SYSTEM AND SIGNAL EXPERIMENT

Template for the EBU5374Coursework:A Formal Report Name:Yaxing Wu
Student number:09212786
College email:jp92786@
TABLE OF CONTENT ABSTRACT (2)
INTRODUCTION (2)
BACKGROUND THEORY (4)
DESCRIPTION OF THE EXPERIMENT (9)
DISCUSSION (27)
CONCLUSIONS (30)
REFERENCES (30)
ABSTRACT:
In the lecture of Singnal&System,we have already learnt Fourier series and Fourier transform.And in this experiment,we need to use what we learnt to analysis signal waveform so as to master the knowledge better.This experiment will, on completion,help us to understand:
(a)the physical meaning of signal spectrum
(b)that each different waveform has a different spectrum(or spectral density)
(c)that certain standard results exist for standard waveforms
(d)the effects of spectrum limitation on the transmission of signals
In this experiment we will analysis RANDOM,APERIODIC,and PERIODIC signals.We can know each spectrum of these signals and the effects of spectrum limitation on the transmission of signals.Fourier Series Java Applet help us to examine the spectra of those signals and the effect of limiting the bandwidth of a real signal.In order to get the bandwidth,we need a binary representation of an analogue signal,look at an unusual signal and examine noise.However,in order to complete this experiment well,we need to understand some fundamental knowledge,like Fourier Series,odd function,even function,Orthogonality,time domain representation, frequency domain representation etc.
Key Words:Signal,Spectrum,Waveform,Spectrum Limitation,Fourier Series, Fourier Analysis,Fourier Transform,Bandwidth,Noise.
Introduction
Many applications in communications or systems in general are concerned with the propagation of a signal through a network or process.The resultant output signal is dependent on the properties of both the input signal and the network or process which acts on the signal.This experiment is concerned with the analysis of a signal and the exploration of the properties of the signal.
Using Fourier Analysis a real signal existing in the time domain will be characterised by its spectrum in the frequency domain.The spectra of some‘typical’signals will then be examined and it will be shown that the individual spectra enable the bandwidth requirements for the transmission of particular signals to be better understood and perhaps specified.This also involves studying the effects of bandwidth limitation,causing a distortion of the time
In the Applet,the real signal being observed and analysed is represented as a white line.The menu on the right hand side allows you to choose which signal type you want to look at in detail.The red line represents the synthesised signal,i.e.the signal which results when you have only a limited number of spectra of the original signal available to combine them into the resulting,desired signal(e.g.a received signal at the system’s receiver,or a signal at a system’s output).The number of line spectra is set by the“slider bar”labelled NUMBER OF TERMS at the bottom right of the menu.
In this experiment,we need to complete4parts：
Part A is to use Fourier Analysis to observe the effect of limiting the bandwidth of
a real signal.In this part,first of all,we need to click the CLEAR option in the menu, and then the SQUAREWAVE option,then we need to
1)reduce the number of terms and observe the signalsand and record.
2)Moving the cursor over the dot on line spectra plots reveals the magnitude of the component it represents.
3)Increase the number of terms used to synthesise the red signal,observe the cosines and sines.
4)Increase the number of terms further,get the correct value.
5)Sketch the red and white waveforms for a few cases of limited number of spectral terms.
Secondly,click the CLEAR option in the menu,and then the SQUAREWAVE option and then the RECTIFY option.Repeat steps1-5above.
Thirdly,click the CLEAR option in the menu,and then the SQUAREWAVE option and move the NUMBER OF TERMS slider to about½way between left and right hand sides.
1)Click the Phase Shift button1,2,3…10times.After each click observe what is happening to the waveforms.
2)Clicking the Phase Shift button,observing the original signal and record it. Fourthly,Click the CLEAR option in the menu,and then the SAWTOOTH option.
1)Reduce the number of terms to the minimum possible.
2)Moving the cursor over the dot reveals the magnitude of the component it represents.getting the value.
3)increasing the number of terms again,observing sines and cosines.
4)Increasing the number of terms until the synthesised signal represents the original signal sufficiently well.
6)Move the slider to about½way between left and right hand sides,and then click the PHASE SHIFT button1,2,3…10times.After each click observe what is happening to the waveform in the time and frequency domains.
8)Move the slider so far to the left that only the1st harmonic appears.Repeat step6 above.
9)Sketch a few plots of the original signal and the signal synthesised from the limited number of Fourier components for a number of phase shifts of interest.
Part B is to use Fourier analysis to decide on the bandwidth needed to support a binary
representation of an analogue signal.First of all,we need to click the CLEAR option in the menu,and then the SAWTOOTH option again.
1)Increase the number of Fourier terms and see what happens with the synthesised (red)signal.Move the slider as far as is necessary so that the synthesised signal is very close to the original sawtooth waveform.
2)Click the Quantize button3times,and observe what happens to the sawtooth signal.
3)CLEAR the screen and select COSINE.Repeat above
4)CLEAR the screen and select SAWTOOTH again.Repeat steps1-2.
5)Hit the Quantize button only twice.Repeat above.
Part C is to use Fourier Analysis to look at an“unusual”signal
First of all,click the CLEAR option in the menu,and then the TRIANGLE option
1)Increase the number of terms until get a very good representation of the original signal.
2)Click the Clip button1,2,3…15times,and observe what happens to the original signal and its spectra.
3)Decrease the number of line spectra,until there are only a handful of harmonics left in the synthesised(red)signal.Notice that the red signal is a very poor representation of the original white signal.
4)draw the original signal against the signal synthesised from the
limited number of harmonics
Part D is to use Fourier Analysis to examine noise.
First of all,move NUMBER OF TERMS slider to the far left hand side.And Click the CLEAR option in the menu,and then the NOISE option
1)Click to move the slider to the right,one click at a time,and observe what happens to the cosine/sine spectra that result.
2)Keep going until the slider gets all of the way to the right hand side.Observe the representation of the noise.
After all these steps,we need towrite the answer of the questions which are on the lab sheet in our lab report.and in this time we can understand the physical meaning of signal spectrum.Identify each different waveform has a different spectral density,certain standard results exist for standard waveforms.And we know that the effects of spectrum limitation on the transmission of signals
background theory
1.even function
definition：Suppose a function y=f（x）have a domain D，for any x∈D,and
f(-x)=f(x)，this function is called even function.
Properties
1、even functions satisfy f(x)=f(-x).For instance:y=x*x，y=Cosx
2、If we know the graph of the even function,the graph should be symmetric about y axis.
3、the domain of the even function must be symmetric about origin，otherwise it’s not
a even function.
Figure1even function
2.Odd function
definition：Suppose a function y=f（x）have a domain D，for any x∈D,and f(-x)=-f(x)，this function called odd function.
Properties:
1.odd functions satisfy f(x)=-f(-x).For instance:y=3x，y=sinx
2.If we know the graph of the odd function ,the graph should be symmetric about origin.
3.the domain of the odd function must be symmetric about origin ，otherwise it’s not a odd even function.
4.f(0)=0
5.odd function*even function=even function
6.even function*even function=even function
7odd function*odd function=even function
3.3.Fourier Fourier serie series s Fourier series decomposes any periodic function or periodic signal into the sum of a (possibly infinite)set of sines and cosines functions (or complex exponentials).The study of Fourier series is a branch of Fourier
analysis.
Figure 3fourier representation
a0=1/T
∫−2
/2/)(T T dt
t x an=2/T dt
nwt t x T T )cos()(2/2/∫−bn=2/T dt
nwt t x T T )sin()(2/2/∫−----original (sine and cosine components)X(t)=a0+
∑∑∞=∞=+11)
sin()cos(n n nwt bn nwt
an
4.Fourier transform
The Fourier transform is a mathematical operation that decomposes a signal into its constituent frequencies.Thus the Fourier transform of a musical chord is a mathematical representation of the amplitudes of the individual notes that make it up.The original signal depends on time,and therefore is called the time domain representation of the signal,whereas the Fourier transform depends on frequency and is called the frequency domain representation of the signal.The term Fourier transform refers both to the frequency domain representation of the signal and the process that transforms the signal to its frequency domain representation.[1]
The Fourier Transform (FT)is defined as:X(jw)=
dt
t x e jwt
t t )(.−∞=−∞=∫The conditions of FT:A signal is said to have a Fourier transform in the ordinary sense if the integral in the following equation converges .
1.if x(t)is “well behaved”.It means that the signal has a finite number of discotnuities,maxima,and minima within any finite interval of time.
2.if the x(t)is absolutely integrable ∞
<∫
dt t x )(5.5.Orthogonality
Orthogonality Orithogonality is fundamental to almost everything that is subsequent in signals and systems theory.the definitions are:
Discrete signal:
If the product of two signals average to zera over period T,then those signals are ORTHOGONAL in the interval T
1/N ∑−==1
00][].[N N n b n a Continous signal:
If two product of two signals integrates to zero over the period T,hen those signals are ORTHOGONAL in the interval T
∫−=2
/2/)().(,,0)().(/1T T t b t a where dt t b t a T
maybe odd function
It’s can simplify Fourier Transform
6.Types of signals
The study of signals and their waveforms is central to circuit theory,telecommunications,control engineering and instrumentation.The signals themselves exist and propagate in the time domain.
These signals can be classified as periodic,aperiodic or random,and each can be analysed by particular mathematical techniques.
RANDOM signals by their nature cannot be characterised by a limited number of precise measures and their mathematical description is achieved by assembling measures over a large number of tests or trials so that the signals may then be analysed by statistical methods. APERIODIC functions may occur in communications and as inputs to systems.They are also used for transient testing of dynamic systems.
PERIODIC signals play a major role in practical systems.A periodic signal is based on a periodic waveform-the sinusoid-which provides the foundation for many of the techniques used in systems and circuit theory.Provided that a periodic signal satisfies certain conditions of continuity and convergence(which are usually satisfied by physical signals),then that periodic signal can be decomposed using Fourier Analysis into an infinite sum of sinusoids having frequencies which are multiples of the basic or fundamental frequency of the original periodic waveform.Each multiple of the basic frequency,or harmonic,contributes with individual amplitude and phase to the reconstruction of the original waveform.
If the original real periodic signal in the time domain,x(t),is defined by
x(t)=x(t+(-)n.T)
whereΤis the fundamental period of x(t)and‘n’is any integer[2]
Any periodic signal,x(t),whose period is T,can be represented by the appropriate sum of sin
and cos components:X(t)=a0+∑∑
∞
=
∞
=
+
11
)
sin(
)
cos(
n n
nwt
bn
nwt
an
,where
a0=1/T
∫
−
2/
2/
)(
T
T
dt
t x
an=2/T
dt
nwt
t x
T
T
)
cos(
)(
2/
2/
∫
−
bn=2/T
dt
nwt
t x
T
T
)
sin(
)(
2/
2/
∫
−
7.Bits per sample
2nL
n means the bits per sample and L means the number of levels
8.Bit Rate
Bit rate means the number of bits being transmitted per seconds.
9.9.Noise
Noise
In common use,the word noise means any unwanted sound.In both analog and digital electronics,noise is an unwanted perturbation to a wanted signal;it is called noise as a generalisation of the audible noise heard when listening to a weak radio transmission.Signal noise is heard as acoustic noise if played through a loudspeaker; it manifests as'snow'on a television or video image.Noise can block,distort,change or interfere with the meaning of a message in human,animal and electronic communication.[3]
10.10.Quantization
Quantization In mathematics and digital signal processing,is the process of mapping input values that are members of some relatively large set of admissible input values to output values that are members of a smaller countable set of output values.The set of possible input values may be infinitely large,and may possibly be continuous and therefore uncountable (such as the set of all real numbers,or all real numbers within some limited range).The set of possible output values may be finite or countably infinite.A device or algorithmic function that performs quantization is called a quantizer.The analysis of quantization involves studying the amount of data (typically measured in digits or bits or bit rate )that is used to represent the output of the quantizer,and studying the loss of precision that is introduced by the quantization process (which is referred to as the distortion ).The general field of such study of rate and distortion is known as rate–distortion theory .[4]
11.Time domain representation
Time domain is a term used to describe the analysis of mathematical functions,or physical signals,with respect to time.In the time domain,the signal or function's value is known for all real numbers,for the case of continuous time,or at various separate instants in the case of discrete time.An oscilloscope is a tool commonly used to visualize real-world signals in the time domain.Speaking non-technically,a time domain graph shows how a signal changes over time,whereas a frequency domain graph shows how much of the signal lies within each given frequency band over a range of frequencies.[5]
12.12.Frequency
Frequency domain representation frequency domain is a term used to describe the domain for analysis of mathematical functions or signals with respect to frequency,rather than time.Speaking non-technically,a time-domain graph shows how a signal changes over time,whereas a frequency-domain graph shows how much of the signal lies within each given frequency band over a range of frequencies.A frequency-domain representation can also include information on the phase shift that must be applied to each sinusoid in order to be able to recombine the frequency components to recover the original time signal.A given function or signal can be converted between the time and frequency domains with a pair of mathematical operators called a transform.An example is the Fourier transform,which decomposes a function into the sum of a (potentially infinite)number of sine wave frequency components.The 'spectrum'of frequency components is the frequency domain representation of the signal.The inverse Fourier transform converts the frequency domain function back to a time function.A spectrum analyzer is the tool commonly used to visualize real-world signals in the frequency domain.[5]
Figure4an example of how time domain transform into frequency domain
13.Bandwidth(signal processing)or analog bandwidth,frequency bandwidth or radio bandwidth:a measure of the width of a range of frequencies,measured in hertz[6]
EXPERIMENT
A)To use Fourier Analysis to observe the effect of limiting the bandwidth of a
real signal.
i)Click the CLEAR option in the menu,and then the SQUAREWAVE option
1)Reduce the number of terms to0by moving the slider to the far left hand side.
2)Note that the synthesised signal,i.e.the red line is flat.State the reason for this in your Lab logbook.
According to Fourier serier,a period signal,x(t),whose period is T,can be represented by the appropriate sun of sine and cosine components.
,we can know as n=0,and the synthesized signal can be presented like
X(t)=a0+∑an.cos(nwt)+∑bn.sin(nwt),in fact n=0,So x(t)=a0
Because x(t)is a odd function,from Orthogonality
We can easy to know that
a0=1/T
∫−2
/2/)(T T dt
t x where a0=0so the synthesized signal is always equal to zero,so it’s flat.
Note that there is a single white dot under cosines,and none at all under sines.Statethe reason for this in your Lab logbook.Because a0=0,so there is a white dot on cosine
3)Moving the cursor over the dot on line spectra plots reveals the magnitude of the component it represents.It should have a value of zero.State the reason for this in your Lab logbook.
Because the dot represent a0,so the value is zero.
4)Increase the number of terms used to synthesise the red signal.Now at least 2new components should now be represented,the sine one has a non-zero value,and the cosine one has a zero e Fourier Series analysis that we learnt in class to check that these are the correct values.Do the calculations in your Lab logbook.
According to Fourier Serier,the square waveform can be represented as follows,where n is the number of terms,x(t)is the synthesized
signal.
X(t)=a0+∑∑∞=∞
=+11)sin()cos(n n nwt bn nwt an ,where
a0=1/T
∫−2
/2/)(T T dt
t x
an=2/T
dt
nwt t x T T )cos()(2
/2/∫−bn=2/T
dt nwt t x T T )sin()(2
/2
/∫−As we proved above,a0=0
x(t)is represented by at least two components which have sine and cosine represented.
And when n=0,a0=0,so there is a zero value in cosine.
bn=T /2[∫∫−−+2
/00
2/)sin()sin(T T dt nwt A dt nwt A ]=]cos 1[/2ππn n A −=⎩⎨⎧==even n odd n n A ,0,/4πSo the sine one has none zero value when n is odd.
5)Increase the number of terms further,and use Fourier Series analysis to check that these are correct values.You can do the first 5terms.Do the calculations in your Lab
Logbook.
an=2/T
dt nwt t x T T )cos()(2
/2/∫−,and x(t)is odd function,cos(nwt)is even function,so
x(t)cos(nwt)is odd function.
According to Orthogonality,an is always equal to 0
a0=0a1=0a2=0a3=0a4=0a5=0
And bn=⎩⎨⎧==even
n odd
n n A ,0,/4πb1=4A/πb2=0b3=4A/3πb4=0b5=4A/5π
According to Applet,a 0=a 1=a 2=a 3=a 4=a 5=0
b 1=1.27,b 2=0,b 3=0.42,b 4=0,b 5=0.25
It's correct
6)Sketch the red and white waveforms in your Lab logbook for a few cases of limited number of spectral terms.
7)Think about the implications of the synthesised signal versus the original signal.Of course the synthesised signal is the bandwidth limited version of the original signal.If asked “At what point do you think the synthesised signal looks enough like the
original signal to be an acceptable compromise”what issues would you have to think about,as an engineer?Explain these in your Lab logbook.
Digital signals has an unlimited bandwidth
Transmission media has a limited bandwidth
Selection transmission media relies on the cost of investment and the quality of the transmission signal.
ii)Click the CLEAR option in the menu,and then the SQUAREWAVE option and then the RECTIFY option.Repeat steps 1)-6)above.
1)Reduce the number of terms to 0by moving the slider to the far left hand side.
2)-3)As (i),according to Fourier series ,∫∫∫=+==−T
T T T T A dt Adt T dt t x T a 2
/2/02/2/2/]0[/1)(/10And when n=0,x(t)=a0=A/2
The white dot on sine represent A/2
4)The calculation is as follow:
⎩⎨⎧===−=+====+
==∫∫∫∫∫∫−−even
n odd
n n A n n A dt dt nwt A T dt nwt t x T bn n n A dt dt nwt A T dt nwt t x T an T T
T T T T
T T T T ,0,/2)]cos(1[/]
0)sin([/2)sin()(/20
)][sin(/]0)cos([/2)cos()(/22/02
/2/2/2
/2/2/2/0πππππSo the sine one has none zero value when n is odd.
The cosine one always is the 0except a0
5)
a0=A/2a1=0a2=0a3=0a4=0a5=0
And bn=⎩⎨⎧==even
n odd
n n A ,0,/2πb1=2A/πb2=0b3=2A/3πb4=b5=2A/5π
According to Applet,a 0=0.5,a 1=a 2=a 3=a 4=a 5=0
b 1=0.63,b 2=0,b 3=0.21,b 4=0,b 5=0.13
correct
6)sketch
iii)Click the CLEAR option in the menu,and then the SQUAREWAVE option and movethe NUMBER OF TERMS slider to about ½way between left and right hand sides.
1)Click the Phase Shift button 1,2,3…10times.After each click observe what is happening to the waveforms.
2)By clicking the Phase Shift button,what are you doing to the original signal i.e.waveform?Note your answer in the log book.
The original signal shift one unit to the left when clicking Phase Shift button once.
3)How does the phase shift in time domain translate into the frequency domain representation of the signal?Remember what we did in the class when we talked about the Fourier Transform for real,non-periodic signals and the Fourier Transform time shifting property.Note your answer into the log book.
X(jw)=∫
−dt t x jwt e )(.=∫
−dt
t x wt j wt )()).sin()(cos(Implement the FT X(t)Im
Re )(),sin(*),cos(*j jw X imaginary
wt real wt +=⎩⎨⎧=The conditions of FT
A signal is said to have a Fourier transform in the ordinary sense if the integral
in the following equation converges
1.if x(t)is “well behaved”.It means that the signal has a finite number
of discotnuities,maxima,and minima within any finite interval of
Time.
2.if the x(t)is absolutely integrable ∞
<∫dt t x )(Time Shifting property
If x(t)is a continuous function,the time-shifted signal is defined as y(t)=x(t-t0).if t0>0,the signal is shifted to the right,and if t0<0,the signal is shifted to the left.If x(t))(w x ↔,w then for any positive or negative real number c,
X(t-c)jwc
e jw X −↔)(Note that i
f c>0,then x(t-c)is a c-second right shift of x(t)
if c<0,then x(t-c)is a(-c)-second left shift of x(t)
Thus the above transform pair is valid for both left and right shift of x(t)
4)Note your answers to the following questions in your Lab logbook:
a.What is happening to the signal in the time domain after ~10clicks?
The signal in the time domain shift a half of period to the left
b.What happens to the line spectra after about 10clicks?Do you recognise this to be the original pattern of the signal spectrum?
The line spectra rotate 180degree
c.Explain why this occurs (i.e.signal representation in the frequency domain –the pattern of line spectra -is similar to the pattern before the 10phase shifts).Because shift half period,so in this time ,synthesized signal is -x(t)=x(-t)now
And a0=1/T
∫−−2
/2/)(T T dt t x =0
an=2/T
dt nwt t x T T )cos()(2
/2/∫−−=0(accurate is -0)
bn=2/T ])sin()([2
/2/dt nwt t x T T ∫−−=]cos 1[/2ππn n A −−=⎩⎨⎧==−even n odd n n A ,0,/4πAnd before the shift,we know from (i)-4),bn=⎩⎨⎧==even
n odd
n n A ,0,/4πSo the sine and cosine rotate 180degree (change from positive to negative)with no change in pattern
d.What are the units for phase shift?For the given example,give an expression or calculate the approximate value of the single phase shift that occurs when you press the PHASE SHIFT button onc
e.
10unite equal to T/2,so 0ne unit equal to T/20
5)Has the requirement for bandwidth to transmit the synthesised signal been altered as the original signal is phase-shifted i.e.delayed?Note your answer in the log book.No ,because the pattern of the signal has no change.
6)How are time shift Δt and phase shift related?In other words,given the angular frequency ωof a single sinusoidal component of a signal,what phase shift
corresponds to time shift Δt?Write the expression in your log book.
Let the synthesized signal is x(t-t ∆),t ∆>0
X(t t ∆+)=a0+))(sin())(cos(11
∑∑∞=∞=∆−+∆−n n t t nw bn t t nw an ,where
a0=1/T
∫∆∆+−∆−t T t T dt t t x 2/2/)(=1/T ][2/2/∫∫∆∆+−∆+∆+−t t T t
T t Adt Adt =0
an=2/T
dt t t nw t t x t
T t T ))(cos()(2/2/∆−∆−∫∆+∆+−According to origonality ,x(t-t ∆)cos(nw(t-t ∆))is
odd function,an=0
bn=2/T
dt
t t nw t t x t
T t T ))(sin()(2/2/∆−∆−∫∆+∆+−=2/T ]
))(sin())(sin([
2/2/∫∫∆∆+−∆+∆∆−+∆−−t
t T t T t dt t t nw A dt t t nw A =]cos 1[/2ππn n A −=⎩
⎨⎧==even n odd n n A ,0,/4πthere is no change in an,bn,so time shift Δt and phase shift in sines and cosines is Δt/T*2π=w t
∆7)Consider a single Fourier component,e.g.si(t)=A.sin(ωt).If the component is delayed or advanced by Δt,show how that translates into phase shift,and explain how
that might affect the magnitude of the component si(t)at any time t.[This will explain the changes in line spectra you noticed in point (3)above.]HINT:A delayed
waveform will now be described as Asin[ω(t +Δt)].Work this out in your logbook.dt e t x jw X jwt ∫−=)()(=dt
e wt A jwt ∫−)sin(Asin(w(t+t ∆))dw
e jw X e dw e jw X jwt t jw t t jw ))((2/1)(2/1)(∫∫∆∆+==ππ)
()(sin(jw X e t t w A t jw ∆↔∆−Because of 1=∆−t jw e so there is no change in magnitude of frequency domain of si(t)
In time domain,the signal is shift,so new signal's magnitude at any time t is the same as the original signal's at t+t
∆8)Given the answers in (6)and (7)above,what would happen if all different frequency components were delayed by the same amount of time Δt?Would each of them experience the same phase shift?What is the possible importance of this shift for a communications system?Note your answers in the log book.
We know that time shift Δt and phase shift in sines and cosines is Δt/T*2π=w t ∆Thus all different frequency components were delayed by the same amount of time Δt ,signal in time domain shift w t /∆,if they have same w,they experience the same phase shift
All signal components are delayed when passing through a device such as an amplifier or a loudspeaker.The signal delay can be (and often is)different for different frequencies.The delay variation means that signals consisting of different frequency components suffer delay (or time)distortion.A small delay variation is usually not a problem,but larger delays can cause trouble such as poor fidelity and intersymbol interference.
iv)Click the CLEAR option in the menu,and then the SAWTOOTH option.This will create
a sawtooth-shaped waveform for Fourier analysis.
1)Reduce the number of terms to the minimum possible.
2)Note that the resulting synthesised signal is flat.State the reason for this in your Lab
logbook.How many sine and cosine components do you see in the frequency domain plot?State for this the reason in your Lab logbook.
X(t)=a0+∑∑∞=∞
=+11
)sin()cos(n n nwt bn nwt an ,where
a0=1/T
∫−2
/2/)(T T dt
t x an=2/T dt
nwt t x T T )cos()(2/2/∫−bn=2/T
dt nwt t x T T )sin()(2
/2
/∫−,x(t)is the synthesized signal,x(t)=kt,According to orthogonality,x(t)is odd function in a period,so a0=0,Because reduce the number of terms to the minimum possible.We can see n as zero,so x(t)=a0=0,it is a constant line,red line is flat.because of n=0,there is no an or bn,the onlydot we can see in frequency domain is a0in sine component.
3)Moving the cursor over the dot reveals the magnitude of the component it represents.
It should have a value of zero.State the reason for this in your Lab logbook.The dot represent a0=0
4)As you again increase the number of terms and new components appear on the frequency plot,you notice that the sine ones have non-zero values and the cosine。