MECH 373 Instrumentation and Measurements

1. MECH 373Instrumentation and Measurements Lecture 10 Discrete Sampling & Analysis of Time-Varying Signals (Chapter 5) • Sampling-Rate Theorem (review) • Spectral Analysis of Time-Varying Signal • Spectral Analysis uisng the Fourier Transform

2. Nyquist Sampling Theorem A continuous signal can be represented by, and reconstituted from, a set of sample values providing that the number of samples per second is at least twice the highest frequency presented in the signal. is the signal frequency (or the maximum signal frequency if there is more than one frequency in the signal) is the sampling rate

3. Nyquist Frequency and Aliasing (1) High frequency signal to be sampled by a low sampling rate may cause to “fold” the sampled data into a false lower frequency signal. This phenomena is known as aliasing.

4. Aliasing Example

5. Higher Frequency Aliases

6. Nyquist Frequency and Aliasing (2) Definitions: Sampling Time, T: Total measuring time of a signal Sampling Interval Dt: Time between two samples Sample Rate : 1/Dt, the number of samples per second Nyquist Frequency, Fnyq: Maximum frequency that can be captured by a sample interval, t Resolution Bandwidth: Minimum frequency that can be represented by a sample T = N t = N/fs [sec] • The faster you sample, the higher frequency you can represent • The longer you sample, the smaller the frequency represented

7. Frequency Resolution • Duration of the sampleT = N t = N/fs [sec] • Minimum frequency that can be resolved is function of sample length …. “bandwidth resolution” • Must capture a full period of the frequency

8. Alias Frequency A simple method to estimate alias frequencies is by using the folding diagram shown below. where, fNis the Nyquist(folding) frequency equal to half the sampling frequency fS.

9. Aliasing Formulas Alias frequency Folding frequency (Nyquist frequency)

10. Question 2 Given that the measured frequency = 187.5Hz, sampling frequency = 250Hz and, find the alias frequency. Solution: fm / fN = 1.5

11. Alias Frequency

12. Spectral Analysis • When a signal is pure sine wave, determining the frequency is a simple process. However, the general time-varying signal does not have the form of a simple sine wave, as a typical example shown below: • Complicated waveforms can be considered to be constructed of the sum of a set of sine or cosine waves of different frequencies. The process of determining these component frequencies is called spectral analysis.

13. Spectral Analysis • To examine the method of spectral analysis, consider a relative simple waveform, a 1000 Hz sawtooth wave as shown below: • It appears that this wave contains only one single frequency of 1000 Hz. However, it is much more complicated, containing all frequencies that are an odd-integer multiple of 1000, such as, 1000, 3000, 5000 Hz. • The lowest frequency, f0, in the periodic wave is called the fundamental or first harmonic frequency. In the above example, it is 1000 Hz. • The fundamental frequency has period T0and angular frequency ω0, where angular frequency ω = 2πf and f = 1/T, and T is the time period.

14. Spectral Analysis • The method used to determine these component frequencies is known as Fourier-Series Analysis. • Any periodic function f(t) can be represented by the sum of a constant and a series of sine and cosine waves as

15. Spectral Analysis • The function f(t) is considered to be an even function if it has the property that, f(-t) = f(t). • The function f(t) is considered to be an odd function if f(-t) = -f(t). • If f(t) is even, it can be represented entirely with a series of cosine terms, which is know as a Fourier Cosine Series. • If f(t) is odd, it can be represented entirely with a series of sine terms, which is known as a Fourier Sine Series. • Many functions are neither even nor odd and require both sine and cosine terms. • The sawtooth wave shown in the previous figure is an odd function and thus, only bnneeds to be computed. Once bnis known, the values of b for any n value can be computed, for example, b1, b2, b3, etc. • The first seven values of the coefficient b for the sawtooth wave are given below (either using direct integration or a numerical method):

16. Spectral Analysis • b1, b3, b5, and b7 are the amplitudes of the first, third, fifth and seventh harmonics of the function f(t). These have frequencies of 1000, 3000, 5000 and 7000 Hz, respectively. • It is useful to present the amplitudes of the harmonics on a plot of amplitude versus frequency as shown below:

17. Spectral Analysis • As can be seen, harmonics beyond the fifth have a very low amplitude. Since the energy is proportional to the square of the amplitude, the larger amplitude indicates higher energy content at a particular frequency. That is, the signal is very strong at that particular frequency. • If the amplitude is very low, it indicates the energy is too low at the frequency and thus, its contribution to the overall signal is very small. • The harmonics that have the higher amplitude or higher energy has a significant influence on the overall waveform. This is illustrated in the figure below which shows first and third harmonics and their sum compared with the overall waveform.

18. Example 1 XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

19. Example 2

20. Spectral Analysis • The figure shows that the sum of first and third harmonics fairly well represents the sawtooth wave. The main problem is apparent as a rounding near the peak. This problem could be reduced if higher harmonics e.g. fifth, seventh, etc. are included. • Fourier analysis of this type can be very useful in specifying the frequency response of instruments. • For example, if the experimenter considers the first-plus-third harmonics to be a satisfactory approximation to the sawtooth wave, then the sensing element needs to have an upper frequency limit of 3000 Hz. That is, the frequency response of the instrument should be high enough to capture the 3000 Hz wave. • One problem associated with Fourier-series analysis is that it appears to only be useful for periodic signals. • In fact, this is not the case and there is no requirement that f(t) be periodic to determine Fourier coefficients for data sampled over a finite time. • We could force a general function of time to be periodic simply by duplicating the function in time as shown below.

21. Spectral Analysis • When Fourier coefficient are computed for a function f(t) over a time period T, the resulting Fourier series will have an implicit fundamental angular frequency ω0, equal to 2π/T. • If the resulting Fourier series is used to compute values of f(t) outside the time interval 0-T, it would result in values that would not resemble the original signal. • The analyst must be careful to select a large enough value of T so that all wanted effects can be represented by the resulting Fourier series. • An alternative method of finding the spectral content of signals is called the Fourier Transform, discussed next.

22. Spectral Analysis

23. Spectral Analysis