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MECE 3320
MECE 3320 – Measurements & Instrumentation
Static and Dynamic Characteristics of Signals
Dr. Isaac ChoutapalliDepartment of Mechanical Engineering
University of Texas – Pan American
MECE 3320Signal Concepts
A signal is the physical information about a measured variable being transmitted between a process and a measurement system
Characteristics of Signals
Magnitude - generally refers to the maximum value of a signal
Range - difference between maximum and minimum values of asignal
Amplitude - indicative of signal fluctuations relative to themean
Frequency - describes the time variation of a signal
MECE 3320Signal Concepts Contd.
Characteristics of Signals Contd.
Dynamic - signal is time varying
Static - signal does not change over the time period of interest
Deterministic - signal can be described by an equation (otherthan a Fourier series or integral approximation)
Non-deterministic - describes a signal which has no discerniblepattern of repetition and cannot be described by a simple equation.
MECE 3320Signal Concepts Contd.
Signal characteristics to be considered in a measurement system
Range Temporal variation
MECE 3320Signal Classification
Analog Signal(continuous in time)
Discrete Signal(information available at discrete points in time)
MECE 3320Signal Classification contd.
Quantization: Assigning a single value to a range of magnitudes of a continuoussignal. E.g. a digital watch displays a single numerical value for the entireduration of 1 min until it is updated at the next discrete time step.
MECE 3320Signal Analysis
2
1
2
1
)(
t
t
t
t
dt
dttyyvalueMean
2
1
2
12
1 t
trms dty
ttyvalueRMS
Root Mean Square (RMS) value is an indication of the amount of variation in the dynamic portion of the signal.
N
iiy
NyvalueMean
1
1
2
1
21 t
tirms dty
NyvalueRMS
Continuous signal Discrete signal
MECE 3320Signal Analysis
tty 6cos230)( Consider the function
2
1
2
12 1
2 1
2 1 2 12 1
1
1 2 = 30 sin 66
1 2 = 30 sin 6 sin 66
30 2cos6
t
t
t
ty
t t
t tt t
t t t tt t
t dt
The mean is given by:
The RMS is given by:
yt t
t dt
t tt t t t t
rmst
t
t
t
1 30 2 6
1 900 1206
6 4 112
6 6 12
2 1
2
12
2 1
12
1
2
1
2
cos
sin sin cos
MECE 3320 Effect of Time Period on Mean Value for Non-Deterministic Signal
As the averaging time period becomes long relative to signal period, the resultingvalues will accurately represent the signal.
Non-deterministic signals have a range of frequencies and no single averagingperiod will produce an exact representation of the signal. In such a case, the signalshould be averaged for a period longer than the longest time period contained withinthe signal.
MECE 3320Effect of Subtracting DC Offset for a Dynamic Signal
DC offset may be removed from a dynamic signal to accentuate the characteristicsof the fluctuating component of the signal.
MECE 3320Signal Analysis Contd.
A complex waveform, represented by white light, can be transformed into simplercomponents, represented by the colors in the spectrum.
A very complex signal, even one that is non-deterministic in nature, can beapproximated as an infinite series of sine and cosine function – Fourier Series.
Source: solarsystem.nasa.gov
MECE 3320Signal Amplitude & Frequency
Spring-Mass SystemGoverning Equation: 02
2
kydt
ydm
Solution: mktBtAy /;sincos
The solution can also be expressed as:
2
tan;tan
sincos
*
1*1
22
*
BA
ABBAC
tCtCy
The time period T is given by:f
T 12
Amplitude of oscillation is C Frequency of oscillation is f
MECE 3320Frequency Analysis
Any complex signal can be thought of as made up of sines and cosines of differentamplitudes and time periods, which are added together in an infinite trigonometricseries – Fourier Series.
Modes of vibration for a string plucked at its center
1
2
3
4
MECE 3320Fourier Series and Coefficients
A function y(t) with a period T = 2 / can be represented by a trigonometric series,such that for any t,
2/
2/
2/
2/
2/
2/0
10
sin)(1
cos)(1
)(1
sincos)(
T
Tn
T
Tn
T
T
nnn
dttntyT
B
dttntyT
A
dttyT
A
tnBtnAAty
If y(t) is even function,
1
cos)(n
n tnAty
If y(t) is odd function,
1
sin)(n
n tnBty
MECE 3320Fourier Series Example
0 - -
- -
2
Determine the Euler coefficients of the Fourier series of the function f(x)=x for - x :
1 1= f(x)dx= xdx=02 21 1f(x)cos(nx)dx= xcos(nx) dx
1[ cos( ) sin( )] 0
1 f(x)si
n
n
a
a
xnx nxn n
b
- -
12
1n(nx)dx= xsin(nx) dx
1 2 2[ sin( ) cos( )] cos( ) ( 1)nxnx nx nn n n n
MECE 3320Fourier Series Example
1
1
2( ) ( 1) sin( )
2 12sin( ) sin(2 ) sin(3 ) sin(4 )3 2
n
n
f x nxn
x x x x
3 2 1 0 1 2 34
2
0
2
43.662
3.662
f3 x( )
f5 x( )
f100 x( )
f x( )
3.143.14 x
Three terms
Five terms 100 terms
Real function
MECE 3320Gibbs Phenomenon
The partial sum of a Fourier series shows oscillations near a discontinuitypoint as shown from the previous example. These oscillations do not flatten outeven when the total number of terms used is very large.
As an example, the real function f(x)=x has a value of 3.14 when x=3.14. Onthe other hand, the partial sum solution using 100 terms has a value of 0.318when x=3.14. It is drastically different from the real value.
In general, these oscillations worsen when the number of terms useddecrease. In the previous example, the value of the five terms solution is only0.016 when x=3.14.
Therefore, extreme attention has to be paid when using the Fourier analysison discontinuous functions.
MECE 3320Fourier Transform and Frequency Spectrum
2/
2/
2/
2/
sin)(1
cos)(1
T
Tn
T
Tn
dttntyT
B
dttntyT
A
The Fourier coefficients are given by:
If the period T of the function approaches infinity, the spacing between the frequencycomponents become infinitely small. In such a case, the coefficients An and Bnbecome continuous functions of frequency and can be expressed as:
dtttyB
dtttyA
sin)()(
cos)()(
Source: Cambridge University
MECE 3320Fourier Transform and Frequency Spectrum
dtetyY
ie
dttittyiBAY
ti
i
)()(
sincos gIntroducin
sincos)()()()(
Now, consider a complex number defined by Y().
The above equation can be rewritten as
dtetyfY fti 2)()(
The above equation provides the two-sided Fourier transform of y(t).
The magnitude of Y(f), also called modulus, is given by 22 )(Im)(Re)( fYfYfY
The phase of Y(f) is given by)(Re)(Im
tan)( 1
fYfY
f
Frequencyspectrum
MECE 3320Fourier Transform and Frequency Spectrum
The frequency spectrum is shown below:
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
5 V 0 V 3 V 0 V 1 V1 Hz 2 Hz 3 Hz 4 Hz 5 Hz0 rad 0 rad 0.2 rad 0 rad 0.1 rad
C C C C Cf f f f f
Consider a signal represented by the Fourier series:
( ) 5sin 2 3sin 6 0.2 sin 10 0.1y t t t t
MECE 3320Discrete Fourier Transform
Consider a discrete signal with N data points with y(t) as the source. So, we havey(0), y(1),………, y(N-1).
We know the Fourier transform of y(t) is
Now, each discrete point can be represented by a delayed impulse function, i.e
This implies that integral above is only evaluated at definite points. So, we canrewrite the integral as:
dtetyfY fti 2)()(
)()()( trttytry
tNfififiitN
ftifti eNyeyeyeydtetydtetyfY
)1(2420)1(
0
22 )1(....)2()1()0()()()(
MECE 3320Discrete Fourier Transform
1
0
2)()(N
k
tfkiekyfY Therefore
This can be evaluated for any f. But, we want to evaluate it for the fundamental frequency and its harmonics, i.e. f, 2f, 3f,….., or 1/Nt, 2/Nt, 3/Nt,….., n/Nt Therefore
,....1,0,)()(1
0
2
nekynY
N
k
kNni
In matrix form, it can be represented as
)1(..............
)1()0(
..1..........
11
1..111
)1(..............
)1()0(
)1)(1()1(21
)1(242
12
Ny
yy
WWW
WWWWWW
NY
YY
NNNN
N
N
Ni
eW2
where
MECE 3320Discrete Fourier Transform – Example
Consider a continuous signal:
ttty 4cos3)2
2cos(25)(
dc 1Hz 2Hz
Now, let us sample the signal 4 times, i.e. fs = 4Hz, at t = 0, ¼, ½, ¾. (Put t = kt, k = 0, 1, 2, 3)
MECE 3320Discrete Fourier Transform – Example
So the discrete signal is given by:
tktkky 4cos3)2
2cos(25)(
)3()2()1()0(
111
1111
)3()2()1()0(
963
642
32
yyyy
WWWWWWWWW
YYYY
We can write the Fourier transform as
1
0
21
0
2)()()(
N
k
kniN
k
kNni
ekyekynY
In matrix form, we can rewrite the transform as:
2 wherei
eW
MECE 3320Discrete Fourier Transform – Example
The matrix therefore becomes:
i
i
ii
ii
YYYY
412
420
0848
111111
111111
)3()2()1()0(
The magnitude of the DFT coefficients is shown below:
MECE 3320Inverse Discrete Fourier Transform
The inverse transform of
is
1
0
2)()(N
k
tfkiekyfY
1
0
2
)(1)(N
n
nkN
ienY
Nky
i.e. the inverse matrix is (1/N) times the complex conjugate of the original matrix.
Since the original matrix is symmetrical, the spectrum is symmetrical about N/2.
Therefore F(n) and F(N-n) produce two frequency components (n x 2/t, n ≤ N/2;and n > N/2) one of which is only valid.
The higher of these two frequencies is called the aliasing frequency.
MECE 3320Inverse Discrete Fourier Transform
Because of symmetry, the contribution to y(k) is both from Y(n) and Y(N-n).Therefore
nkN
inkN
ikiknNN
i
N
k
knNN
i
knNN
inkN
i
n
eeee
ekynNYky
enNYenYN
ky
222)(2
1
0
)(2
)(22
But
)()(),(
)()(1)(
Therefore F*(n) = F(N-n) i.e. the complex conjugate.
MECE 3320Inverse Discrete Fourier Transform
Substituting into equation for yn(k),
)]}(arg[)2cos{()(2)(
2sin)}(Im{2cos)}(Re{2)(..
)()(1)(2
*2
nYtktN
nYN
kyor
nkN
nYnkN
nYN
kyei
enYenYN
ky
n
n
nkN
inkN
i
n
This represents a samples sine/cosine wave at a frequency of 1/Nt Hz and amagnitude of 2Y(n)/N.
MECE 3320Interpretation of Example
1. Y(0) = 20 represents a d.c value of Y(0)/N = 20/4 = 5.
2. Y(1) = -4i represents a fundamental component with peak amplitude 2Y(1)/N =2x4/4 = 2 with phase given by arg[Y(1)] = -90 deg, i.e. 2cos(k/2 – 90).
3. Y(2) = 12 (n = N/2) implies a component -3cos(k).
In typical applications, N is generally 1024 or greater.
- Y(n) has 1024 components: 513 to 1023 are complex conjugates of 1 to 511
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