1–1 Introduction to Mathematical Modeling of Signals and Systems Mathematical Representation of Signals • Signals represent or encode information – In communications applications the information is almost always encoded – In the probing of medical and other physical systems, where signals occur naturally, the information is not pur- posefully encoded – In human speech we create a waveform as a function of time when we force air across our vocal cords and through our vocal tract ! #$%&'()'*+ ),- %'*.+&/+0 /)+ -'1*0 (&+--1&+ 2&'# /)+ .'%,3 /&,%/ $*/' ,* +3+%/&$%,3 -$4*,3 /),/ .,&$+- '.+& /$#+5 !
Transcript
1–1
Introduction to Mathematical Modeling of Signals and Systems
Mathematical Representation of Signals
• Signals represent or encode information
– In communications applications the information is almostalways encoded
– In the probing of medical and other physical systems,where signals occur naturally, the information is not pur-posefully encoded
– In human speech we create a waveform as a function oftime when we force air across our vocal cords and throughour vocal tract
Introduction to Mathematical Modeling of Signals and Systems
1-2
• Signals, such as the above speech signal, are continuousfunctions of time, and denoted as a continuous-time signal
• The independent variable in this case is time, t, but could beanother variable of interest, e.g., position, depth, temperature,pressure
• The mathematical notation for the speech signal recorded bythe microphone might be
• In order to process this signal by computer means, we maysample this signal at regular interval , resulting in
(1.2)
• The signal is known as a discrete-time signal, and isthe sampling period
– Note that the independent variable of the sampled signal isthe integer sequence
– Discrete-time signals can only be evaluated at integer val-ues
% "
&%% ' % '&%=
% ' &%
' ! & " & !––
1–3
• The speech waveform is an example of a one-dimensionalsignal, but we may have more that one dimension
• An image, say a photograph, is an example of a two-dimen-sional signal, being a function of two spatial variables, e.g.
• If the image is put into motion, as in a movie or video, wenow have a three-dimensional image, where the third inde-pendent variable is time,
– Note: movies and videos are shot in frames, so actuallytime is discretized, e.g., (often fps)
• To manipulate an image on a computer we need to sample theimage, and create a two-dimensional discrete-time signal
(1.3)
where m and n takes on integer values, and and repre-sent the horizontal and vertical sampling periods respectively
Mathematical Representation of Systems
• In mathematical modeling terms a system is a function thattransforms or maps the input signal/sequence, to a new out-put signal/sequence
(1.4)
where the subscripts c and d denote continuous and discretesystem operators
( ! )
! ) "
" '&% & &% '"=
( * ' ( * ! ' )=
! )
) " &+ ! "=
) ' &, ! '=
Introduction to Mathematical Modeling of Signals and Systems
1–4
• Because we are at present viewing the system as a pure math-ematical model, the notion of a system seems abstract anddistant
• Consider the microphone as a system which converts soundpressure from the vocal tract into an electrical signal
• Once the speech waveform is in an electrical waveform for-mat, we might want to form the square of the signal as a firststep in finding the energy of the signal, i.e.,
(1.5)
• The squarer system also exists for discrete-time signals, andin fact is easier to implement, since all we need to do is mul-tiply each signal sample by itself
) " ! " !=
) " ! " !=
!"-61,&+&"-7-/+#
1–5
(1.6)
• If we send through a second system known as a digitalfilter, we can form an estimate of the signal energy
– This is a future topic for this course
Thinking About Systems
• Engineers like to use block diagrams to visualize systems
• Low level systems are often interconnected to form largersystems or subsystems
• Consider the squaring system
• The ideal sampling operation, described earlier as a means toconvert a continuous-time signal to a discrete-times signal isrepresented in block diagram form as an ideal C-to-D con-verter
Introduction to Mathematical Modeling of Signals and Systems
1–6
• A more complex system, depicted as a collection of subsys-tem blocks, is a system that records and then plays back anaudio source using a compact disk (CD) storage medium
• The optical disk reader shown above is actually a high-levelblock, as it is composed of many lower-level subsystems,e.g.,
– Laser, on a sliding carriage, to illuminate the CD
– An optical detector on the same sliding carriage
– A servo control system positions the carriage to follow thetrack over the disk
– A servo speed control to maintain a constant linear veloc-ity as 1/0 data is read from different portions of the disk
– more ...
The Next Step
• Basic signals, composed of linear combinations of trigono-metric functions of time will be studied next
• We also consider complex number representations as a meansto simplify the combining of more than one sinusoidal signal
2�–1
Sinusoids• A general class of signals used for modeling the inter-
action of signals in systems, are based on the trigono-metric functions sine and cosine
• The general mathematical form of a single sinusoidal signalis
(2.1)
where denotes the amplitude, is the frequency in radi-ans/s (radian frequency), and is the phase in radians
– The arguments of and are in radians
• We will spend considerable time working with sinusoidal sig-nals, and hopefully the various modeling applications pre-sented in this course will make their usefulness clear
Example:
• The pattern repeats every
• This time interval is known as the period of
x t A 0t +cos=
A 0
cos sin
x t 10 2 440 t 0.4–cos=
1 440 0.00227 2.27ms= =
x t
Review of Sine and Cosine Functions
2�–2
• The text discusses how a tuning fork, used in tuning musicalinstruments, produces a sound wave that closely resembles asingle sinusoid signal
– In particular the pitch A above middle C has an oscillationfrequency of 440 hertz
Review of Sine and Cosine Functions• Trigonometric functions were first encountered in your K–12
math courses
• The typical scenario to explain sine and cosine functions isdepicted below
– The right-triangle formed in the first quadrant has sides oflength x and y, and hypotenuse of length r
– The angle has cosine defined as x/r and sine defined as y/r
– The above graphic also shows how a point of distance rand angle in the first quadrant of the x–y plane is related
Review of Sine and Cosine Functions
2�–3
to the x and y coordinates of the point via sin( ) and cos( ),e.g.,
(2.2)
• Moving beyond the definitions and geometry interpretations,we now consider the signal/waveform properties
– The function plots are identical in shape, with the sine plotshifted to the right relative to the cosine plot by
– This is expected since a well known trig identity states that
(2.3)
– We also observe that both waveforms repeat every radi-ans; read period =
– Additionally the amplitude of each ranges from -1 and 1
• A few key function properties and trigonometric identities
x y r r sincos=
2
sin 2–cos=
22
Review of Sine and Cosine Functions
2�–4
are given in the following tables
• For more properties consult a math handbook
Table 2.1: Some sine and cosine properties
Property Equation
Equivalence or
Periodicity , when k is an integer; holds for sine also
Evenness of cosine
Oddness of sine
Table 2.2: Some trigonometric identities
Number Equation
1
2
3
4
5
6
7
sin 2–cos=cos 2+sin=
2 k–cos cos=
–cos cos=
–sin sin–=
sin2 cos2+ 1=
2cos cos2 sin2–=
2sin 2 cossin=
sin sincoscossin=
cos sinsincoscos=
cos2 12--- 1 2cos+=
sin2 12--- 1 2cos–=
Review of Sine and Cosine Functions
2�–5
• The relationship between sine and cosine show up in calculustoo, in particular
(2.4)
– This says that the slope at any point on the sine curve is thecosine, and the slope at any point on the cosine curve is thenegative of the sine
Example: Prove Identity #6 Using Identities #1 and #2
• If we add the left side of 1 to the right side of 2 we get
(2.5)
Example: Find an expression for in terms of ,, and using #5
• Let and , then write out #5 under both signchoices
(2.6)
or
(2.7)
dsind
-------------- and dcosd
---------------cos sin–= =
2cos2 1 2cos+=
or cos2 12--- 1 2cos+=
8cos 9cos7cos cos
8= =
8 +cos 8 8 sinsin–coscos=
8 –cos 8 coscos 8 sinsin+=
9 7cos+cos 2 8 coscos=+
8cos 9 7cos+cos2cos
-------------------------------------=
Review of Complex Numbers
2�–6
Review of Complex Numbers• See Appendix A of the text for more information
• A complex number is an ordered pair of real numbers1
denoted
– The first number, x, is called the real part, while the secondnumber, y, is called the imaginary part
– For algebraic manipulation purposes we write where ; electrical engi-
neers typically use j since i is often used to denote current
Note:
• The rectangular form of a complex number is as definedabove,
• The corresponding polar form is
1.Tom M. Apostle, Mathematical Analysis, second edition, Addison Wesley, p. 15, 1974.
– MATLAB is also consistent with all of the above, startingwith the fact that i and j are predefined to be
• To convert from polar to rectangular we can use simple trigo-nometry to show that
(2.24)
• Similarly we can show that rectangular to polar conversion is
(2.25)
z1 x12 y1
2+ (magnitude)=
z1 tan 1– y1 x1 (angle)=
z1* x1 jy1 (complex conjugate)–=
1–
polar
rectangular
x rcos=
y r sin=
r x2 y2+=
tan 1– y x note add outside Q1 & Q4=
Review of Complex Numbers
2�–10
Example: Rect to Polar and Polar to Rect
• Consider
– In MATLAB we simply enter the numbers directly and thenneed to use the functions abs() and angle() to convert
>> z1 = 2 + j*5
z1 = 2.0000e+00 + 5.0000e+00i
>> [abs(z1) angle(z1)]
ans = 5.3852e+00 1.1903e+00 % mag & phase in rad
– Using say a TI-89 calculator is similar
• Consider
– In MATLAB we simply enter the numbers directly as acomplex exponential
>> z2 = 2*exp(j*45*pi/180)
z2 = 1.4142e+00 + 1.4142e+00i
z1 2 j5+=
z2 2 45=
Review of Complex Numbers
2�–11
– Using the TI-89 we can directly enter the polar form usingthe angle notation or using a complex exponential
Example: Complex Arithmetic
• Consider and
• Find >> z1 = 1+j*7;>> z2 = -4-j*9;>> z1+z2
ans = -3.0000e+00 - 2.0000e+00i
– Using the TI-89 we obtain
z1 1 j7+= z2 4– j9–=
z1 z2+
Review of Complex Numbers
2�–12
• Find >> z1*z2
ans = 5.9000e+01 - 3.7000e+01i
– Using the TI-89 we obtain
• Find >> z1/z2
ans =-6.9072e-01 - 1.9588e-01i
Euler’s Formula: A special mathematical result, of specialimportance to electrical engineers, is the fact that
(2.26)
z1z2
z1 z2
TI-89Results
ej j sin+cos=
Sinusoidal Signals
2�–13
• Turning (2.26) around yields (inverse Euler formulas)
and (2.27)
• It also follows that
(2.28)
Sinusoidal Signals• A general sinusoidal function of time is written as
(2.29)
where in the second form
• Since it follows that swings between , sothe amplitude of is A
• The phase shift in radians is , so if we are given a sine sig-nal (instead of the cosine version), we see via the equivalenceproperty that
(2.30)
which implies that
• Engineers often prefer the second form of (2.8) where isthe oscillation frequency in cycles/s.
sin ej e j––2j
----------------------= cos ej e j–+2
----------------------=
z x jy+ rcos jr sin+= =
x t A 0t +cos A 2 f0t +cos= =
0 2 f0=
cos 1 x t Ax t
x t A 0t +sin A 0t 2–+cos= =
2–=
f0
02------ rad/s
rad----------- f0 sec
1–=
Sinusoidal Signals
2�–14
Example:
• Clearly, , cycles/s, and rad
• Since this signal is periodic, the time interval between max-ima, minima, and zero crossings, for example, are identical
Relation of Frequency to Period
• A signal is periodic if we can write
(2.31)
where the smallest satisfying (2.10) is the period
• For a single sinusoid we can relate to by considering
(2.32)
• From the periodicity property of cosine, equality is main-tained if , so we need to have
x t 20 2 40 t 0.4–cos=
A 20= f0 40= 0.4–=
140------ 0.025s=
25ms=
MaximaInterval(period)
x t T0+ x t=
T0T0 f0
x t T0+ x t=
A 0 t T0+ +cos A 0t +cos=
0t 0T0+ +cos 0t +cos=
2 kcos cos=
Sinusoidal Signals
2�–15
(2.33)
• So we see that and are reciprocals, with the units of being time and the units of inverse time or cycles per sec-ond, as stated earlier
– In honor of Heinrich Hertz, who first demonstrated theexistence of radio waves, cycles per second is replacedwith Hertz (Hz)
0T0 2 T02
0------= =
or 2 f0 T0 T01f0----=
T0 f0 T0f0
Sinusoidal Signals
2�–16
Example: with , 100, and 0 Hz
• The inverse relationship between time and frequency will beexplored through out this course
Period doubles asfrequency halves
A constant signalas the oscillationfrequency is zero
5 2 f0tcos f0 200=
Sinusoidal Signals
2�–17
Phase Shift and Time Shift
• We know that the phase shift parameter in the sinusoid movesthe waveform left or right on the time axis
• To formally understand why this is, we will first form anunderstanding of time-shifting in general
• Consider a triangularly shaped signal having piece wise con-tinuous definition
(2.34)
• Now we wish to consider the signal
• As a starting point we note that is active over just theinterval , so with we have
(2.35)
which means that is active over
• The piece wise definition of can be obtained by directsubstitution of everywhere appears in (2.34)
s t
2t, 0 t 1 213--- 4 2t– , 1 2 t 2
0, otherwise
=
0 1 2 3t
-1
1s t
2t13--- 4 2t–
12---
x1 t s t 2–=
s t0 t 2 t t 2–
0 t 2– 2 2 t 4
x1 t 2 t 4
x1 tt 2– t
Sinusoidal Signals
2�–18
(2.36)
• In summary we see that the original signal is moved tothe right by 2 s
Example: Plot
• With we expect that the signal will shift to the leftby one second
x1 t
2 t 2– , 0 t 2– 1 213--- 4 2 t 2–– , 1 2 t 2– 2
0, otherwise
=
2t 4,– 2 t 5 213--- 8 2t– , 5 2 t 4
0, otherwise
=
1 2 3 4t
0
1x1 t s t 2–=
52---
2 t 2–13--- 8 2t–
s t
s t 1+
t t 1+
0 1 2 3t
-1
1s t 1+
12---–
Sinusoidal Signals
2�–19
• The new equations are obtained as before
(2.37)
so
(2.38)
• Modeling time shifted signals shows up frequently
• In general terms we say that
(2.39)
is delayed in time relative to if , and advanced intime relative to if
• A cosine signal has positive peak located at
• If this signal is delayed by the peak shifts to the right andthe corresponding phase shift is negative
• Consider
0 t 1+ 2 1– t 1
s t 1+
2 t 1+ , 0 t 1+ 1 213--- 4 2 t 1+– , 1 2 t 1+ 2
0, otherwise
=
2t 2,+ 1– t 1– 213--- 2 2t– , 1– 2 t 1
0, otherwise
=
x1 t s t t1–=
s t t1 0s t t1 0
t 0=
t1
x0 t A 0tcos=
Sinusoidal Signals
2�–20
(2.40)
which implies that in terms of phase shift we have
• For a given phase shift we can turn the above analysis aroundand solve for the time delay via
(2.41)
• Since , we can also write the phase shift in termsof the period
(2.42)
– An important point to note here is that both cosine and sineare mod functions, meaning that phase is only uniqueon a interval, say or
Example: Suppose ms and ms
• Direct substitution into (2.21) results in
(2.43)
• We need to reduce this value modulo to the interval by adding (or subtracting as needed) multiples of
• The result is the reduced phase value
x0 t t1– A 0 t t1–cos=
A 0t 0t1–cos=
0t1–=
t10
------–2 f0-----------–= =
T0 1 f0=
2 f0t1– 2t1T0-----–= =
22 ( ]– (0 2 ]
t1 10= T0 3=
2 103------– 20
3------– 6.6667–= = =
2( ]– 2
Sampling and Plotting Sinusoids
2�–21
(2.44)
• Does this result make sense?
• A time delay of 10 ms with a period of 3 ms means that wehave delayed the sinusoid three full periods plus 1 ms
• A 1 ms delay is 1/3 of a period, with half of a period corre-sponding to rad, so a delay of 1/3 period is a phase shift of
; agrees with the above analysis
• The value of phase shift that lies on the interval isknown as the principle value
Sampling and Plotting Sinusoids• When plotting sinusoidal signals using computer tools, we
are also faced with the fact that only a discrete-time version
203------– 6+ 20 18+
3------------------– 2
3---– 0.6667–= = = =
2 3– 0.6667–=
t (ms)
Actual Delay of 10 msModulo the perioddelay of 1 ms
Blue = no delayRed = 10 ms Delay
–
Sampling and Plotting Sinusoids
2�–22
of
may be generated and plotted
• This fact holds true whether we are using MATLAB, C, Math-ematica, Excel, or any other computational tool
• When we need to realize that sample spacing needsto be small enough relative to the frequency such thatwhen plotted by connecting the dots (linear interpolation),the waveform picture is not too distorted
– In Chapter 4 we will discuss sampling theory, which willtell us the maximum sample spacing (minimum samplingrate which is ), such that the sequence
can be used to perfectly reconstruct from
• For now we are more concerned with having a good plotappearance relative to the expected sinusoidal shape
• A reasonable plot can be created with about 10 samples perperiod, that is with
• We will now consider several MATLAB example plots >> t = 0:1/(5*3):1; x = 15*cos(2*pi*3*t-.5*pi);>> subplot(311)>> plot(t,x,'.-'); grid>> xlabel('Time in seconds')>> ylabel('Amplitude')>> t = 0:1/(10*3):1; x = 15*cos(2*pi*3*t-.5*pi);>> subplot(312)
x t A 2 f0t +cos=
t nTsf0
1 Tsx n x nTs= x t
x n
Ts 1 10f0 T0 10=
Sampling and Plotting Sinusoids
2�–23
>> plot(t,x,'.-'); grid>> xlabel('Time in seconds')>> ylabel('Amplitude')>> t = 0:1/(50*3):1; x = 15*cos(2*pi*3*t-.5*pi);>> subplot(313)>> plot(t,x,'.-'); grid>> xlabel('Time in seconds')>> ylabel('Amplitude')>> print -depsc -tiff sampled_cosine.eps
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1!20
!10
0
10
20
Time in seconds
Am
plitu
de
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1!20
!10
0
10
20
Time in seconds
Am
plitu
de
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1!20
!10
0
10
20
Time in seconds
Am
plitu
de
f0 = 3 Hz, A = 15, = - /2
TsT05
------=
Ts1T010
---------=
TsT050------=
5 Samplesper period
10 Samplesper period
50 Samplesper period
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
3:3;
Complex Exponentials and PhasorsModeling signals as pure sinusoids is not that common. We typi-cally have more that one sinusoid present. Manipulating multiplesinusoids is actually easier when we form a complex exponentialrepresentation.
Complex Exponential Signals
• Motivated by Euler’s formula above, and the earlier defini-tion of a cosine signal, we define the complex exponentialsignal as
(2.44)
where and
• Note that using Euler’s formula
(2.45)
• We see that the complex sinusoid has amplitude A, phaseshift , and frequency rad/s
– Note in particular that
(2.46)
! " #$% !" +
=
! " #= ! " ! ""#$ !" += =
! " #$% !" +
=
# !" +%&' %# !" +'()+=
!
*+ ! " # !" +%&'=
,- ! " # !" +'()=
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
3:3<
– The result of (2.46) is what ultimately motivates us to con-sider the complex exponential signal
• We can always write
(2.47)
The Rotating Phasor Interpretation
• Complex numbers in polar form can be easily multiplied as
(2.48)
• For the case of
(2.49)
& " *+ #$% !" +
# !" +%&'= =
!. '/$% / '0$
% 0 '/'0$% / 0+
= =
!/ !0
! " #$% !" +
=
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
3:34
we can write
(2.50)
where
• The complex amplitude is called the phasor, as it is thegain and phase value applied to the time varying component
to form
– This is common terminology is electrical engineering cir-cuit theory
• The time varying term has unit magnitude and rotatescounter clockwise in the complex plane at a rate of rad/s( rotations/s)
– The time duration for one rotation is the period
• The combination (product) of the fixed phasor and results in a rotating phasor
– For positive frequency the rotation is counter clock-wise, and for negative frequency the rotation is clockwise
! " #$% $% !" ($
% !"= =
( #$%=
(
$ % !" ! "
$ % !"
!)!
*! / )!=
( $ % !"
!
!"
#$
!"
%&'()(*"+,"-."/01
2"34)(*"+,"-."/01
!"#$#%&'()*$+",+
" "
" !" +=
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
3:3=
Example:
• Plot a series of snap shots of the rotating phasor when = 1/8 (note s)
• The above is valid since the real and imaginary parts addindependently, that is
(2.54)
and the same holds for the imaginary part
• Secondly, a real sinusoid can always be written in terms of acomplex sinusoid via
(2.55)
Proof:
#,$j ,
, /=
-(,
, /=
-( #$%= =
*+ (,, /=
-
*+ (,, /=
-
=
# !" +%&' *+ #$% !" +
=
#, !" ,+%&', /=
-*+ #,$
% !" ,+
, /=
-=
*+ #,$% ,
, /=
-$% !"
=
*+ #$% $% !"=
*+ #$% !" +
=
# !" +%&'=
+&55&6'78,&$79:;<4>
01-."2(?//,+,"*
3:A5
Example: Phasor Addition Rule in Action
• Consider the sum
(2.56)
• The frequency of the sinusoids is 15 Hz
• Using phasor notation we can write that
(2.57)
so in the phasor addition rule
(2.58)
• We perform the complex addition and conversion back topolar form using the TI-89
so
(2.59)
& " &/ " &0 "+=
156= .! " .6 /7!+%&'8 950 .! " 7! /7!+%&'+
&/ " *+ 156$%.6 /7! $%.! "=
&0 " *+ 950$%7! /7! $%.! "=
(/ 156$%.6 /7!= (0 950$%7! /7!=
(
8()8#":(")'
( (/ (0+ 15;.<16 %;5<9/9/+= =
/!57679$%<05;<!0 /7!=
01-."2(?//,+,"*
3:A3
• Finally,
(2.60)
• We can check this by directly plotting the waveform in MAT-LAB
XX#*#D#:K?"AY:N?YBK:@9C
XX#Q?#D#F@YN&.%AZ:N)(N*MZYN)("?H:BC
XX#Q9#D#[@9N&.%AZ:N)(N*MH:N)("?H:BC
XX#Q#D#Q?MQ9C#!),.*#3%(0/#4.,=#10=#,(0-#%*\,-%
• The measured amplitude, 10.822, is close to the expectedvalue
& " /!57< .! " <05;<!0 /7!+%&'=
!0.06 !0.04 !0.02 0 0.02 0.04 0.06
!10
!5
0
5
10
Time in seconds
Am
plitu
de
x
1(t)
x2(t)
x(t)
?@;A::
B??;CD7$'
&/ "
&0 "
& "
01-."2(?//,+,"*
3:AA
• The location of the peak can be converted to phase via
(2.61)
Summary of Phasor Addition
• When we need to form the sum of sinusoids at the same fre-quency, we obtain the final amplitude A and phase via
(2.62)
where and
(2.63)
Example:
• Find
• From the given we observe that
0– //5<9–<<5<9
---------------- /5/!8#": <.5!0= = =
( (/ (0 (-+ + + #$%= =
(, #,$% ,=
& " #, !" ,+%&', /=
-
=
# !" +%&'=
& " *+ .$% 0 )!" 0
---+
6$% 0 )!" 1
---–
. %0+ $%0 )!"+ +=
( (/ (0 (.+ +=
& "
(/ .$%0---
= (0 6$%– 1---
= (. . %0+=
019.,B.("C(+1&(DE*,*8(F"2G
3:A;
• To perform the complex addition we will work step-by-step
• To add complex numbers we convert to rectangular form
• Now,
• For use in the phasor sum formula we likely need the answerin polar form
Physics of the Tuning ForkThe tuning fork signal generation example discussed earlier wasimportant because it is an example of a physical system thatwhen struck, produces nearly a pure sinusoidal signal.
– By pure we mean a signal composed of a single frequencysinusoid, no other sinusoids at other frequencies, say har-monics (multiples of ) are present
Equations from Laws of PhysicssA 2-D model of the tuning fork is shown below
• When struck the vibration of the metal tine moves air mole-cules to produce a sound wave
• Hooke’s law from physics (springs, etc.) says that the force torestore the tine back to its original position is the sameas the original deformation (striking force), except for a signchange,
(2.64)
where k is the material stiffness constant
• The acceleration produced by the restoring force (Newton’ssecond law) is
E6&7)(/"'7&8)F"7)./(/378&,G
& !=
. ,&–=
019.,B.("C(+1&(DE*,*8(F"2G
3:A4
(2.65)
• To balance the two forces (sum is zero), we must have
(2.66)
General Solution to the Differential Equation
• To solve this equation we can actually guess the solution byinserting a test function of the form
• We now plug this result into (2.66) to obtain
(2.67)
which tells us that we must have
(2.68)
so it must be that
(2.69)
. /0 /10&1"0--------= =
/10&1"0-------- ,& "–=
& " !"%&'=
10& "1"0
--------------- 11"----- ! !"'()–=
!0
!"%&'–=
/10&1"0-------- ,& "–=
/ !0
!"%&'– , !"%&'–=
/ !0– ,–=
!,/----=
019.,B.("C(+1&(DE*,*8(F"2G
3:A=
• This tells us that the oscillation frequency of the tuning forkis related to the ratio of the stiffness constant to the mass
– Greater stiffness means a higher oscillation frequency
– Greater mass means a lower oscillation frequency
• In terms of a real sinusoid the sound wave, to within a phaseshift constant is of the form
(2.70)
• The sound produced by the 440Hz tuning fork was capturedusing MATLAB on a PC with a sound card and microphone
– The results were converted to double precision and savedin a .mat file along with a time axis vector
