static and dynamic characteristics of signal -...
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Static and Dynamic Characteristics of Signal
2141-375 Measurement and Instrumentation
Input-Output Signal Concepts
Engineering tasks in the measurement of physical vari ables(1) Selection of a measurement system(2) Interpreting the output from a measurement system.
Static signal: rangeTime-varying signal: magnitude and the rate of chang e of a variable
Selection of Measurement system based on input sign al range
Signal Classification• Signal : the physical information about a measured variabl e
Sig
nal
Time
Sig
nal
Time
Continuous time, continuous valueDefine for each instant of time and its amplitude may vary continuously with time and assume any value
• Analog signal
Discrete time, continuous valueDefine at discrete instants of time and its amplitude may vary continuously with time and assume any value
• Discrete time signal
Signal Classification
Discrete time, discrete valueDefine at discrete instants of time and its amplitude may assume discrete value
• Digital signal
Time
Sig
nal
Signal Classification: waveform
•Static signal: does not vary in time (or change very slowly).
•Dynamic signal: is defined as a time-dependent signal
•Deterministic signal: varies in time in a predictable manner.
•Periodic signal: the variation of magnitude of the signal repeats at regular intervals
•Non-Periodic signal: does not repeat at regular interval.
•Non-Deterministic signal: cannot be prescribed before it occurs.
signal
Static Dynamic
Determinitic Non-Determinitic
Periodic Non-Periodic
Signal Classification: waveform
Periodic Signal
Mathematic of periodic signal 3,2,1for ),()( =+= nnTtxtx
Where T is the period of the signal x(t)
t (sec)
0
1
2
-1
-20 0.2 0.4 0.6 0.8 1.0 1.2 1.4
x(t)
t (sec)
0.5
1.0
1.5
0
-0.50 0.2 0.4 0.6 0.8 1.0 1.2 1.4
x(t)
t (sec)
0.5
1.0
1.5
0
-0.50 0.2 0.4 0.6 0.8 1.0 1.2 1.4
x(t)
t (sec)
0
1
2
-1
-20 0.2 0.4 0.6 0.8 1.0 1.2 1.4
x(t)
)sin()( φω += tAtx
0
1
2
-1
-20 0.2 0.4 0.6 0.8 1.0 1.2 1.4
t (sec)
Generic sinusoidal signal:
Where A is the amplitude, ω ω ω ω is the radian frequency and φφφφ is the phase
φφφφ = 0
φφφφ = ππππ/3
Sinusoidal Signal
Signal Analysis
• Average or mean value during t 2 – t1
∫
∫≡
2
1
2
1
)(t
t
t
t
dt
dttyy
0t1 t2
t
y
Mean value
0t1 t2
t
y
Average value(dc offset)
Amplitude of theac component
• root-mean-square value during t 2 – t1
∫−≡
2
1
2
12
)(1 t
trms dttytt
y
∑=
=N
iiy
Ny
1
1
∑=
=N
iirms y
Ny
1
21
Analog
Analog
Discrete or digital
Discrete or digital
RMS: Root-Mean-Square
RMS is a measure of a signal’s average power. Insta ntaneous power delivered to a resistor is P= [v(t)]2/R.
[ ] ( )R
Vdttv
TRP rms
Tt
tavg
220
0
)(11
=
= ∫+
[ ]∫+
=Tt
trms dttvT
V0
0
2)(1
An AC voltage with a given RMS value has the same h eating (power) effect as an DC voltage with the same value.
All the following voltage waveforms have the same R MS value, and should indicate 1 V ac on an rms meter
Waveform Sine Triangle Square DCVpeak 1.414 1.733 1 1Vrms 1 1 1 1
0 0.01 0.02 0.03 0.04t HsecL
0
0.5
1
1.5
2
2.5
3
xHtLH
tloV
LSignal Analysis
Ex Given a voltage signal as follows, the ac component has the frequency of 50 Hz, determine the average and rms values of signa l that will be indicated by the meter when (a) S1 is open and (b) S1 is closed
Known signal characteristics
Find average and rms values
T=20 ms
Averageor True
rmsmeter
C
S1
x(t)
System
Signal are always associated with one or more syste ms. A system transforms the input into an output of the system.
SystemInput Output
x(t) y(t)ℑℑℑℑ
Input Output
x(t) y(t)
Block diagram of a system
ℑℑℑℑ is an mathematical operator
Linear, time invariant (LTI) system
LTISystem
Input Output
x(t) y(t)
A linear system is one which is both homogeneous (s cale) and additive
System
SystemInput Output
mx(t) my(t)
Homogeneous system
A homogeneous system is one which a scaled input pr oduces an equally scaled output.
An additive system is one for which
SystemInput Output
x1(t) y1(t)
SystemInput Output
x2(t) y2(t)System
Input Output
x1(t) + x2(t) y1(t) + y2(t)
System
A time invariant system is one for which a delay in the application of the input signal results in the same delay in the output.
SystemInput Output
x1(t) y1(t)
SystemInput Output
x1(t-ττττ) y1(t-ττττ)
Time and Frequency Domain
A signal in frequency domain shows “How much” of a s ignal is associated with a certain frequency.
0 500 1000 1500 2000FrequencyHHzL
0
0.5
1
1.5
2
edutilpm
A
0 0.001 0.002 0.003 0.004tHsecL
-2
-1
0
1
2
xHtL
Time and frequency domain representation of the sig nal.
)2000sin(2)( ttx π=
0 200 400 600 800 1000FrequencyHHzL
0
0.5
1
1.5
2
edutilpm
A
Comparing signals in the time and frequency domain
)1000cos()600sin(21)( tttx ππ ++=
0 0.005 0.01 0.015 0.02tHsecL
-2
-1
0
1
2
3
4
xHtL
Time and Frequency Domain
Fourier Series of Complex Periodic Signal
Where the fundamental frequency
( )∑∞
=
++=1
000 sincos)(n
nn tnBtnAAty ωω
T
πω
20 =
Fourier series for y(t):
A0, An, Bn are constants that depend on y(t) and n
∫−
=2/
2/
0 )(1 T
T
dttyT
A
,...3,2,1 ;cos)(2 2/
2/
0 == ∫−
ntdtntyT
AT
T
n ω ,...3,2,1 ;sin)(2 2/
2/
0 == ∫−
ntdtntyT
BT
T
n ω
y(t)
t
T
Useful Trigonometric Relations
( ) ( )
( ) ( )
( ) ( )
( ) ( )ABBABA
BABABA
BABABA
BABABA
−+=−
−+=+
−+=−
−+=+
2
1sin
2
1sin2coscos
2
1cos
2
1cos2coscos
2
1sin
2
1cos2sinsin
2
1cos
2
1sin2sinsin
( ) ( )[ ]
( ) ( )[ ]
( ) ( )[ ]BABABA
BABABA
BABABA
++−=
++−=
+−−=
sinsin2
1cossin
coscos2
1coscos
coscos2
1sinsin
( )( ) BABABA
BABABA
sinsincoscoscos
sincoscossinsin
m=±
±=±
Useful Trigonometric Relations
Orthogonal principles
02cos2sin1
01
2/1
0
2cos2cos1
00
2/1
0
2sin2sin1
0
0
0
=
==
=
≠
=
==
=
≠
=
∫
∫
∫
dtT
tn
T
tm
T
nm
nm
nm
dtT
tn
T
tm
T
nm
nm
nm
dtT
tn
T
tm
T
T
T
T
ππ
ππ
ππ
Fourier Series of Periodic Signal
A function f(t) is even if it is symmetric about the vertical axis
)()( tftf −=
A function f(t) is odd if it is symmetric about the origin
)()( tftf −−=
Cosine is even function and Sine is odd function
Fourier Cosine Series ( )∑∞
=
+=1
00 cos)(n
n tnAAty ω
Fourier Sine Series ( )∑∞
=
+=1
00 sin)(n
n tnBAty ω
Fourier Series of Periodic Signal
Ex Determine the Fourier series that represents the fu nction as shown below
... 10
10sin
5
4
10
6sin
3
4
10
2sin
4)( +++= tttty
ππ
ππ
ππ
Known T = 10 and A 0 = 0 (Symmetrical)
Find Fourier coefficient An and Bn
∑∞
=
=1
2sin)(
nn T
tnBty
π
Solution Since the function is odd, the Fourier series will contain only sine terms
The resulting Fourier Series is then
-6 -4 -2 0 2 4 6tHsecL
-1
-0.5
0
0.5
1
yHtL
number odd ;4
== nn
Bn π
Fourier Series of Periodic Signal
-6 -4 -2 0 2 4 6tHsecL
-1
-0.5
0
0.5
1
yHtL
-6 -4 -2 0 2 4 6tHsecL
-1
-0.5
0
0.5
1
yHtL
-6 -4 -2 0 2 4 6tHsecL
-1
-0.5
0
0.5
1
yHtL
-6 -4 -2 0 2 4 6tHsecL
-1
-0.5
0
0.5
1
yHtL
10
10sin
5
4
10
6sin
3
4
10
2sin
4)( tttty
ππ
ππ
ππ
++=
tty10
2sin
4)(
ππ
= ttty10
6sin
3
4
10
2sin
4)(
ππ
ππ
+=
10
14sin
7
4
10
10sin
5
4
10
6sin
3
4
10
2sin
4)( ttttty
ππ
ππ
ππ
ππ
+++=
Fourier Series of Periodic Signal
-6 -4 -2 0 2 4 6tHsecL
-1
-0.5
0
0.5
1
yHtL
0 0.5 1 1.5 2Frequency HHzL
0
0.2
0.4
0.6
0.8
1
1.2
edutilpm
A
0 5 10 15 20n
0
0.2
0.4
0.6
0.8
1
1.2
edutilpm
Aro
nB
Fourier Series of Periodic Signal
Ex find the frequency content of the output voltage fr om a full-wave rectifier, If the ac input signal is given by
Known The full-wave signal can be expressed
Find The frequency content of this signal
Solution Since the function is even, the Fourier series will contain only cos terms
ttE π120sin120)( =
tty π120sin120)( =
Fourier Series of Periodic Signal
tnn
tyn
602cos1
41201202)(
2π
ππ ∑ −+
×=
∑∞
=
+=1
0
2cos)(
nn T
tnAAty
π
The resulting Fourier Series is then
π1202
0
×=A
numbereven ;1
2
1
2120=
+
+−
−= n
nnAn π
numbereven =n
f(t)
tTT/20
1
-1
-T/2-T
<<−
<<=
02
1-
201
)(t
T
T t
tf
number odd 2
sin14
)(1
=
= ∑∞
=
ntT
n
ntf
n
ππ
f(t)
tTT/20
1
-T/2-T
<<−−
<<==
02
22
02
2)(
tT
tT
T t t
TtT
tf
number odd 2
cos14
2
1)(
122
=
−= ∑∞
=
ntT
n
ntf
n
ππ
f(t)
TT/20
1
-T/2-Tt
<<−−
<<==
02
2sin
20
2sin2
sin)(t
Tt
T
T t t
TtT
tf π
ππ
( ) numbereven 2
cos1
142)(
22
=
−+= ∑
∞
=
ntT
n
ntf
n
πππ
number odd 2
cossin2
2
1)(
1
=
+= ∑∞
=
ntT
n
n
ntf
n
πππ
f(t)
tTT/20
1
-T/2-T
<<−
<<=
02
0
20
2sin
)(t
T
T t t
Ttf
π
( )numbereven
2
cos1
122sin
2
11)(
22
=
−++= ∑
∞
=
n
tT
n
nt
Ttf
n
ππ
ππ
f(t)
tTT/20
1
-T/2-T
f(t)
TT/20
1
-T/2-Tt
<<−
<<=
02
0
201
)(t
T
T t
tf
<<−<<−
<<−=
24 and
420
441
)(T
tTT
tT
T t
T
tf
number odd 2
sin12
2
1)(
1
=
+= ∑∞
=
ntT
n
ntf
n
ππ
Fourier Transform of Complex Signal
Definition of Fourier Transform of y(t)
dtetyfY fti π2 )()( −∞
∞−∫=
Definition of Inverse Fourier Transform of Y(f)
dfefYty fti π2 )()( ∫∞
∞−=
The Fourier transform is a mathematical function th at transforms a signal from the time domain, y(t) to the frequency domain, Y(f).
)(ty )( fYℑ
Time Domain Frequency Domain
Fourier Transform
Ex Determine the Fourier transform, consider the one-s ided exponential signal.
Find Fourier Transform of x(t)
Solution
Known x(t) = e-7tu(t)
∫∞ −−=ℑ=
0
27)]([)( dteetxfX ftjt π
fjetu t
π27
1])([ 7
+=ℑ −
)()( 7 tuetx t−=
Impulse Function: Review
<<−
=elsewhere0;
2/2/;/1)(
εεεδε
tt
1)( =∫+∞
∞−
dttεδ
The function may be considered as a rectangular pul se of width εεεε and height 1/ εεεε. In the limit εεεε →→→→ 0, the height increase in such a way that the are a is 1
Which implies
tεεεε/2−ε−ε−ε−ε/2
1111////εεεε
δδδδ(t)
Visualization symbol
t
1111
δδδδ(t)
)()( lim0
tt εε
δδ→
=
This leads to the definition
Unit impulse function or Dirac delta function
Impulse Function: Review
dfett ttfj∫+∞
∞−
−=− )(20
0)( πδ
)()()( 00 tfdttttf =−∫+∞
∞−
δ
=∞
=−elsewhere0;
;)( 0
0
ttttδ
t
δδδδ(t)
t0
δδδδ(t-t0)
dfet ftj∫+∞
∞−
= πδ 2)(
Shifting of function by t 0
dteff tffj∫+∞
∞−
−−=− )(20
0)( πδ
dtef ftj∫+∞
∞−
−= πδ 2)(
1)( 2 =∫+∞
∞−
− dtet ftj πδ
Discrete Fourier Transform
Discrete Fourier Transform (DFT) of yr
∑=
−=N
r
Nrkik etry
NfY
1
/2)(2
)( πδ 2,...,2,1
Nk =
)( tryyr δ= Nr ,...,2,1=
where fkfk δ= and N
f
tNf s==
δδ
1
Example: Estimate the amplitude spectrum or frequency content of the discrete data taken from y(t) = 10 sin 2ππππt using a time increment of 0.125 s for the duration of 1 s.
Known: δδδδt = 0.125 s or fs = 8 Hz
Solution:
Discrete Fourier Transform
Discrete Data Set for y(t) = 10 sin2ππππt
Discrete Fourier Transform
r y(r δt )1 7.0712 10.0003 7.0714 0.0005 -7.0716 -10.0007 -7.0718 0.000
0 1 2 3 4Frequency HHzL
0
5
10
15
20
edutilpm
AHV
L
k f k (Hz) Y (f k ) |Y (f k )|
1 1 -10i 102 2 0 03 3 0 04 4 0 0
Discrete Fourier Transform of y(t)
Hz 18
8===
N
ff sδ
Hz kfkfk == δ