Download Sinusoids

Sinusoids
Chapter
• A general class of signals used for modeling the interaction of signals in systems, are based on the trigonometric functions sine and cosine
2
• The general mathematical form of a single sinusoidal signal
is
x t = A cos Z 0 t + I (2.1)
where A denotes the amplitude, Z 0 is the frequency in radians/s (radian frequency), and I 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 > 2S 440 t – 0.4S @
• The pattern repeats every 1 e 440 = 0.00227 = 2.27ms
• This time interval is known as the period of x t 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
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 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 T 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 T 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 T – S e 2 (2.3)
– We also observe that both waveforms repeat every 2S radians; read period = 2S
– 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
Equation
Equivalence
sin T = cos T – S e 2 or
cos T = sin T + S e 2 Periodicity
cos T – 2Sk = cos T , when k is an
integer; holds for sine also
Evenness of cosine cos – T = cos T
Oddness of sine
sin – T = – sin T
Table 2.2: Some trigonometric identities
Number
Equation
1
sin T + cos T = 1
2
cos 2T = cos T – sin T
3
sin 2T = 2 sin T cos T
4
sin D r E = sin D cos E r cos D sin E
5
cos D r E = cos D cos E sin D sin E
6
1
2
cos T = --- 1 + cos 2T 2
7
2
1
sin T = --- 1 – cos 2T 2
2
2
2
2
• 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 cos T
d sin T
-------------- = cos T and --------------- = – sin T
dT
dT
(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
2
2cos T = 1 + cos 2T
2
1
or cos T = --- 1 + cos 2T 2
(2.5)
Example: Find an expression for cos 8T in terms of cos 9T ,
cos 7T , and cos T using #5
• Let D = 8T and E = T , then write out #5 under both sign
choices
cos 8T + T = cos 8T cos T – sin 8T sin T
cos 8T – T = cos 8T cos T + sin 8T sin T
+
cos 9T + cos 7T = 2 cos 8T cos T
(2.6)
cos 9T + cos 7T
cos 8T = ------------------------------------2 cos T
(2.7)
or
ECE 2610 Signals and Systems
2––5
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 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 x y = x + iy = x + jy where i = j = – 1 ; electrical engineers typically use j since i is often used to denote current
Note:
–1 u –1 = –1 Ÿ j u j = –1
• The rectangular form of a complex number is as defined
above,
z = x y = x + jy
• The corresponding polar form is
z = re
jT
= 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 + j5 , z = 4 – j3 , z = – 5 + j0 , z = – 3 – j3
ECE 2610 Signals and Systems
2––7
Review of Complex Numbers
Example:: z = 2 ‘45q , z = 3 ‘150q , & z = 3 ‘– 80 q
• For complex numbers z 1 = x 1 + jy 1 and z 2 = x 2 + jy 2 we
define/calculate
z 1 + z 2 = x 1 + x 2 + j y 1 + y 2 (sum)
z 1 – z 2 = x 1 – x 2 + j y 1 – y 2 (difference)
z 1 z 2 = x 1 x 2 – y 1 y 2 + j x 1 y 2 + y 1 x 2 (product)
x1 x2 + y1 y2 – j x1 y2 – y1 x2 z1
---- = ------------------------------------------------------------------------- (quotient)
z2
x2 + y 2
2
ECE 2610 Signals and Systems
2
2––8
Review of Complex Numbers
z1 =
2
2
x 1 + y 1 (magnitude)
–1
‘z 1 = tan y 1 e x 1 (angle)
z *1 = x 1 – jy 1 (complex conjugate)
– MATLAB 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 =
2
x +y
2
–1
(2.25)
T = tan y e x note add r S outside Q1 & Q4
ECE 2610 Signals and Systems
2––9
Review of Complex Numbers
Example: Rect to Polar and Polar to Rect
• Consider z 1 = 2 + j5
– In MATLAB 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 ‘45q
– In MATLAB 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 + j7 and z 2 = – 4 – j9
• Find z 1 + z 2
>> z1 = 1+j*7;
>> z2 = -4-j*9;
>> z1+z2
ans = -3.0000e+00 - 2.0000e+00i
– Using the TI-89 we obtain
ECE 2610 Signals and Systems
2––11
Review of Complex Numbers
• Find z 1 z 2
>> z1*z2
ans = 5.9000e+01 - 3.7000e+01i
– Using the TI-89 we obtain
• Find z 1 e z 2
>> z1/z2
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
jT
ECE 2610 Signals and Systems
= cos T + j sin T
(2.26)
2––12
Sinusoidal Signals
• Turning (2.26) around yields (inverse Euler formulas)
– jT
jT
jT
– jT
–e e +e
sin T = e--------------------and cos T = ---------------------2j
2
(2.27)
• It also follows that
z = x + jy = r cos T + jr sin T
(2.28)
Sinusoidal Signals
• A general sinusoidal function of time is written as
x t = A cos Z 0 t + I = A cos 2Sf 0 t + I (2.29)
where in the second form Z 0 = 2Sf 0
• Since cos T d 1 it follows that x t swings between r A , so
the amplitude of x t is A
• The phase shift in radians is I , so if we are given a sine signal (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 + Ic – S e 2 (2.30)
which implies that I = Ic – S e 2
• Engineers often prefer the second form of (2.8) where f 0 is
the oscillation frequency in cycles/s; why?
Z 0 § rad/s·
–1
------ ----------- = f 0 sec 2S © rad ¹
ECE 2610 Signals and Systems
2––13
Sinusoidal Signals
Example: x t = 20 cos > 2S 40 t – 0.4S @
• Clearly, A = 20 , f 0 = 40 cycles/s, and I = – 0.4S rad
Maxima
Interval
(period)
1- = 0.025s
----40
= 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
x t + T0 = x t (2.31)
where the smallest T 0 satisfying (2.10) is the period
• For a single sinusoid we can relate T 0 to f 0 by considering
x t + T0 = x t A cos Z 0 t + T 0 + I = A cos Z 0 t + I (2.32)
cos Z 0 t + I + Z 0 T 0 = cos Z 0 t + I • From the periodicity property of cosine, equality is maintained if cos T r 2Sk = cos T , so we need to have
ECE 2610 Signals and Systems
2––14
Sinusoidal Signals
2S
Z 0 T 0 = 2S œ T 0 = -----Z0
1or 2Sf 0 T 0 œ T 0 = --f0
(2.33)
• So we see that T 0 and f 0 are reciprocals, with the units of T 0
being time and the units of f 0 inverse time or cycles per second, as stated earlier
– 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 2Sf 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
­
°
°
st = ®
°
°
¯
st
2t,
0dtd1e2
1-- 4 – 2t , 1 e 2 d t d 2
3
0,
otherwise
1
1
--- 4 – 2t 3
2t
-1
(2.34)
0
1--2
1
2
3
t
• 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 d t d 4
(2.35)
which means that x 1 t is active over 2 d t d 4
• The piece wise definition of x 1 t can be obtained by direct
substitution of t – 2 everywhere t appears in (2.34)
ECE 2610 Signals and Systems
2––17
Sinusoidal Signals
­
°
°
x1 t = ®
°
°
¯
­
°
°
= ®
°
°
¯
2 t – 2 ,
0 d t – 2 d 1 e 2
1-- 4 – 2 t – 2 , 1 e 2 d t – 2 d 2
3
0,
otherwise
(2.36)
2t – 4,
2dtd5e2
1-- 8 – 2t , 5 e 2 d t d 4
3
0,
otherwise
x1 t = s t – 2 1
1
--- 8 – 2t 3
2t – 2
0
1
2
5--2
3
4
t
• 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
st + 1
1
-1
1
– --2
0
ECE 2610 Signals and Systems
1
2
3
t
2––18
Sinusoidal Signals
• The new equations are obtained as before
0 d t + 1 d 2 Ÿ –1 d t d 1
(2.37)
so
­
°
°
st + 1 = ®
°
°
¯
­
°
°
= ®
°
°
¯
2 t + 1 ,
0 d t + 1 d 1 e 2
1-- 4 – 2 t + 1 , 1 e 2 d t + 1 d 2
3
0,
otherwise
(2.38)
2t + 2,
–1 d t d –1 e 2
1-- 2 – 2t , – 1 e 2 d t d 1
3
0,
otherwise
• Modeling time shifted signals shows up frequently
• In general terms we say that
x1 t = s t – t1 (2.39)
is delayed in time relative to s t if t 1 ! 0 , and advanced in
time relative to s t if t 1 0
• A cosine signal has positive peak located at t = 0
• If this signal is delayed by t 1 the peak shifts to the right and
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 t – t 1 = A cos > Z 0 t – t 1 @
= A cos > Z 0 t – Z 0 t 1 @
(2.40)
which implies that in terms of phase shift we have
I = –Z0 t1
• For a given phase shift we can turn the above analysis around
and solve for the time delay via
I
I
t 1 = – ------ = – ----------Z0
2Sf 0
(2.41)
• Since T 0 = 1 e f 0 , we can also write the phase shift in terms
of the period
t1 ·
§
I = – 2Sf 0 t 1 = – 2S ----© T 0¹
(2.42)
– An important point to note here is that both cosine and sine
are mod 2S functions, meaning that phase is only unique
on a 2S interval, say ( – S S] or (0 2S]
Example: Suppose t 1 = 10 ms and T 0 = 3 ms
• Direct substitution into (2.21) results in
20
10
I = – 2S § ------· = – ------ S = – 6.6667S
© 3¹
3
(2.43)
• We need to reduce this value modulo 2S to the interval
( – S S] by adding (or subtracting as needed) multiples of 2S
• The result is the reduced phase value
ECE 2610 Signals and Systems
2––20
Sampling and Plotting Sinusoids
20
20 + 18
2
I = – ------ S + 6S = – ------------------ S = – --- S = – 0.6667S
3
3
3
(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 corresponding to S rad, so a delay of 1/3 period is a phase shift of
– 2 e 3 S = – 0.6667S ; 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 I d S is
known 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
ECE 2610 Signals and Systems
2––21
Sampling and Plotting Sinusoids
of
x t = A cos 2Sf 0 t + I may be generated and plotted
• This fact holds true whether we are using MATLAB, C, Mathematica, Excel, or any other computational tool
• When t o nT s we need to realize that sample spacing needs
to be small enough relative to the frequency f 0 such that
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 1 e T s ), such that the sequence
x > n @ = x nT s can be used to perfectly reconstruct x t from x > n @
• 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 | 1 e 10f 0 = T 0 e 10
• 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)
ECE 2610 Signals and Systems
2––22
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
f0 = 3 Hz, A = 15, I = -S/2
Amplitude
20
T
T s = -----05
10
0
5 Samples
−10 per period
−20
0
0.1
0.2
0.3
Amplitude
20
0.7
0.8
0.9
1
0.7
0.8
0.9
1
0.7
0.8
0.9
1
1T
T s = --------010
10
0
−10
−20
10 Samples
per period
0
0.1
0.2
0.3
20
Amplitude
0.4
0.5
0.6
Time in seconds
0.4
0.5
0.6
Time in seconds
T
T s = -----050
10
0
−10 50 Samples
per period
−20
0
0.1
0.2
0.3
ECE 2610 Signals and Systems
0.4
0.5
0.6
Time in seconds
2––23
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
Complex Exponentials and Phasors
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
! " = #$
% Z! " + I (2.44)
where ! " = # and ‘! " = "#$ ^ ! " ` = Z ! " + I
• Note that using Euler’s formula
%Z " + I
! " = #$ !
= # %&' Z ! " + I + %# '() Z ! " + I (2.45)
• We see that the complex sinusoid has amplitude A, phase
shift I , and frequency Z ! rad/s
– Note in particular that
*+ ^ ! " ` = # %&' Z ! " + I ,- ^ ! " ` = # '() Z ! " + I )!)(3456(7,8*-%.(-*/(79.+&#.
(2.46)
3:3;
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
– The result of (2.46) is what ultimately motivates us to consider 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
˜ '0 $
!/
%T 0
­
®
¯
%T /
­
®
¯
!. = '/ $
= '/ '0 $
!0
% T/ + T0 (2.48)
• For the case of
! " = #$
)!)(3456(7,8*-%.(-*/(79.+&#.
% Z! " + I (2.49)
3:3<
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
we can write
%I
! " = #$ ˜ $
where ( = #$
%Z ! "
= ($
%Z ! "
(2.50)
%I
• 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
%Z "
• The time varying term $ ! has unit magnitude and rotates
counter clockwise in the complex plane at a rate of Z ! rad/s
( ) ! rotations/s)
– The time duration for one rotation is the period * ! = / e ) !
%Z ! "
• The combination (product) of the fixed phasor ( and $
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"
!"
!"#$#%&'()*$+",+
)!)(3456(7,8*-%.(-*/(79.+&#.
T"
!"
T " = Z! " + I
3:34
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
Example: ! " = +23 > 0S" – S e 1 @
• 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=
)!)(3456(7,8*-%.(-*/(79.+&#.
3:3=
!"#$%&'()'$"*&*+,-%.(-*/(01-."2.
Time = 0.0000 s
Time = 0.1250 s
1
1
0
−1
$
−2
−1
0
–% S e 1
1
2
0
−1
$
−2
Time = 0.2500 s
1
$
0
−1
−2
0
%S e 1
2
−1
−2
0
−1
−1
2
−2
Time = 0.7500 s
1
0
0
)!)(3456(7,8*-%.(-*/(79.+&#.
2
0
2
Time = 0.8750 s
1
0
0
Time = 0.6250 s
0
−2
2
0
1
−1
1
1
1
0
0
Time = 0.3750 s
Time = 0.5000 s
−2
−1
%!
2
−1
−2
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 = # § -----------------------------------------------------·
©
¹
0
/ %Z ! " / 4 –%Z! "
= --- ($
+ --- ( $
0
0
(2.51)
/
/ 4
= --- ! " + --- ! " 0
0
= *+ ^ ! " `
Phasor Addition
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 , =
,=/
+&55&6'78,&$79:;<=>
-
¦
*+ ^ # , $
% Z! " + I, `
,=/
­§ %I ,· %Z ! " ½
= *+ ® ¨
#, $ ¸ $ ¾
¹
¯©, = /
¿
¦
%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" + .6S e /7! 8 + 950 %&' .!S" + 7!S e /7! (2.56)
• The frequency of the sinusoids is 15 Hz
• Using phasor notation we can write that
& / " = *+ ^ 156$
%.6S e /7!
& 0 " = *+ ^ 950$
%7!S e /7!
˜$
%.!S"
˜$
%.!S"
`
`
(2.57)
so in the phasor addition rule
( / = 156$
%.6S 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
(
I8()8#":(")'
so
( = ( / + ( 0 = 15;.<16 + %;5<9/9/
= /!57679$
)!)(3456(7,8*-%.(-*/(79.+&#.
%<05;<!0S e /7!
(2.59)
3:A5
01-."2(?//,+,"*
• Finally,
& " = /!57< %&' .!S" + <05;<!0S e /7! (2.60)
• We can check this by directly plotting the waveform in MATLAB
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::
x1(t)
10
x2(t)
&"
x(t)
Amplitude
5
&/ " 0
B??;CD7$'
−5
&0 " −10
−0.06
−0.04
−0.02
0
0.02
Time in seconds
0.04
0.06
• The measured amplitude, 10.822, is close to the expected
value
)!)(3456(7,8*-%.(-*/(79.+&#.
3:A3
01-."2(?//,+,"*
• The location of the peak can be converted to phase via
– //5<9
I = – 0 S § ----------------· = /5/!8#": = <.5!0q
© <<5<9 ¹
(2.61)
Summary of Phasor Addition
• When we need to form the sum of sinusoids at the same frequency, we obtain the final amplitude A and phase I via
( = ( / + ( 0 + } + ( - = #$
where ( , = # , $
%I ,
%I
(2.62)
and
-
&" =
¦ #, %&' Z! " + I, ,=/
(2.63)
= # %&' Z ! " + I Example:
­
& " = *+ ® .$
¯
S
% § 0S) ! " + ---·
©
0¹
+ 6$
S
% § 0S) ! " – ---·
©
1¹
+ . + %0 $
%0S) ! " ½
¾
¿
• Find ( = ( / + ( 0 + ( .
• From the given & " we observe that
S
% --0
( / = .$ ( 0 = 6$
)!)(3456(7,8*-%.(-*/(79.+&#.
S
– % --1
( . = . + %0
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
S
( / = . %&' --- + %. '() --- = %.
0
0
S
S
( 0 = 6 %&' § – ---· + %6 '() § – ---· = .56.66 – %.56.66
© 1¹
© 1¹
( . = . + %0
• Now,
( = %. + .56.66 – %.56.66 + . + %0 = <56.66 + %/51<16 • For use in the phasor sum formula we likely need the answer
in polar form
( =
0
0
<56.66 + /51<16 ‘ "=") /51<16
---------------<56.66
= <5<;9< ‘!500!1 = <5<;9<$
%!500!1
Physics of the Tuning Fork
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 = / -------01"
(2.65)
• To balance the two forces (sum is zero), we must have
0
1 &
/ -------0- = – ,& " 1"
(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
11-------------& " - = --- – Z ! '() Z ! " 0
1"
1"
0
= – Z ! %&' Z ! "
• We now plug this result into (2.66) to obtain
0
1 &
/ -------0- = – ,& " 1"
(2.67)
0
– / Z ! %&' Z ! " = – , %&' Z ! "
which tells us that we must have
0
– /Z ! = – ,
(2.68)
,Z ! = r --/
(2.69)
so it must be that
)!)(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 MATLAB 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
Amplitude
1
0.5
0
−0.5
−1
0
2
H&&$7IJKJ;@:L7'
1
Amplitude
1
3
4
5
6
Time in seconds
7
8
9
10
0.5
0
−0.5
−1
4
4.002 4.004 4.006 4.008 4.01 4.012 4.014 4.016 4.018
Time in seconds
)!)(3456(7,8*-%.(-*/(79.+&#.
4.02
3:A=
019.,B.("C(+1&(DE*,*8(F"2G
• How pure is the signal produced by the tuning fork?
• In Chapter 3 of the text we begin a study of spectrum representation
• The zoom of the captured signal looks like a single sinusoid,
but spectral analysis can be more revealing
• Consider the use of MATLAB’s power spectral density function )%=AQ8G++*8+%1U)B (text Chapter 13)
XX#)%=AQ89]?98H:::B
40
JJ@7MN78./O4$"/)457)./(/378&,G7P()0F
Power Spectrum Magnitude (dB)
20
AA@7MN7'"0&/O7F4,$&/(0
0
Q)F",7F4,$&/(0'
−20
−40
−60
R."7)&7')4,)B.P7),4/'("/)
−80
0
500
1000
)!)(3456(7,8*-%.(-*/(79.+&#.
1500
2000
2500
Frequency (Hz)
3000
3500
4000
3:A>
019.,B.("C(+1&(DE*,*8(F"2G
• A time-frequency plot can be obtained using the MATLAB’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%-.
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 multiple
sinusoids model is only an approximation
• Modeling still has great value in system design
)!)(3456(7,8*-%.(-*/(79.+&#.
3:;6