HW 07_2023-solutions copy

MEEN 260 Introduction to Engineering Experimentation Homework 7: Frequency Analysis and Fourier Series Assigned: Monday, March 27 th , 2023 Due: Sunday, April 2nd at 11:59 pm Note: Upload one file to Canvas, a PDF document of your work. Either type or write neatly. Each problem must have the following: • Briefly restate the problem in your own words (summarize what is known , what you are asked to find , and the solution including equations you used in symbolic form, and show the data/values that you used in these equations.) • Box final answers with appropriate significant digits and units. Problem #1: A period signal y(t) has the following function: !(#) = 3 sin+2-# + - 3 / 0 + 7 cos(5-#) − 5 sin+13-# + - 4 / 0 Before continuing with the analysis, it is useful to plot the function of the signal to visualize its characteristics: For this signal: a) determine the fundamental time interval ( T o ) and frequency ( f o ). Note that the fundamental frequency is defined by the time at which the entire signal repeats. -20 -15 -10 -5 0 5 10 15 20 0 0.5 1 1.5 2 2.5 Signal Amplitude (no units) Time (sec) Input signal y(t)

Circular frequency is converted to cycle frequency by: 2 p f = w By inspection of the circular frequency values, we can see there are three cyclic frequencies present: 1 Hz, 5/2 = 2.5 Hz and 13/2 = 6.5 Hz We can convert each frequency into a time interval or “wavelength”, T = 1/f: 1 sec, 2/5 sec, 2/13 sec The shortest time length that would contain a whole number of intervals or wavelengths of all three waves is 2 seconds: (2 for the 1 Hz term, 5 for the 2.5 Hz term, and 13 for the 6.5 Hz term. By careful inspection of the plot of the signal, we can also verify the repeat interval for the signal is 2 seconds. T o = 2 sec, f o = 0.5 Hz b) compute the DC-offset DC-offset can be found by computing the average value of the signal over one cycle interval; e.g. 0 to 2 seconds. We could integrate the signal over one interval and divide by interval length, i.e. However, this signal is composed of three sine/cosine waves and does not include any additional terms. Since each sine/cosine wave will have an average value of zero over any interval length having a whole number of cycles, the average value of the sum of the sine/cosine waves will be zero. DC-offset = 0 c) compute the peak-to-peak amplitude Peak-to-peak values can be read to a high degree of accuracy by inspection of the figure, or more precisely computed by identifying local maximum and minimum values. To find a local max or minimum, we set the derivative of the signal = 0 8! 8# = 0 = 6- cos+2-# + - 3 / 0 − 35- sin(5-#) − 65- cos+13-# + - 4 / 0

Before solving this equation, we can notice from the figure that a global minimum appears to occur around t = 0.6 seconds and a global maximum around t = 1.2 seconds. Using a trial-and-error process (for example ‘Solver’ or ‘Goal-Seek’ in Excel) to find the values of the time that makes the derivative zero near these initial guesses yields: t min = 0.62835 seconds, y(t min ) = -14.035 t max = 1.17658 seconds, y(t max ) = 13.980 Peak-to-Peak Amplitude = 13.980 – (-14.035) = 28.015 d) compute the RMS amplitude The RMS value is defined as: The equation evaluated for a single sine or cosine wave shows that the RMS value is related to the amplitude ( A ) by: ; !"# = < √2 / The RMS level of multiple sine waves with amplitudes A, B, C… in the same signal can be combined by root-square-sum: ; !"#$%&'()*+, = > < - 2 / + ? - 2 / + @ - 2 / … ; !"#$%&'()*+, = > 3 - 2 / + 7 - 2 / + (−5) - 2 / X RMS = 6.4420 e) write a new equation for the equivalent signal where none of the sine/cosine terms have phase angle, i.e. all arguments are either 2 p t , 5 p t , or 13 p t . To eliminate the phase angles, we can use the trigonometric identities:

For the first term, 3 sin+2-# + - 3 / 0 = 3 sin(2-#) cos+ - 3 / 0 + 3cos(2-#) sin ( - 3 / ) 3 sin+2-# + - 3 / 0 = 3 sin(2-#) ∙ 1 2 / + 3cos(2-#) ∙ C √3 2 / D 3 sin+2-# + - 3 / 0 = 3 2 Esin(2-#) + √3 cos(2-#)F For the third term, 5 sin+13-# + - 4 / 0 = 5 sin(13-#) cos+ - 4 / 0 + 5cos(13-#) sin ( - 4 / ) 5 sin+13-# + - 4 / 0 = 5 sin(13-#) ∙ √2 2 / + 5cos(13-#) ∙ √2 2 / 5 sin+13-# + - 4 / 0 = 5√2 2 [sin(13-#) + cos(13-#)] Re-assembling the equation with replaced terms, I(J) = K L EMNO(LPJ) + √K QRM(LPJ)F +S QRM(TPJ) − T√L L [MNO(UKPJ) + QRM(UKPJ)] Problem #2: A period signal y(t) has the following function: !(#) = 3 sin(6#) − 7 cos(6#) For this signal: a) Determine a new equation for the equivalent signal involving only a sine function (with phase angle). By inspection of your equation state: i) DC-offset, ii) the peak- to-peak amplitude, iii) the RMS amplitude. To combine the sine/cosine waves, we use the trigonometric identities: