• A general class of signals used for modeling the interaction of signals in systems, are based on the trigonometric functions sine and cosine
Chapter
2
• The general mathematical form of a single sinusoidal signal is x t = A cos Z
0 t + I (2.1) where denotes the amplitude
I
, Z
0
is the frequency in radians/s (radian frequency), and is the phase in radians
– The arguments of cos and sin are in radians
• We will spend considerable time working with sinusoidal signals, and hopefully the various modeling applications presented in this course will make their usefulness clear
Example: x t = 10 cos > 2 S 440 t – 0.4
S @
• The pattern repeats every = 0.00227
=
• This time interval is known as the period of x t
2.27ms
ECE 2610 Signal and Systems 2––1

Review of Sine and Cosine Functions
• The text discusses how a tuning fork, used in tuning musical instruments, produces a sound wave that closely resembles a single sinusoid signal
– In particular the pitch A above middle C has an oscillation frequency of 440 hertz
• Trigonometric functions were first encountered in your K–12 math courses
• The typical scenario to explain sine and cosine functions is depicted below
– The right-triangle formed in the first quadrant has sides of length 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 r and angle in the first quadrant of the x – y plane is related
ECE 2610 Signals and Systems 2––2

Review of Sine and Cosine Functions to the x and y coordinates of the point via sin( ) and cos( ), e.g., x y = r cos T r sin T (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 plot shifted to the right relative to the cosine plot by S e 2
– This is expected since a well known trig identity states that sin T = cos – e 2 (2.3)
– We also observe that both waveforms repeat every 2 S radians; read period = 2 S
– Additionally the amplitude of each ranges from -1 and 1
• A few key function properties and trigonometric identities
ECE 2610 Signals and Systems 2––3

Review of Sine and Cosine Functions are given in the following tables
Table 2.1: Some sine and cosine properties
Property
Equivalence
Periodicity
Evenness of cosine
Oddness of sine
Equation sin T cos T
=
= cos sin
– e 2
+ e 2
or cos T – 2 S k = cos T , when k is an integer; holds for sine also cos – T = cos T sin – T = – sin T
Number
1
Table 2.2: Some trigonometric identities
Equation sin
2
T + cos
2
T = 1
2
3
4
5 cos 2 T sin 2 T
=
= cos
2
T – sin
2
T
2 sin T cos T sin cos
=
= sin D cos E r cos D sin E cos D cos E sin D sin E
6
7 cos
2
T =
1
2
1 + cos 2 T sin
2
T =
1
2
1 – cos 2 T
• For more properties consult a math handbook
ECE 2610 Signals and Systems 2––4

Review of Sine and Cosine Functions
• The relationship between sine and cosine show up in calculus too, in particular d sin d T
T
= cos T and d cos T d T
= – sin T (2.4)
– This says that the slope at any point on the sine curve is the cosine, and the slope at any point on the cosine curve is the negative 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
2cos
2
T = 1 + cos 2 T or cos
2
T =
1
2
1 + cos 2 T
(2.5)
Example: Find an expression for cos 7 T , and cos T using #5 cos 8 T in terms of cos 9 T ,
• Let D = 8 T and E = T , then write out #5 under both sign choices cos 8 T T
+ cos cos
9 T +
8 T T cos 7 T
=
=
= cos 8 T cos T – cos 8 T cos T +
2 cos 8 T cos T sin 8 T sin T sin 8 T sin T (2.6) or cos 8 T = cos 9 T + cos cos
2 T
7 T
(2.7)
ECE 2610 Signals and Systems 2––5

Review of Complex Numbers
• See Appendix A of the text for more information
• A complex number is an ordered pair of real numbers
1 denoted z = x y
– The first number, x , is called the real part, while the second number, y , is called the imaginary part
– For algebraic manipulation purposes we write
= = where i = j = – 1 x y
; electrical engineers typically use j since i is often used to denote current
Note: – 1 u – 1 = – 1 Ÿ = – 1
• The rectangular form of a complex number is as defined above, z = x y = x jy
• The corresponding polar form is z = re j T
= r ‘ T = z e j arg z
1.Tom M. Apostle, Mathematical Analysis , second edition, Addison Wesley, p. 15, 1974.
ECE 2610 Signals and Systems 2––6

Review of Complex Numbers
• We can plot a complex number as a vector x y
Example:: z = 2 + j 5 , z = 4 – j 3 , z = – 5 + j 0 , z = – 3 – j 3
ECE 2610 Signals and Systems 2––7

Review of Complex Numbers
Example:: z = 2 45 q , z = 3 150 q , & z = 3 ‘ – 80 q
• For complex numbers z
1 define/calculate
= z
1
+ z
2 z
1
– z
2 z
1 z
2 z z
1
2
=
=
=
= x
1
+ jy
1
and z
2
= x
2
+ jy
2
we x
1
+ x
2
+ j y
1
+ y
2
(sum) x
1
– x
2
+ j y x
1 x
2
– y
1 y
2
1
+
– y j x
1
2
(difference) y
2
+ y
1 x
2
(product) x x + y y – j x y – y x
------------------------------------------------------------------------ (quotient) x 2
2
+ y 2
2
ECE 2610 Signals and Systems 2––8

Review of Complex Numbers z
1
‘ z
1
=
= x
2
1
+ y
2
1
(magnitude) tan
– 1 y
1 e x
1
(angle) z *
1
= x
1
– jy
1
(complex conjugate)
– M ATLAB is also consistent with all of the above, starting with the fact that i and j are predefined to be – 1 rectangular polar
• To convert from polar to rectangular we can use simple trigonometry to show that x = r cos T y = r sin T
(2.24)
• Similarly we can show that rectangular to polar conversion is r = x
2
T = tan
– 1
+ y
2 y x note add r S outside Q1 & Q4
(2.25)
ECE 2610 Signals and Systems 2––9

Review of Complex Numbers
Example: Rect to Polar and Polar to Rect
• Consider z
1
= 2 + j 5
– In M ATLAB we simply enter the numbers directly and then need 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 z
2
= 2 45 q
– In M ATLAB we simply enter the numbers directly as a complex exponential
>> z2 = 2*exp(j*45*pi/180) z2 = 1.4142e+00 + 1.4142e+00i
ECE 2610 Signals and Systems 2––10

Review of Complex Numbers
– Using the TI-89 we can directly enter the polar form using the angle notation or using a complex exponential
Example: Complex Arithmetic
• Consider z
1
= 1 + j 7 and z
2
• Find z
1
+ z
2
>> z1 = 1+j*7;
>> z2 = -4-j*9;
>> z1+z2
= – 4 – j 9 ans = -3.0000e+00 - 2.0000e+00i
– Using the TI-89 we obtain
ECE 2610 Signals and Systems 2––11

• Find z
1 z
2
>> z1*z2 ans = 5.9000e+01 - 3.7000e+01i
– Using the TI-89 we obtain
Review of Complex Numbers
• Find z
1 e z
>> z1/z2
2 ans =-6.9072e-01 - 1.9588e-01i
TI-89
Results
Euler’s Formula: A special mathematical result, of special importance to electrical engineers, is the fact that e j T
= cos T + j sin T (2.26)
ECE 2610 Signals and Systems 2––12

Sinusoidal Signals
• Turning (2.26) around yields ( inverse Euler formulas ) sin T = j T
–
2 j e
– j T
and cos T =
• It also follows that z = x jy = r cos T + jr sin T j T
+
2 e
– j T
(2.27)
(2.28)
• A general sinusoidal function of time is written as x t = A cos Z
0 t + I = A cos 2 S f
0 t + I (2.29) where in the second form Z
0
• Since
= 2 S f
0 cos T d 1 it follows that x t swings between r A , so the amplitude of x t is A nal (instead of the cosine version), we see via the equivalence property that x t = A sin Z
0 t + Ic = A cos Z
0 t + – e 2 (2.30) which implies that I = – e 2 the oscillation frequency in cycles/s; why?
Z
------
0
2 S
= f
0 sec
– 1
ECE 2610 Signals and Systems 2––13

Sinusoidal Signals
Example: x t = 20 cos > 2 S 40 t – 0.4
S @
• Clearly, A = 20 , f
0
= 40 cycles/s, and I = – 0.4
S rad
1
0.025
s
40
=
Maxima
Interval
(period)
= 25ms
• Since this signal is periodic , the time interval between maxima, minima, and zero crossings, for example, are identical
Relation of Frequency to Period
• A signal is periodic if we can write
0
= x t (2.31) where the smallest T
0
satisfying (2.10) is the period
• For a single sinusoid we can relate T
0 f
0
0
= x t
A cos Z
0 cos t T
0
+
Z
0 t I Z
0
T
0
=
=
A cos Z
0 t + I cos Z
0 t + I
(2.32)
• From the periodicity property of cosine, equality is maintained if cos T r 2 S k = cos T , so we need to have
ECE 2610 Signals and Systems 2––14

Sinusoidal Signals
Z
0
T
0
= 2 S œ T
0
= ------
0
(2.33)
or 2 S f
0
T
0
œ T
0
= f
0
• So we see that T
0 ond, as stated earlier f
0 f
0
T
0 being time and the units of inverse time or cycles per sec-
– In honor of Heinrich Hertz, who first demonstrated the existence of radio waves, cycles per second is replaced with Hertz (Hz)
ECE 2610 Signals and Systems 2––15

Sinusoidal Signals
Example: 5 cos 2 S f
0 t with f
0
= 200 , 100, and 0 Hz
Period doubles as frequency halves
A constant signal as the oscillation frequency is zero
• The inverse relationship between time and frequency will be explored through out this course
ECE 2610 Signals and Systems 2––16

Sinusoidal Signals
Phase Shift and Time Shift
• We know that the phase shift parameter in the sinusoid moves the waveform left or right on the time axis
• To formally understand why this is, we will first form an understanding of time-shifting in general
• Consider a triangularly shaped signal having piece wise continuous definition s t =
­
°
°
®
°
°
¯
2 t , 0 d d 1 2
1
3
4 2 t , 1 2 d d 2
0, otherwise s t
1
(2.34)
2 t
1
3
– t t
-1 0 1 2 3
2
• Now we wish to consider the signal x
1 t = s t – 2
• As a starting point we note that s t is active over just the interval 0 d t d 2 , so with t o t – 2 we have
0 d t – 2 d 2 Ÿ 2 t 4 (2.35) which means that x
1 t is active over 2 t 4
• The piece wise definition of t – 2 x
1 t t can be obtained by direct substitution of everywhere appears in (2.34)
ECE 2610 Signals and Systems 2––17

Sinusoidal Signals x
1 t =
­
°
°
®
°
°
¯
2 t – 2 ,
1
3
– t – 2
0,
0 d t d
– 2 otherwise d t – 2 d 2
(2.36) x
1 t = s t – 2
1
=
­
°
°
®
°
°
¯
2 t – 4, 2 d d 5 2
1
3
8 2 t , 5 2 d d 4
0, otherwise
2 t – 2
1
3
– t t
0 1 2 3 4
2
• In summary we see that the original signal s t is moved to the right by 2 s
Example: Plot s t + 1
• With t o t + 1 we expect that the signal will shift to the left by one second s t + 1
1
2 3 t
-1 –
2
0
ECE 2610 Signals and Systems
1
2––18

Sinusoidal Signals
• The new equations are obtained as before
0 d t + 1 d 2 Ÿ – 1 d d 1 so s t + 1 =
­
°
°
®
°
°
¯
2 t + 1 ,
1
3
– t + 1
0,
0 d t + 1 otherwise d d t + 1 d 2
=
­
°
°
®
°
°
¯
2 t + 2, – 1 d d – 1 e 2
1
3
2 2 t , 1 e 2 t 1
0, otherwise
(2.37)
(2.38)
• Modeling time shifted signals shows up frequently
• In general terms we say that x
1 t = –
1
(2.39) is delayed in time relative to s t if t
1 time relative to s t if t
1
0
!
0 , and advanced in
• A cosine signal has positive peak located at t = 0 the corresponding phase shift is negative
• Consider x
0 t = A cos Z
0 t
ECE 2610 Signals and Systems 2––19

Sinusoidal Signals x
0 1
=
=
A cos > Z
0 1
@
A cos > Z
0 t – Z
0 t
1
@
(2.40) which implies that in terms of phase shift we have
I = – Z
0 t
1
• For a given phase shift we can turn the above analysis around and solve for the time delay via t
1
= –
Z
0
= –
2 S f
0
(2.41)
• Since T
0
= of the period
1 e f
0
, we can also write the phase shift in terms
I = – 2 S f
0 t
1
= – 2 S t
----(2.42)
– An important point to note here is that both cosine and sine are mod 2 S functions, meaning that phase is only unique on a 2 S interval, say ( – S S ] or (0 2 S ]
Example: Suppose t
1
= 10 ms and T
0
= 3
• Direct substitution into (2.21) results in ms
I = – 2 S
§ ·
3
= –
20
3
S = – 6.6667
S (2.43)
• We need to reduce this value modulo 2 S to the interval
( – S S ] by adding (or subtracting as needed) multiples of 2 S
• The result is the reduced phase value
ECE 2610 Signals and Systems 2––20

Sampling and Plotting Sinusoids
I = –
20
3
S + 6 S = –
20 18
3
S = –
2
3
S = – 0.6667
S (2.44)
• Does this result make sense?
• A time delay of 10 ms with a period of 3 ms means that we have 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-
– S = – 0.6667
S ; agrees with the above analysis
Modulo the period delay of 1 ms
Actual Delay of 10 ms t (ms)
Blue = no delay
Red = 10 ms Delay
• The value of phase shift that lies on the interval – S d known as the principle value
is
• When plotting sinusoidal signals using computer tools, we are also faced with the fact that only a discrete-time version
ECE 2610 Signals and Systems 2––21

Sampling and Plotting Sinusoids of x t = A cos 2 S f
0 t + I may be generated and plotted
• This fact holds true whether we are using M ATLAB , C, Mathematica, Excel, or any other computational tool
• When t o nT s
we need to realize that sample spacing needs f
0 when plotted by connecting the dots ( linear interpolation ), the waveform picture is not too distorted
– In Chapter 4 we will discuss sampling theory , which will tell us the maximum sample spacing (minimum sampling rate which is x n from
= x nT s
1 e T s
), such that the sequence
can be used to perfectly reconstruct x t
• For now we are more concerned with having a good plot appearance relative to the expected sinusoidal shape
• A reasonable plot can be created with about 10 samples per period, that is with T s
| e f
0
= T
0 e 10
• We will now consider several M ATLAB 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)
ECE 2610 Signals and Systems 2––22

ECE 2610 Signals and Systems
Sampling and Plotting Sinusoids
>> 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
= 3 Hz, A = 15, I = S /2
20 f
0
10
T s
=
T
------
5
0
−10
5 Samples per period
−20
0 0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time in seconds
0.8
0.9
20
10
T s
=
1 T
10
0
−10
10 Samples per period
−20
0 0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time in seconds
0.8
0.9
20
10
T s
=
T
------
50
0
−10
−20
0
50 Samples per period
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time in seconds
0.8
0.9
1
1
1
2––23

!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
Modeling signals as pure sinusoids is not that common. We typically have more that one sinusoid present. Manipulating multiple sinusoids is actually easier when we form a complex exponential representation.
Complex Exponential Signals
• Motivated by Euler’s formula above, and the earlier definition of a cosine signal, we define the complex exponential signal as where ! " = #
! " = #$
% Z
!
" + I
and ‘ ! " = "#$ ^ ! " ` = Z
!
" + I
• Note that using Euler’s formula
! " = #$
% Z
!
" + I
= # %&' Z
!
" + I + %# '() Z
!
" + I
(2.44)
(2.45)
• We see that the complex sinusoid has amplitude A , phase shift , and frequency Z
!
rad/s
– Note in particular that
*+ ^ ! " ` = # %&' Z
!
" + I
,^ ! " ` = # '() Z
!
" + I
(2.46)
)!)(3456(7,8*-%.(-*/(79.+&#.
3:3;

!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
– The result of (2.46) is what ultimately motivates us to con-
sider the complex exponential signal
• We can always write
& " = *+ ^ #$
% Z
!
" + I
` = # %&' Z
!
" + I (2.47)
The Rotating Phasor Interpretation
• Complex numbers in polar form can be easily multiplied as
!
.
= '
/
$
% T
­ ® ¯
/ ˜ '
0
$
% T
­ ® ¯
0
!
/
!
0
= '
/
'
0
$
% T
/
+ T
0
(2.48)
• For the case of
! " = #$
% Z
!
" + I
(2.49)
)!)(3456(7,8*-%.(-*/(79.+&#.
3:3<

!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
we can write
! " = #$
% I where ( = #$
% I
˜ $
% Z
!
"
= ($
% Z
!
"
(2.50)
• The complex amplitude is called the phasor , as it is the gain and phase value applied to the time varying component
$
% Z
!
"
to form ! "
– This is common terminology is electrical engineering circuit theory
• The time varying term $
% Z
!
"
has unit magnitude and rotates counter clockwise in the complex plane at a rate of Z
!
rad/s
– The time duration for one rotation is the period *
!
= / e )
$
% Z
!
!
" results in a rotating phasor
– For positive frequency Z
!
the rotation is counter clockwise, and for negative frequency the rotation is clockwise
#$
%&'()(*"
+,"-."/01
2"34)(*"
+,"-."/01
T "
!"
T "
!"
!"#$#%&'()*$+",+
T " = Z
!
" + I
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3:34

!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
Example: ! " = +23 > 0 S " – S e 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=
)!)(3456(7,8*-%.(-*/(79.+&#.
3:3=

Time = 0.0000 s
1
0
− 1
− 2 − 1 0 1
$
– % S e 1
2
1
0
− 1
− 2
Time = 0.2500 s
$
% S e 1
0 2
1
0
− 1
− 2
Time = 0.5000 s
0 2
1
0
− 1
− 2
Time = 0.7500 s
0 2
)!)(3456(7,8*-%.(-*/(79.+&#.
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
Time = 0.1250 s
1
0
− 1
− 2 − 1 0 1
$
% !
2
1
0
− 1
− 2
Time = 0.3750 s
0 2
1
0
− 1
− 2
Time = 0.6250 s
0 2
1
0
− 1
− 2
Time = 0.8750 s
0 2
3:3>

01-."2(?//,+,"*
• The inverse Euler formulas can be used to see that a cosine signal is composed of positive and negative frequency exponentials
# %&' Z
!
" + I =
=
=
=
#
§
©
$
% Z
!
" + I
+ $
– % Z
!
" + I
0
·
¹
/
0
($
% Z
!
"
+
/
0
(
4
$
– % Z
!
"
/
0
! " +
*+ ^ ! " `
/
0
!
4
"
(2.51)
We often have to deal with multiple sinusoids. When the sinusoids are at the same frequency, we can derive a formula of the form
-
¦
, = /
#
,
%&' Z
!
" + I
,
= # %&' Z
!
" + I (2.52)
At present we have only the trig identities to aid us, and this approach becomes very messy for large N .
Phasor Addition Rule
• We know that when complex numbers are added we must add real and imaginary parts separately
• Consider the sum
)!)(3456(7,8*-%.(-*/(79.+&#.
3:3@

01-."2(?//,+,"*
-
¦
, = /
#
,
$
I
, =
-
¦
(
,
{ (
, = /
= #$
% I
(2.53)
• The above is valid since the real and imaginary parts add independently, that is
*+
­
®
¯
-
¦
, = /
(
,
½
¾
¿
=
-
¦
, = /
*+ (
,
(2.54) and the same holds for the imaginary part
• Secondly, a real sinusoid can always be written in terms of a complex sinusoid via
# %&' Z
!
" + I = *+ ^ #$
% Z
!
" + I
` (2.55)
Proof:
-
¦
, = /
#
,
%&' Z
!
" + I
,
=
-
¦
, = /
*+ ^ #
,
$
% Z
!
" + I
, `
+&55&6'78,&$79:;<=>
= *+
­
®
¯
¨
©
§ -
¦
, = /
#
,
$
% I
, ¸
¹
·
$
% Z
!
"
½
¾
¿
=
=
=
*+ ^ #$
% I
$
% Z
!
"
`
*+ ^ #$
% Z
!
" + I
# %&' Z
!
" + I
`
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A6

01-."2(?//,+,"*
Example: Phasor Addition Rule in Action
• Consider the sum
& " = &
/
" + &
0
"
= 156 %&' .!
S " + .6
S e /7!
+ %&' .!
S " + 7!
S e /7!
• The frequency of the sinusoids is 15 Hz
• Using phasor notation we can write that
&
/
" = ^ $
% .6
S e /7!
˜ $
% .!
S "
`
&
0
" = ^ $
% 7!
S e /7!
˜ $
% .!
S "
`
(2.56)
(2.57) so in the phasor addition rule
(
/
= 156 $
% .6
S e /7!
(
0
= 950 $
% 7!
S e /7!
(2.58)
• We perform the complex addition and conversion back to polar form using the TI-89
(
I 8()8#":(")' so
( = (
/
+ (
0
=
= 15;.<16 +
/!57679
$
% <05;<!0
S e /7!
% ;5<9/9/
)!)(3456(7,8*-%.(-*/(79.+&#.
(2.59)
3:A5

01-."2(?//,+,"*
• Finally,
& " = /!57< %&' .!
S " + <05;<!0
S e /7!
(2.60)
• We can check this by directly plotting the waveform in M AT -
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-#%*\,-%
?@;A::
10
& " x
1
(t) x
2
(t) x(t)
5
&
/
"
0
B??;CD7$'
− 5
&
0
"
− 10
− 0.06
− 0.04
− 0.02
0
Time in seconds
0.02
0.04
0.06
• The measured amplitude, 10.822, is close to the expected value
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A3

01-."2(?//,+,"*
• The location of the peak can be converted to phase via
I = – 0 S
§
©
–
<<5<9
¹
·
= /5/!8#": = <.5!0
q (2.61)
Summary of Phasor Addition
• When we need to form the sum of sinusoids at the same frequency, we obtain the final amplitude A and phase via where (
,
( = (
= #
,
$
/
% I
+ (
0
+
}
+ (
, and
-
& " =
=
= #$
% I
-
¦
#
,
%&'
, = /
# %&' Z
!
Z
!
" + I
" + I
,
(2.62)
(2.63)
Example:
& " =
­
®
¯
$
%
§
0 S )
!
" +
0
+ 6 $
%
§
0 S )
!
" –
1
+ .
+ % 0 $
% 0 S )
!
"
½
¾
¿
• Find ( = (
/
+ (
0
+ (
.
• From the given & " we observe that
(
/
= .
$
%
0
(
0
= 6 $
– %
1
(
.
= .
+ % 0
)!)(3456(7,8*-%.(-*/(79.+&#.
3:AA

019.,B.("C(+1&(DE*,*8(F"2G
• To perform the complex addition we will work step-by-step
• To add complex numbers we convert to rectangular form
(
/
= .
%&'
S
0
+ % .
'()
0
= % .
(
0
= 6 %&'
S
1
+ % 6 '()
S
1
= .56.66
– % .56.66
(
.
= .
+ % 0
• Now,
( = % .
+ .56.66
– % .56.66
= <56.66
+ % /51<16
+ .
+ % 0
• For use in the phasor sum formula we likely need the answer in polar form
( =
=
<56.66
0
+ /51<16
0
‘ "=")
<56.66
‘ = <5<;9< $
% !500!1
The tuning fork signal generation example discussed earlier was important because it is an example of a physical system that when struck, produces nearly a pure sinusoidal signal.
– By pure we mean a signal composed of a single frequency sinusoid, no other sinusoids at other frequencies, say harmonics (multiples of ) are present
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A;

019.,B.("C(+1&(DE*,*8(F"2G
Equations from Laws of Physicss
A 2-D model of the tuning fork is shown below
E6&7)(/"'7&8
)F"7)./(/378&,G
• When struck the vibration of the metal tine moves air molecules to produce a sound wave
• Hooke’s law from physics (springs, etc.) says that the force to restore the tine back to its original & = !
position is the same as the original deformation (striking force), except for a sign change,
.
= – ,& (2.64) where k is the material stiffness constant
• The acceleration produced by the restoring force (Newton’s second law) is
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A<

019.,B.("C(+1&(DE*,*8(F"2G
.
= /0 = /
1
0
1"
0
(2.65)
• To balance the two forces (sum is zero), we must have
/
0
&
1"
0
= – ,& " (2.66)
General Solution to the Differential Equation
• To solve this equation we can actually guess the solution by inserting a test function of the form & " = %&' Z
!
"
0
& "
1"
0
=
1
1"
– Z
!
'() Z
!
"
= – Z
0
!
%&' Z
!
"
• We now plug this result into (2.66) to obtain
/
1
0
1"
0
– / Z
0
!
%&' Z
!
"
=
=
– ,& "
– , %&' Z
!
"
(2.67) which tells us that we must have
– / Z
0
!
= – , so it must be that
(2.68)
Z
!
= r
/
(2.69)
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A4

019.,B.("C(+1&(DE*,*8(F"2G
• This tells us that the oscillation frequency of the tuning fork is 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 phase shift constant is of the form
& " = # %&'
§
©
,
/
" + I
·
¹
(2.70)
• The sound produced by the 440Hz tuning fork was captured using M ATLAB on a PC with a sound card and microphone
– The results were converted to double precision and saved in a .mat file along with a time axis vector
XX#,.1=#*30(0/+.'6
XX#),.*A*8QB
1
0.5
0
− 0.5
− 1
0 1 2
H&&$7IJKJ;@:L7'
3
1
4 5 6
Time in seconds
7 8 9 10
0.5
0
− 0.5
− 1
4 4.002 4.004 4.006 4.008 4.01 4.012 4.014 4.016 4.018 4.02
Time in seconds
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A=

019.,B.("C(+1&(DE*,*8(F"2G
• How pure is the signal produced by the tuning fork?
• In Chapter 3 of the text we begin a study of spectrum representation
• The zoom of the captured signal looks like a single sinusoid, but spectral analysis can be more revealing
• Consider the use of M ATLAB ’s power spectral density function )%=AQ8G++*8+%1U)B (text Chapter 13)
XX#)%=AQ89]?98H:::B
40
JJ@7MN78./O4$"/)457)./(/378&,G7P()0F
20
AA@7MN7'"0&/O7F4,$&/(0
0
Q)F",7F4,$&/(0'
− 20
− 40
− 60
− 80
0 500
R."7)&7')4,)B.P7),4/'("/)
1000 1500 2000
Frequency (Hz)
2500 3000 3500 4000
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A>

019.,B.("C(+1&(DE*,*8(F"2G
• A time-frequency plot can be obtained using the M ATLAB ’s spectrogram function (text Chapter 13)
XX#%)-&*'./'1UAQ89]?:8Y:8ST8H:::B
EF"7JJ@7MN78./O4$"/)457,"$4(/'7'),&/3
EF"7F(3F",7F4,$&/(0'784O"7(/7)($"
Listening to Tones
• To play the tuning fork sound on the PC speakers using Matlab we type
XX#%.30=AQ8H:::B where the second argument sets the sampling frequency for playback
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A@

D,#&(7,8*-%.H(I"2&(D1-*(F"2#E%-.
• The signal modeling of this chapter has focused on single sinusoids
• In practice real signals are far more complex, even a multiple sinusoids model is only an approximation
• Modeling still has great value in system design
)!)(3456(7,8*-%.(-*/(79.+&#.
3:;6