Spfirst l01 Ev
Transcript of Spfirst l01 Ev
-
7/21/2019 Spfirst l01 Ev
1/20
Copyright Monash University 2009
Signal
ProcessingFirst
Lecture1
Sinusoids
-
7/21/2019 Spfirst l01 Ev
2/20
Copyright Monash University 2009
READING ASSIGNMENTS ThisLecture:
Chapter2,
pp.
917
AppendixA:
Complex
Numbers
AppendixB:MATLAB
Chapter1:Introduction
2
-
7/21/2019 Spfirst l01 Ev
3/20
Copyright Monash University 2009
CONVERGING FIELDS
3
EECmpE
Math
Applications
Physics
Computer
Science
-
7/21/2019 Spfirst l01 Ev
4/20
Copyright Monash University 2009
COURSE OBJECTIVE Studentswillbeableto:
Understandmathematical descriptions
of
signalprocessingalgorithms andexpress
those
algorithms
as
computer
implementations (MATLAB)
Whatareyourobjectives?
4
-
7/21/2019 Spfirst l01 Ev
5/20
Copyright Monash University 2009
WHY USE DSP ? Mathematicalabstractions leadto
generalizationand
discovery
of
new
processingtechniques
Computerimplementationsareflexible
Applicationsprovideaphysical context
5
-
7/21/2019 Spfirst l01 Ev
6/20
Copyright Monash University 2009
Fourier Everywhere Telecommunications
Sound&Music
CDROM,Digital
Video
FourierOptics
XrayCrystallography
ProteinStructure
&
DNA
ComputerizedTomography
MagneticResonanceImaging:MRI
Radioastronomy MechanicalEngineering Ref:Prestini,TheEvolutionofAppliedHarmonicAnalysis
6
-
7/21/2019 Spfirst l01 Ev
7/20
Copyright Monash University 2009
LECTURE OBJECTIVES Writegeneralformulaforasinusoidal
waveform,or
signal
Fromtheformula,plotthesinusoidversus
time
Whatsasignal?
Itsafunction of
time,
x(t)
or
s(t)
inthemathematicalsense
7
-
7/21/2019 Spfirst l01 Ev
8/20
Copyright Monash University 2009
TUNING FORK EXAMPLE CDROMdemo
Ais
at
440
Hertz
(Hz)
WaveformisaSINUSOIDALSIGNAL
Computerplot
looks
like
asine
wave
Thisshouldbethemathematicalformula:
8
))440(2cos( tA
-
7/21/2019 Spfirst l01 Ev
9/20
Copyright Monash University 2009
TUNING FORK A-440 Waveform
9
ms3.2
85.515.8
T
Hz435
3.2/1000
/1
Tf
Time (sec)
-
7/21/2019 Spfirst l01 Ev
10/20
Copyright Monash University 2009
SPEECH EXAMPLE Morecomplicatedsignal(BAT.WAV)
Waveformx(t) is
NOT
aSinusoid
Theorywilltellus
x(t) is
approximately
asum
of
sinusoids
FOURIERANALYSIS
Breakx(t) intoitssinusoidalcomponents
Calledthe
FREQUENCY
SPECTRUM
10
-
7/21/2019 Spfirst l01 Ev
11/20
Copyright Monash University 2009
Speech Signal: BAT
Nearly
Periodic in
Vowel
RegionPeriodis(Approximately)T=0.0065sec
11
-
7/21/2019 Spfirst l01 Ev
12/20
Copyright Monash University 2009
DIGITIZE the WAVEFORM x[n] isaSAMPLEDSINUSOID
Alist
of
numbers
stored
in
memory
Sampleat11,025samplespersecond
Called
the
SAMPLING
RATE
of
the
A/DTimebetweensamplesis
1/11025=90.7microsec
Outputvia
D/A
hardware
(at
Fsamp)
12
-
7/21/2019 Spfirst l01 Ev
13/20
Copyright Monash University 2009
STORING DIGITAL SOUND x[n] isaSAMPLEDSINUSOID
Alist
of
numbers
stored
in
memory
CDrateis44,100samplespersecond
16bit
samples
Stereouses2channels
Numberofbytesfor1minuteis
2X(16/8)X60X44100=10.584Mbytes
13
-
7/21/2019 Spfirst l01 Ev
14/20
Copyright Monash University 2009
SINES and COSINES AlwaysusetheCOSINEFORM
Sineisaspecialcase:
14
))440(2cos( tA
)cos()sin(2
tt
-
7/21/2019 Spfirst l01 Ev
15/20
Copyright Monash University 2009
SINUSOIDAL SIGNAL
FREQUENCY
Radians/sec
Hertz(cycles/sec)
PERIOD (in
sec)
AMPLITUDE
Magnitude
PHASE
15
A tcos( )
( )2 f
Tf
1 2
-
7/21/2019 Spfirst l01 Ev
16/20
Copyright Monash University 2009
EXAMPLE of SINUSOID GiventheFormula
Makeaplot
16
)2.13.0cos(5 t
-
7/21/2019 Spfirst l01 Ev
17/20
Copyright Monash University 2009
PLOT COSINE SIGNAL
FormuladefinesA,,and
17
5 0 3 12cos( . . ) t
A 5
0.3
1.2
-
7/21/2019 Spfirst l01 Ev
18/20
Copyright Monash University 2009
PLOTTING COSINE SIGNAL from
the FORMULA
Determine period:
Determineapeak locationbysolving
Zero crossingisT/4beforeorafter
Positive&NegativepeaksspacedbyT/218
)2.13.0cos(5 t
3/203.0/2/2 T
0)2.13.0(0)( tt
-
7/21/2019 Spfirst l01 Ev
19/20
Copyright Monash University 2009
PLOT the SINUSOID
UseT=20/3andthepeaklocationatt=4
19
)2.13.0cos(5 t
320
-
7/21/2019 Spfirst l01 Ev
20/20
Copyright Monash University 2009
TRIG FUNCTIONS CircularFunctions
CommonValues
sin(k)=0
cos(0)=1
cos(2k)=1andcos((2k+1) )=1
cos((k+0.5)
)=0
20