mece 3320 – measurements & instrumentation static...

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 Classification contd.

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


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


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)


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


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.


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.


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

mece 3320 – measurements & instrumentation static...

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