chap02 - PROBLEM 2.1 FIND Define signal and provide...

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PROBLEM 2.1 FIND: Define signal and provide examples of static and dynamic input signals to measurement systems. SOLUTION: A signal is information in motion from one place to another, such as between stages of a measurement system. Signals have a variety of forms, including electrical and mechanical. Examples of static signals are: 1. weight, such as weighing merchandise, etc. 2. body temperature, over the time period of interest 3. length or height, such as the length of a board or a person's height Examples of dynamic signals: 1. input to an automobile speed control 2. input to a stereo amplifier from a component such as a CD player 3. output signal to a printer from a computer
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PROBLEM 2.2 FIND: List the important characteristics of signals and define each. SOLUTION: 1. Magnitude - generally refers to the maximum value of a signal 2. Range - difference between maximum and minimum values of a signal 3. Amplitude - indicative of signal fluctuations relative to the mean 4. Frequency - describes the time variation of a signal 5. Dynamic - signal is time varying 6. Static - signal does not change over the time period of interest 7. Deterministic - signal can be described by an equation (other than a Fourier series or integral approximation) 8. Non-deterministic - describes a signal which has no discernible pattern of repetition andcannot be described by a simple equation. COMMENT: A random signal, or stochastic noise, represents a truly non- deterministic signal. However, chaotic systems produce signals that appear random, but are truly deterministic. An example would be the velocity in a turbulent fluid flow, that may appear random, but is actually governed by the Navier-Stokes equations.
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PROBLEM 2.3 KNOWN: y t t ( ) cos = + 30 2 6 π FIND: y y rms and for the time periods t 1 to t 2 listed below a) 0 to 0.1 s b) 0.4 to 0.5 s c) 0 to 1/3 s d) 0 to 20 s SOLUTION: For the function y ( t ) 2 1 2 1 1 ( ) t t y y t dt t t = - and [ ] 2 1 2 2 1 1 ( ) t rms t y y t dt t t = - Thus in general, ( 29 ( 29 ( 29 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 = 30 sin 6 6 1 2 = 30 sin 6 sin 6 6 30 2cos6 t t t t y t t t t t t t t t t t t t dt π π π π π π = - + - - + - - + and ( 29 y t t t dt t t t t t t t rms t t t t = - + = - + + + 1 30 2 6 1 900 120 6 6 4 1 12 6 6 1 2 2 1 2 1 2 2 1 1 2 1 2 1 2 cos sin sin cos π π π π π π
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The resulting values are a) 31.01 31.02 rms y y = = b) 28.99 29.00 rms y y = = c) 30 30.03 rms y y = = d) 30 30.03 rms y y = = COMMENT: The average and rms values for the time period 0 to 20 seconds represents the long-term average behavior of the signal. The values which result in parts a) and b) are accurate over the specified time periods, and for a measured signal may have specific significance. If we examined the period 0 to 1/3, it would represent one complete cycle of the simple periodic signal and results in average and rms values which accurately represent the long-term behavior of the signal.
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