2168i -0.8172 - 0.3385i -0.5715 + 0.3819i


0.0000 -0.4480 - 0.2994i -0.4997 + 0.2070i -0.0204 + 0.1023i


-0.2466 - 0.2466i -0.4416 + 0.0878i -0.0631 + 0.1523i -0.1257 - 0.1881i


-0.3877 + 0.0000i -0.1147 + 0.1716i -0.0525 - 0.1267i -0.3344 - 0.0665i


-0.1696 + 0.1696i -0.0127 - 0.0637i -0.2803 - 0.1161i -0.2252 + 0.1505i


0.0000 -0.2252 - 0.1505i -0.2803 + 0.1161i -0.0127 + 0.0637i


-0.1696 - 0.1696i -0.3344 + 0.0665i -0.0525 + 0.1267i -0.1147 - 0.1716i


-0.3877 - 0.0000i -0.1257 + 0.1881i -0.0631 - 0.1523i -0.4416 - 0.0878i


-0.2466 + 0.2466i -0.0204 - 0.1023i -0.4997 - 0.2070i -0.4480 + 0.2994i


0.0000 -0.5715 - 0.3819i -0.8172 + 0.3385i -0.0431 + 0.2168i


-0.6893 - 0.6893i -1.6747 + 0.3331i -0.3396 + 0.8199i -1.0389 - 1.5549i


-5.7677 - 0.0000i -5.0387 + 7.5409i 3.8338 + 9.2556i 7.7556 + 1.5427i


2.5586 - 2.5586i 0.1512 + 0.7603i 2.9208 + 1.2098i 2.1828 - 1.4585I

Figure 11: The Sequence M is the FFT output from FFTSEQ




17

Note that the original values of the sequence s1 are not altered. Although a frequency

resolution of 25 Hz was requested, the actual value returned is chosen to be one that

makes N a power of 2 and yet is better than the requested resolution.

m =

Columns 1 through 7

0 0.5878 0.9511 0.9511 0.5878 0.0000 -0.5878


-0.9511 -0.9511 -0.5878 0.0000 0.5878 0.9511 0.9511


0.5878 0.0000 -0.5878 -0.9511 -0.9511 -0.5878 0.0000


0 0 0 0 0 0 0


0 0 0 0 0 0 0


0 0 0 0 0 0 0


0 0 0 0 0 0 0


0 0 0 0 0 0 0


0 0 0 0 0 0 0

Column 64

0

Figure 12: The Sequence m is the zero-padded version of the sequence s1




18

df =

15.6250

» f=[0:df:df*length(m)-1]-500

f =

Columns 1 through 7

-500.0000 -484.3750 -468.7500 -453.1250 -437.5000 -421.8750 -406.2500


-390.6250 -375.0000 -359.3750 -343.7500 -328.1250 -312.5000 -296.8750


-281.2500 -265.6250 -250.0000 -234.3750 -218.7500 -203.1250 -187.5000


-171.8750 -156.2500 -140.6250 -125.0000 -109.3750 -93.7500 -78.1250


-62.5000 -46.8750 -31.2500 -15.6250 0 15.6250 31.2500


46.8750 62.5000 78.1250 93.7500 109.3750 125.0000 140.6250


156.2500 171.8750 187.5000 203.1250 218.7500 234.3750 250.0000


265.6250 281.2500 296.8750 312.5000 328.1250 343.7500 359.3750


375.0000 390.6250 406.2500 421.8750 437.5000 453.1250 468.7500

Column 64

484.3750

Figure 13: The adjusted frequency variable with improved resolution




19

We can now plot the spectrum with a higher resolution as shown in figure 14.

Exercise 6:

Instead of improving the frequency resolution by zero-padding, increase N by creatingmore samples of the waveform but without zero-padding. You can do this by using more

cycles of the wave. Based on figure 2, increase N and plot the corresponding spectrum.

Compare it with figure 14. Which is the better method to improve the frequencyresolution?

» plot(f,fftshift(abs(M)/10))

» grid

» axis([-300 300 0 1.5])»

Figure 14: Higher Resolution Plot by using Zero-Padding




20

Modeling Amplitude Modulation:

A carrier, )2sin( t f E cc π ⋅ is modulated by the following signal:

)2sin()2sin()2sin()(2 332211 t f E t f E t f E t s mmmmm π π π ⋅+⋅+⋅= (2)

Where

E m1 = 5 V and f m1 = 400 Hz E m2 = 2 V and f m2 = 250 Hz (3)

E m3 = 8 V and f m3 = 125 Hz

Exercise 7:

(a) Plot the waveform s2.

(b) Plot the spectrum of s2.

NB: You must choose the appropriate sampling frequency.

The AM process results in the following signal:

)2sin())(2()(3 t f t s E t s cc π ⋅+= (4)

Where f c = 500 Hz and Ec = 10 V.

Exercise 8:

(c) Plot the waveform s3.

(d) Plot the spectrum of s3.

Modeling the FM Process:

The frequency modulation process is mathematically represented by:

∫ +⋅= dt t vK t f E t V m I ccFM )(2cos)( π (5)

where K I is the deviation sensitivity having the unitsvolt

radians sec/.

And vm(t) is the modulating signal. Let the modulating signal be given by:




21

)2cos()2cos()2cos()( 332211 t f E t f E t f E t v mmmmmmm π π π ++= (6)

Exercise 9:

(a) Substitute equation (6) into equation (5) and obtain the equation for VFM(t) by

performing the integration.

(b) Plot the time domain representation of the FM waveform, VFM(t). Assume K I = 10,

Ec =10 V and f c = 500 Hz. Use the amplitudes and frequencies given in the AM

process of exercise 7.

(c) Plot the spectrum of VFM(t). Vary the value of K I and plot the spectrum again.Describe the effect of K I on the modulated signal.

Effects of the fiber material on the propagation of light waves.

Light propagates as an electromagnetic field through optic fibers. The equations of propagation for the electric and magnetic intensities are:

( )( ) y ym

x xm

t e H (z,t)t e E (z,t)

azHazE

z

z

⋅−⋅⋅=⋅−⋅⋅= −

−

β ω β ω

α

α

coscos

(1)

The light sources used are however do not generate a single wavelength but a band of

wavelengths about a central wavelength. The difference between the highest and lowest

wavelengths generated is referred to as the spread. Typical centre wavelength used is1550 nm with a spread of up to ± 500 nm.

Exercise 10 :

Using MATLAB, create a simulation of the electric intensity (E) propagation for variousvalues of α and β for a single fixed wavelength. Present plots showing how the electric

intensity varies with distance. Describe the effect of the parameters of α and β on the

waveform’s propagation.




22

Exercise 11 :

Simulate the spectral content of a laser source by generating a light pulse using a square

wave with its first seven harmonics. Pass the pulse through an optic fiber which causes a

phase delay according to:

λ φ 22.0 e= radians

Plot the output pulse for various combinations of the first seven harmonics. Assume noother propagation effects.

Fiber Connection and Coupling Losses

There are several types of misalignment possibilities that exist between a fiber optic

source and a fiber optic receiver. In figure 15 it shows three typical cases of mechanical

misalignments.

Figure 15

The lateral or axial offset condition shown above is further demonstrated below, where

we show the motivation for the mathematical model to predict the insertion loss due tothis type of misalignment.

Figure 16 Axial/ Lateral Offset Condition




23

To illustrate the effects of axial misalignment, consider the case of 2 identical fibers of radii a. Suppose that their axes are offset by a separation d at the common junction, as

shown in the figure 15a. Also, assume that there is a uniform modal power distribution in

the emitting fiber. Since the numerical aperture is constant across the end faces of the twofibers, the optical power coupled from one fiber into the other is proportional to the

common area, A, common to the two fiber cores.

Defining the offset = d and the fiber radius = a, the form of common area is given by:

21

2212

4)

2(cos2 ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−= − d

ad a

d a Acomm (1)

The efficiency of coupling is given by:

2πη

a

Acomm= (2)

The resulting insertion loss in dB is given by:

Loss = -10log(η) (3)

Using MATLAB, investigate efficiency of coupling using equations (1) to (3) by plottinghow the efficiency varies with the parameters a and d.

HINT: investigate the variation of η with d for various fiber sizes, a. Take typical core

radius to vary between 1μm to 100μm.

An alternative theory to calculate the insertion loss was proposed by D. Marcuse. Insteadof modeling the physical overlap of the fibers themselves, he modeled the overlap of the

radiation patterns in terms of the Mode Field Diameters (MFD) of the transmitting and

receiving fibers. This distribution is the Gaussian pattern shown in Figure 17(b).




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In this model, the transmission offset is given by:

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+

−

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+=

22

21

22)1(

2

2

2

1

212)(

ww

d

off eww

wwd T (4)

where w1 and w2 are the MDF of the transmitting and receiving fibers respectively. Thecorresponding loss is then given by:

))(log(10)( d T d T off loss −= (5)

The MDF is calculated using:

( )65.1 879.2619.165.02 −− ++= V V aw

Where V is the normalized frequency parameter given by:

2

2

2

1λ

π2nn

aV −⎟

⎠

⎞⎜⎝

⎛ = (7)

where:- a - core radius

n1 - core refractive index

n2 - cladding refractive index

λ - wavelength

Since we are modeling single mode fibers, ensure that V < 2.405 so that only the

fundamental mode is transmitted.

corePower radiation distribution

(a)

Core diameter

Mode Field Diameter

Power radiation distribution

(b)

Figure 17 Mode Field Diameter for single mode fiber




25

Generally, when splicing, the transmitting and receiving fibers are the same and so have

the same parameters, thus w1 = w2.

Using equation (4) to (7), evaluate the alternative model for lateral misalignment.

Exercise 12 :

1. Select a range of appropriate values to vary the misalignments. For example, since

we have chosen 8 μm core (radius a = 4μm), an appropriate range would be 0 to 8

microns. In the case of 0 micros, the fibers are aligned. In the case, when d = 8μm, the fibers are completely misaligned.

2. Pick appropriate grid spacing for each of the ranges. You need sufficient points to

clearly observe the trend lines.3. Program your model using MATLAB.

4. Construct plots to visualize your results.

5. You may keep the refractive indices and wavelengths fixed at any representative

values (e.g. n1 = 1.5, n2 = 1.493 and λ = 1.55 nm). You may do further

experimentation by choosing other values to see how these affect the insertionloss e.g. how high or low a wavelength can you use before the insertion loss

renders splicing of cables undesired?

6. Make comparison between Kieser’s model and Marcuse’s model.

~~~END OF MANUAL~~~

ENG317 Jan 2011 Lab Manual

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