Sinusoids Chapter

ECE 2610 Signal and Systems 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

Chapter

2

Review of Sine and Cosine Functions

ECE 2610 Signals and Systems 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



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



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–=



• 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


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.

z x y=

x yx iy+= x jy+= i j 1–= =

1– 1– 1–= j j 1–=

z x y x jy+= =

z rej r z e j zarg= = =



• We can plot a complex number as a vector

Example:: , , ,

x y

z 2 j5+= z 4 j3–= z 5– j0+= z 3– j3–=



Example:: , , &

• For complex numbers and wedefine/calculate

z 2 45= z 3 150= z 3 80–=

z1 x1 jy1+= z2 x2 jy2+=

z1 z2+ x1 x2+ j y1 y2+ (sum)+=

z1 z2– x1 x2– j y1 y2– (difference)+=

z1z2 x1x2 y1y2– j x1y2 y1x2+ (product)+=

z1z2----

x1x2 y1y2+ j x1y2 y1x2––

x22 y2

2+------------------------------------------------------------------------- (quotient)=



– 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=



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=



– 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+



• 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


• 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; why?

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


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


(2.33)

• 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


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


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


(2.36)

• 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


• 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


(2.40)

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


(2.44)

• 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

–



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=



>> 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.

)!)(3456(7,8*-%.(-*/(79.+&#. 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.

)!)(3456(7,8*-%.(-*/(79.+&#. 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.

)!)(3456(7,8*-%.(-*/(79.+&#. 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.

)!)(3456(7,8*-%.(-*/(79.+&#. 3:3=

Example:

• Plot a series of snap shots of the rotating phasor when = 1/8 (note s)

!""""""""""""""""""""""""""""""""""""""""""""""""""""""

!#$#%&'()*#+(,-#+.'#/-0-'1*(0/#1#%-23-0&-#.+#'.*1*(0/#

!#)41%.'#%01)#%4.*%

!

!#51'6#7(&6-'*8#$3/3%*#9::;

!""""""""""""""""""""""""""""""""""""""""""""""""""""""

#

!#/-*#*4-#+.&3%#.+#+(/3'-#<(0=.<#>?#.'#&'-1*-#

!#(+#0.*#&'-1*-=@

+(/3'-A?B#

&,+A?BC#!#&,-1'#+(/3'-#<(0=.<#>?

#

$#D#?@:C#+:#D#?C#)4(#D#E)("FC

G#D#HC#!#&'-1*-#H#I-&*.'#),.*%

J%#D#?"HC

+.'#0#D#:KGE?

####%3L),.*AF8980M?B

####*#D#:K?"9::K?C

####),.*A&.%A9N)(N*B8%(0A9N)(N*B8O6KOB

####4.,=#.0

####P#D#$N-Q)ARNA9N)(N+:N0NJ%M)4(BBC

####),.*AS:8'-1,APBT8S:8(U1/APBT8OV(0-7(=*4O8?B

####!#(0%(=-#%)'(0*+#&'-1*-%#1#+.'U1**-=#%*'(0/

####*(*,-A%)'(0*+AOJ(U-#D#!?@F+#%O80NJ%BBC

####1Q(%A?@?NSE$#$#E$#$TBC#1Q(%#-231,C

####),.*A'-1,APB8(U1/APB8O'@O8O51'6-'W(P-O8?HB

####4.,=#.++

-0=

! " 0 " 1–+23=

*+*! /=

!"#$%&'()'$"*&*+,-%.(-*/(01-."2.

)!)(3456(7,8*-%.(-*/(79.+&#. 3:3>

−2 −1 0 1 2−1

0

1Time = 0.0000 s

−2 −1 0 1 2−1

0

1Time = 0.1250 s

−2 0 2−1

0

1Time = 0.2500 s

−2 0 2−1

0

1Time = 0.3750 s

−2 0 2−1

0

1Time = 0.5000 s

−2 0 2−1

0

1Time = 0.6250 s

−2 0 2−1

0

1Time = 0.7500 s

−2 0 2−1

0

1Time = 0.8750 s

$ %– 1 $%!

$% 1

01-."2(?//,+,"*

)!)(3456(7,8*-%.(-*/(79.+&#. 3:3@

• The inverse Euler formulas can be used to see that a cosinesignal is composed of positive and negative frequency expo-nentials

(2.51)

Phasor AdditionWe often have to deal with multiple sinusoids. When the sinu-soids are at the same frequency, we can derive a formula of theform

(2.52)

At present we have only the trig identities to aid us, and thisapproach becomes very messy for large N.

Phasor Addition Rule

• We know that when complex numbers are added we must addreal and imaginary parts separately

• Consider the sum

# !" +%&' # $% !" +

$% !" +–

+0

-----------------------------------------------------=

/0---($

% !" /0---(4$

% !"–+=

/0---! " /

0---!4 "+=

*+ ! "=

#, !" ,+%&', /=

-

# !" +%&'=

01-."2(?//,+,"*

)!)(3456(7,8*-%.(-*/(79.+&#. 3:A6

(2.53)

• 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:

#,$,

, /=

-(,

, /=

-( #$%= =

*+ (,, /=

-

*+ (,, /=

-

=

# !" +%&' *+ #$% !" +

=

#, !" ,+%&', /=

-*+ #,$

% !" ,+

, /=

-=

*+ #,$% ,

, /=

-$% !"

=

*+ #$% $% !"=

*+ #$% !" +

=

# !" +%&'=

+&55&6'78,&$79:;<=>

01-."2(?//,+,"*

)!)(3456(7,8*-%.(-*/(79.+&#. 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(?//,+,"*

)!)(3456(7,8*-%.(-*/(79.+&#. 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(?//,+,"*

)!)(3456(7,8*-%.(-*/(79.+&#. 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

)!)(3456(7,8*-%.(-*/(79.+&#. 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

(/ .0---%&' %.

0---'()+ %.= =

(0 61---–%&' %6

1---–'()+ .56.66 %.56.66–= =

(. . %0+=

( %. .56.66 %.56.66– . %0++ +=

<56.66 %/51<16+=

( <56.660 /51<160+ /51<16<56.66----------------"=")=

<5<;9< !500!1 <5<;9<$%!500!1==

)!

019.,B.("C(+1&(DE*,*8(F"2G

)!)(3456(7,8*-%.(-*/(79.+&#. 3:A<

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

)!)(3456(7,8*-%.(-*/(79.+&#. 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

)!)(3456(7,8*-%.(-*/(79.+&#. 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

XX#,.1=#*30(0/+.'6

XX#),.*A*8QB

& " # ,/----" +%&'=

0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1

Time in seconds

Am

plitu

de

4 4.002 4.004 4.006 4.008 4.01 4.012 4.014 4.016 4.018 4.02−1

−0.5

0

0.5

1

Time in seconds

Am

plitu

de

H&&$7IJKJ;@:L7'

019.,B.("C(+1&(DE*,*8(F"2G

)!)(3456(7,8*-%.(-*/(79.+&#. 3:A>

• How pure is the signal produced by the tuning fork?

• In Chapter 3 of the text we begin a study of spectrum repre-sentation

• The zoom of the captured signal looks like a single sinusoid,but spectral analysis can be more revealing

• Consider the use of MATLAB’s power spectral density func-tion )%=AQ8G++*8+%1U)B (text Chapter 13)XX#)%=AQ89]?98H:::B

0 500 1000 1500 2000 2500 3000 3500 4000−80

−60

−40

−20

0

20

40

Frequency (Hz)

Pow

er S

pect

rum

Mag

nitu

de (

dB)

JJ@7MN78./O4$"/)457)./(/378&,G7P()0F

AA@7MN7'"0&/O7F4,$&/(0

Q)F",7F4,$&/(0'

R."7)&7')4,)B.P7),4/'("/)

019.,B.("C(+1&(DE*,*8(F"2G

)!)(3456(7,8*-%.(-*/(79.+&#. 3:A@

• A time-frequency plot can be obtained using the MATLAB’sspectrogram function (text Chapter 13) XX#%)-&*'./'1UAQ89]?:8Y:8ST8H:::B

Listening to Tones

• To play the tuning fork sound on the PC speakers using Mat-lab we typeXX#%.30=AQ8H:::B

where the second argument sets the sampling frequency forplayback

EF"7JJ@7MN78./O4$"/)457,"$4(/'7'),&/3

EF"7F(3F",7F4,$&/(0'784O"7(/7)($"

D,#&(7,8*-%.H(I"2&(D1-*(F"2#E%-.

)!)(3456(7,8*-%.(-*/(79.+&#. 3:;6

Time Signals: More Than Formulas• The signal modeling of this chapter has focused on single

sinusoids

• In practice real signals are far more complex, even a multiplesinusoids model is only an approximation

• Modeling still has great value in system design

Documents

Sinusoids Chapter