1. ME 322: Instrumentation Lecture 21
March 9, 2015
Professor Miles Greiner
Spectral analysis, Min, max and resolution frequencies, Aliasing,

2. 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.

3. 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?

4. 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?

5. Fourier Transform
n = 0
n = 1
n = 2
sine
cosine V
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 ∞ 𝑎 𝑛 𝑐𝑜𝑠 2𝜋 𝑓 𝑛 𝑡 + 𝑏 𝑛 𝑠𝑖𝑛 2𝜋 𝑓 𝑛 𝑡 , 𝑓 𝑛 = 𝑛 𝑇 1
Only modes with an integer number of oscillations over the total sampling time T1 are used.
Discrete (not continuous) frequencies: 𝑓 𝑛 = 𝑛 𝑇 1 , n = 0, 1, 2, … ∞ (integers)
The coefficient’s an and bn quantify the relative importance (energy content) and phase of each mode (wave).
The root-mean-square (RMS) coefficient 𝑉 𝑟𝑚𝑠 𝑛 = ( 𝑎 𝑛 𝑏 𝑛 2 ) 2 for each mode frequency 𝑓 𝑛 quantifies its total energy content (both sine and cosine waves)

6. a Cylinder in Cross Flow
Examples (ME 322 Labs)
Time Domain
Frequency Domain
Function Generator
100 Hz sine wave
Damped Vibrating
Cantilever Beam
Unsteady air Speed
Downstream from
a Cylinder in Cross Flow
Real signal may have a narrow or wide spectrum of energetic modes

7. 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
𝑓 𝑛 = 𝑛 𝑇 1 =𝑛 𝑓 1 (T1=nTn)
What is the frequency resolution?
Smallest change in frequency that can be detected
𝑓 𝑛+1 − 𝑓 𝑛 =(𝑛+1) 𝑓 1 −𝑛 𝑓 1 = 𝑓 1
Increasing the total sampling time T1 reduces the lowest detectable frequency and improves frequency resolution,

8. 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?

9. Lab 8 Aliasing Spreadsheet Example
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?

10. How to predict indicated (or Alias) Frequency?
𝑓 𝑁 = 𝑓 𝑆 2
Maximum frequency that can
be accurately measured using
sampling frequency fS .
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.

11. 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 ?
𝑓 𝑁 = 𝑓 𝑆 2
𝒇 𝒎 /𝒇 𝑵
𝒇 𝒂 /𝒇 𝑵

12. 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

13. 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.

14. How to find 𝑉 𝑟𝑚𝑠 vs fn?
For 𝑉 𝑡 = 𝑛=0 ∞ 𝑎 𝑛 𝑐𝑜𝑠 2𝜋 𝑓 𝑛 𝑡 + 𝑏 𝑛 𝑠𝑖𝑛 2𝜋 𝑓 𝑛 𝑡
𝑎 𝑛 = 2 𝑇 𝑇 1 𝑉 𝑡 𝑐𝑜𝑠 2𝜋 𝑛 𝑇 1 𝑡 𝑑𝑡 (cosine transform)
𝑏 𝑛 = 2 𝑇 𝑇 1 𝑉 𝑡 𝑠𝑖𝑛 2𝜋 𝑛 𝑇 1 𝑡 𝑑𝑡 (sine transform)
𝑉 𝑟𝑚𝑠 𝑛 = ( 𝑎 𝑛 𝑏 𝑛 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 𝑇 1 ) < (fN = fS/2)

15. Fourier Transfer Example
Lab 8 site:
Dependence of coefficient b (sine transform) on weigh function frequency and phase shift
Dependence of Vrms on weight function frequency, but not phase shift.

16. Lab 8: Time Varying Voltage Signals
Digital Scope
Function Generator
NI myDAQ
fM = 100 Hz
VPP = ±1 to ± 4 V
Sine wave
Triangle wave
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)

17. Estimate Maximum Slope
VPP
VPP P P
Sine wave
𝑉 𝑡 = 𝑉 𝑃𝑃 2 𝑠𝑖𝑛 2𝜋𝑓𝑡+φ
𝑑𝑉 𝑑𝑡 =2𝜋𝑓 𝑉 𝑃𝑃 2 𝑐𝑜𝑠 2𝜋𝑓𝑡+φ
𝑑𝑉 𝑑𝑡 𝑚𝑎𝑥 =𝜋𝑓 𝑉 𝑃𝑃
Triangle Wave
𝑑𝑉 𝑑𝑡 𝑚𝑎𝑥 = 𝑉 𝑝𝑝 𝑃 2 =2 𝑉 𝑝𝑝 𝑓

18. Fig. 3 Sine Wave and Derivative Based on Different Time Steps
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.

19. Fig. 4 Sawtooth Wave and Derivative Based on Different Time Steps
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.

20. Fig. 5 Measured Spectral Content of 100 Hz Sine Wave for Different Sampling Frequencies
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.

21. Table 2 Peak Frequency versus Sampling Frequency
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.

22. Table 3 Signal and Indicated Frequency Data
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.

23. Figure 6 Dimensionless Indicated Frequency versus Signal Frequency
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.

24. Figure 2 VI Block Diagram

25. Figure 1 VI Front Panel

26. Lab 8 Sample Data
Calculate Derivatives
Plot using secondary axes
Frequency Domain Plot
Lowest finite frequency f1 = 1/T1

28. Effect of Random Noise on Differentiation
Measured voltage has Real and Noise components
VM = VR+VN
𝑑 𝑉 𝑀 𝑑𝑡 = 𝑉 𝑅 − 𝑉 𝑁 𝑡+ − 𝑉 𝑅 − 𝑉 𝑁 𝑡− 2∆ 𝑡 𝐷 = ∆ 𝑉 𝑅 +∆ 𝑉 𝑁 2∆ 𝑡 𝐷
∆ 𝑉 𝑅 = 𝑉 𝑅 𝑡+ − 𝑉 𝑅 𝑡− = 𝑑 𝑉 𝑅 𝑑𝑡 2∆ 𝑡 𝐷
∆ 𝑉 𝑁 = 𝑉 𝑁 𝑡+ − 𝑉 𝑁 𝑡− ≈ 𝑤 𝑉
𝑑 𝑉 𝑀 𝑑𝑡 = 𝑑 𝑉 𝑅 𝑑𝑡 + 𝑊 𝑉 2∆ 𝑡 𝐷
For small ∆ 𝑡 𝐷 , 𝑊 𝑉 2∆ 𝑡 𝐷 is large and random
Want 𝑑 𝑉 𝑅 𝑑𝑡 ≫ 𝑊 𝑉 2∆ 𝑡 𝐷
wV decreases as FS gets smaller and N increases
Want ∆ 𝑡 𝐷 to be large enough to avoid random error but small enough to capture real events
RF, IRE, other errors, Random, but does not increase with ∆ 𝑡 𝐷

29. Time Dependent Data How to find 𝑑𝑇 𝑑𝑡 𝑜𝑟 𝑑𝑉 𝑑𝑡 from 𝑇 𝑡 or 𝑉 𝑡
1st order numerical differentiation (center difference)
𝑑𝑉 𝑑𝑡 𝑡 = 𝑙𝑖𝑚 ∆𝑡→0 𝑉 𝑡+∆ 𝑡 𝐷 −𝑉 𝑡−∆ 𝑡 𝐷 𝑡+∆ 𝑡 𝐷 + 𝑡−∆ 𝑡 𝐷
∆ 𝑡 𝐷 = differentiation time
∆𝑡 𝑠 ≡ sampling time
∆𝑡 𝐷 =𝑚∆ 𝑡 𝑠 m = 1, 2, 3, … i
T = (∆ts)i V
𝒅𝑽 𝒅𝒕 1
∆ts 2
2∆ts