ME 322: Instrumentation
Lecture 21
March 9, 2015
Professor Miles Greiner
Spectral analysis, Min, max and resolution
frequencies, Aliasing,
Announcements/Reminders
• HW 8 Due Friday
– Then Spring Break!
• This week in lab:
– Lab 7 Boiling Water Temperature in Reno
• Please fully participate in each lab and
complete the Lab Preparation Problems
– For the final you will repeat one of the last 3 labs,
solo, including performing the measurements, and
writing Excel, LabVIEW and PowerPoint.
A/D Converters
• Can be used to measure a long series of very rapidly
changing voltage
• Useful for measuring time-dependent voltage
signals and assessment of their dynamic properties
– Rates of Change (derivatives) and
– Frequency Content (Spectral Analysis)
• What can go wrong?
– Last time we showed that small random errors (RF
noise, IRE) can strongly affect calculation of derivatives
• So: Make derivative time-step long enough so that the real
signal changes by a larger amount than the random noise.
• What is Frequency Content?
Spectral Analysis
• Evaluates energy content associated with different
frequency components within a signal
• Use to evaluate
– Tonal Content (music)
• You hear notes, not time-varying pressure
– Dominant or natural frequencies
• car vibration, beam or bell ringing
– System response (Vibration Analysis)
• Resonance
• Spectral analysis transforms a signal from the
time-domain V(t) to the frequency-domain,
VRMS(f)
– What does this mean?
Fourier Transform
n=1
n=0
n=2
sine
V
cosine
0
t
T1
• Any function V(t), over interval 0 < t < T1, may be decomposed into
an infinite sum of sine and cosine waves
– 𝑉 𝑡 =
∞
𝑛=0
𝑛
𝑎𝑛 𝑐𝑜𝑠 2𝜋𝑓𝑛 𝑡 + 𝑏𝑛 𝑠𝑖𝑛 2𝜋𝑓𝑛 𝑡 , 𝑓𝑛 = 𝑇
1
– Only modes with an integer number of oscillations over the total sampling time T1 are
used.
𝑛
– Discrete (not continuous) frequencies: 𝑓𝑛 = 𝑇 , n = 0, 1, 2, … ∞ (integers)
1
– The coefficient’s an and bn quantify the relative importance (energy content) and phase of
each mode (wave).
– The root-mean-square (RMS) coefficient 𝑉𝑟𝑚𝑠
𝑛
=
(𝑎𝑛 2 + 𝑏𝑛 2 ) 2 for each mode
frequency 𝑓𝑛 quantifies its total energy content (both sine and cosine waves)
Examples (ME 322 Labs)
Frequency Domain
Time Domain
Function Generator
100 Hz sine wave
0.14
0.5
0.12
t1 = 1.14 sec, a1 = 0.314 g
0.3
0.1
t2 = 5.88 sec, a2 = 0.152 g
arms [g's]
Damped Vibrating
Cantilever Beam
Dimensinoless Acceleration, g
0.4
0.2
0.08
0.1
0.06
0
-0.1
0.04
-0.2
0.02
-0.3
-0.4
0
-0.5
0
0
2
4
6
8
10
10
20
30
f [Hz]
40
50
60
Time t [sec]
Unsteady air Speed
Downstream from
a Cylinder in Cross
Flow
• Real signal may have a narrow or wide spectrum of
energetic modes
What is the lowest Frequency mode that can be
observed during measurement time T1
• Example
– If we measure outdoor temperature for one hour, can we
observe variations that require a day to repeat?
• The lowest (finite) observable frequency is f1 = 1/T1
• The only other frequencies that can be detected are
– 𝑓𝑛 =
𝑛
𝑇1
= 𝑛𝑓1 (T1=nTn)
• What is the frequency resolution?
– Smallest change in frequency that can be detected
– 𝑓𝑛+1 − 𝑓𝑛 = (𝑛 + 1)𝑓1 − 𝑛𝑓1 = 𝑓1
• Increasing the total sampling time T1 reduces the lowest
detectable frequency and improves frequency
resolution,
Sampling Rate Theory
• What discrete sampling rate fS must be used to
accurately observe a sinusoidal signal of frequency
fM?
• Must be greater than fM, but much how larger?
Lab 8 Aliasing Spreadsheet Example
• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH322Instrumentati
on/Labs/Lab%2008%20Unsteady%20Voltage/Lab8Index.htm
• Measured sine wave, fm = 10 Hz
– V(t) = (1volt)sin[2p(10Hz)(t+tshift)]
• Total sampling time, T1 = 1 sec
• How many peaks to you expect to observe in
one second?
• How large does the sampling rate fS need to be
to capture this many peaks?
How to predict indicated (or Alias) Frequency?
𝑓𝑆
2
Maximum frequency that can
be accurately measured using
sampling frequency fS .
𝑓𝑁 =
• 3 Frequencies:
– fm being measured; fs Sampling frequency; fa indicated frequency
• fa = fm if fs > 2fm
• Otherwise using folding chart on page 106
– Let fN = fs/2 be the maximum frequency that can be accurately observed
using sampling frequency fs.
Problem 5.26 (p. 127)
• A 1-kHz sine wave signal is sampled at 1.5 kHz.
What would be the lowest expected alias frequency?
• ID: Is fs > 2fm ?
𝑓𝑆
𝑓𝑁 =
2
𝒇𝒎 /𝒇𝑵
𝒇𝒂 /𝒇𝑵
A more practical example
• Using a sampling frequency of 48,000 Hz, a peak
in the spectral plot is observed at 18,000 Hz.
• What are the lowest 4 values of fm that can cause
this?
• ID: what is known? fs and fa
18,000
Upper and Lower Frequency Limits
• If a signal is sampled at a rate of fS for a total time of
T1 what are the highest and lowest frequencies that
can be accurately detected?
– (f1 = 1/T1) < f < (fN = fS/2)
• To reduce lowest frequency (and increase frequency
resolution), increase total sampling time T1
• To observe higher frequencies, increase the
sampling rate fS.
How to find 𝑉𝑟𝑚𝑠 vs fn?
• For 𝑉 𝑡 =
– 𝑎𝑛 =
– 𝑏𝑛 =
–
𝑉𝑟𝑚𝑠
2
𝑇1
2
𝑇1
𝑛
∞
𝑛=0
𝑇1
𝑉
0
𝑇1
𝑉
0
=
𝑎𝑛 𝑐𝑜𝑠 2𝜋𝑓𝑛 𝑡 + 𝑏𝑛 𝑠𝑖𝑛 2𝜋𝑓𝑛 𝑡
𝑡 𝑐𝑜𝑠
𝑡 𝑠𝑖𝑛
𝑛
2𝜋 𝑡
𝑇1
𝑛
2𝜋 𝑡
𝑇1
𝑑𝑡 (cosine transform)
𝑑𝑡 (sine transform)
(𝑎𝑛 2 + 𝑏𝑛 2 ) 2
• How to evaluate these integrals?
– For simple V(t), in closed form
– For complex or discretely-sampled signals
• Numerically (trapezoid or other methods)
• Appendix A, pp 450-2
• LabVIEW Spectral Measurement VI does this
• for (f1 = 1/T1) < (fn =
𝑛
𝑇1
) < (fN = fS/2)
Fourier Transfer Example
• Lab 8 site:
– http://wolfweb.unr.edu/homepage/greiner/teaching/MECH322Instrume
ntation/Labs/Lab%2008%20Unsteady%20Voltage/Lab8Index.htm
• Dependence of coefficient b (sine transform)
on weigh function frequency and phase shift
• Dependence of Vrms on weight function
frequency, but not phase shift.
Lab 8: Time Varying Voltage Signals
Digital
Scope
Function
Generator
fM = 100 Hz
VPP = ±1 to ± 4 V
Sine wave
Triangle wave
NI
myDAQ
fS = 100 or 48,000 Hz
Total Sampling time
T1 = 0.04 sec
4 cycles
192,000 samples
• Produce sine and triangle waves with fm = 100 Hz, VPP = ±1-4 V
– Sample both at fS = 48,000 Hz and numerically differentiate with two
different differentiation time steps
• Evaluate Spectral Content of sine wave at four different sampling
frequencies fS = 5000, 300, 150 and 70 Hz (note: some < 2 fm )
• Sample singles between 10,000 Hz < fM < 100,000 Hz using fS =
48,000 Hz (fa compare to folding chart)
Estimate Maximum Slope
VPP
VPP
P
P
• Triangle Wave
• Sine wave
𝑉𝑃𝑃
𝑠𝑖𝑛 2𝜋𝑓𝑡 + φ
2
𝑉
2𝜋𝑓 𝑃𝑃 𝑐𝑜𝑠 2𝜋𝑓𝑡 +
2
– 𝑉 𝑡 =
–
–
𝑑𝑉
=
𝑑𝑡
𝑑𝑉
𝑑𝑡 𝑚𝑎𝑥
= 𝜋𝑓𝑉𝑃𝑃
–
φ
𝑑𝑉
𝑑𝑡 𝑚𝑎𝑥
=
𝑉𝑝𝑝
𝑃
2
= 2𝑉𝑝𝑝 𝑓
Fig. 3 Sine Wave and Derivative Based on
Different
Time
Steps
0.8
800
V(t)
0.6
600
dV/dtIdeal,Max
0.4
V [Volts]
0.2
200
dV/dtm=1
0
0
-0.2
-200
-0.4
dV/dtm=10
-0.6
-400
dV/dtIdeal,Min
-0.8
-600
-1
0
0.005
dV/dt [Volts/sec]
400
0.01
0.015
-800
0.02
t [sec]
• dV/dt1 (Dt=0.000,020,8 sec) is nosier than dV/dt10 (Dt=0.000,208 sec)
• The maximum slope from the finite difference method is slightly
larger than the ideal value. This may be because the actual wave was
not a pure sinusoidal.
Fig. 4 Sawtooth Wave and Derivative Based on
Different Time Steps
1
500
400
V(t)
dV/dtIdeal,Max
0.6
0.4
dV/dtm=1
V [Volts]
0.2
300
200
dV/dtm=10
100
0
0
-0.2
-100
-0.4
-200
-0.6
dV/dt [Volts/sec]
0.8
-300
-0.8
dV/dtIdeal,Min
-1
-400
-1.2
-500
0.02
0
0.005
0.01
t [sec]
0.015
• dV/dt1 is again nosier than dV/dt10
• dV/dt1 responds to the step change in slope more accurately
than dV/dt10
• The maximum slope from the finite difference method is larger
than the ideal value.
Fig. 5 Measured Spectral Content of 100 Hz Sine Wave for
Different Sampling Frequencies
0.6
fs = 150 Hz
fs = 70 Hz
0.5
VRMS [Volts]
fs = 300 Hz
fs = 5000 Hz
0.4
0.3
0.2
0.1
0
0
20
40
60
80
100
120
140
160
180
200
frecuency f [Hz]
• The measured peak frequency fP equals the maximum signal
frequency fM = 100 Hz when the sampling frequency fS is greater
than 2fM
• fs = 70 and 150 Hz do not give accurate indications of the peak
frequency.
Table 2 Peak Frequency versus Sampling
Frequency
S am pling F requency, f s [H z]
P eak S pectral F requency, f p [H z]
5000
300
150
70
100
100
50
30
• For fS > 2fM = 200 Hz the measured peak is
close to fM.
• For fS < 2fM the measured peak is close to the
magnitude of fM–fS.
• The results are in agreement with sampling
theory.
Table 3 Signal and Indicated Frequency Data
fm [Hz]
0
9910
19540
23120
30190
40510
47320
50180
61200
71800
72400
79800
89500
95400
99700
fa [Hz]
0
9925
19575
23125
17800
7475
675
2175
13275
23850
23575
16125
6475
475
3725
fm/fN
0.00
0.41
0.81
0.96
1.26
1.69
1.97
2.09
2.55
2.99
3.02
3.33
3.73
3.98
4.15
fa/fN
0.00
0.41
0.82
0.96
0.74
0.31
0.03
0.09
0.55
0.99
0.98
0.67
0.27
0.02
0.16
• This table shows the dimensional and dimensionless signal frequency
fm (measured by scope) and frequency indicated by spectral analysis,
fa.
• For a sampling frequency of fS = 48,000 Hz, the folding frequency is
fN = 24,000 Hz.
Figure 6 Dimensionless Indicated Frequency versus
Signal Frequency
1.00
0.90
0.80
0.70
fa/fN
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
fm/fN
• The characteristics of this plot are similar to those of
the textbook folding plot
• For each indicated frequency fa, there are many
possible signal frequencies, fm.
Figure 2 VI Block Diagram
Statistics
This Express VI produces the following measurements:
Time of Maximum
Convert to Dynamic Data
Double Click the Icon and select “Single Waveform” (it is located at the
bottom of the list)
Convert from Dynamic Data
Double Click the Icon and select “1D array of scalars – single waveform”
Figure 1 VI Front Panel
Lab 8 Sample Data
• http://wolfweb.unr.edu/homepage/greiner/teac
hing/MECH322Instrumentation/Labs/Lab%20
08%20Unsteady%20Voltage/Lab8Index.htm
• Calculate Derivatives
• Plot using secondary axes
• Frequency Domain Plot
– Lowest finite frequency f1 = 1/T1
Effect of Random Noise on Differentiation
• Measured voltage has Real and Noise components
– VM = VR+VN
–
𝑑𝑉𝑀
𝑑𝑡
=
𝑉𝑅 −𝑉𝑁 𝑡+ − 𝑉𝑅 −𝑉𝑁 𝑡−
2∆𝑡𝐷
• ∆𝑉𝑅 = 𝑉𝑅𝑡+ − 𝑉𝑅𝑡− =
𝑑𝑉𝑅
2∆𝑡𝐷
𝑑𝑡
• ∆𝑉𝑁 = 𝑉𝑁𝑡+ − 𝑉𝑁𝑡− ≈ 𝑤𝑉
•
•
•
𝑑𝑉𝑀
𝑑𝑡
𝑑𝑉𝑅
𝑑𝑡
𝑊𝑉
=
+
2∆𝑡𝐷
𝑊𝑉
For small ∆𝑡𝐷 ,
2∆𝑡𝐷
𝑑𝑉𝑅
𝑊𝑉
Want
≫
𝑑𝑡
2∆𝑡𝐷
=
∆𝑉𝑅 +∆𝑉𝑁
2∆𝑡𝐷
RF, IRE, other errors,
Random, but does not
increase with ∆𝑡𝐷
is large and random
– wV decreases as FS gets smaller and N increases
– Want ∆𝑡𝐷 to be large enough to avoid random error but small
enough to capture real events
Time Dependent Data
How to find
𝑑𝑇
𝑑𝑡
𝑜𝑟
𝑑𝑉
𝑑𝑡
from 𝑇 𝑡 or 𝑉 𝑡
1st order numerical differentiation (center difference)
𝑑𝑉
𝑙𝑖𝑚 𝑉 𝑡 + ∆𝑡𝐷 − 𝑉 𝑡 − ∆𝑡𝐷
𝑡 =
𝑑𝑡
∆𝑡 → 0 𝑡 + ∆𝑡𝐷 + 𝑡 − ∆𝑡𝐷
∆𝑡𝐷 = differentiation time
i
T = (∆ts)i
0
0
1
∆ts
2
2∆ts
V
∆𝑡𝑠 ≡ sampling time
∆𝑡𝐷 = 𝑚∆𝑡𝑠
m = 1, 2, 3, …
𝒅𝑽
𝒅𝒕