Mechanical Engineering - Measurement

Multiple choice test

8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 1/12 Meas 2021 July 18 Part1 Every question can have more than a single good answer Ciao Luca, quando invierai il modulo, il proprietario potrà vedere il tuo nome e l'indirizzo di posta elettronica. 1 The FT of a real odd function is (2 punti) An imaginary odd function A complex function A real odd function An imaginary odd function An imaginary even function 2 Let’s consider y: (4 punti) 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 2/12It is an autocorrelation It is a spectrum It is always an odd function It is always an even function It is a convolution It is defined in the frequency domain It is a cross correlation Its value is always higher than the product of the RMS values of x and y It is defined in the time domain 3 In a measurement process I have a sine with random phase and I have 1000 records. Three samples are given. Which of the following do apply? (4 punti) 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 3/12If I apply a trigger at .7V with positive slope, I get a good estimate of the sine wave If I average the records I get the best estimate of the sine wave cleaning it from noise Continous trigger is not the best choice for the actual situation If I average the full spectra (vector averaging) of each record I get the best estimate in the frequency domain of the sine wave cleaning it from noise If I apply a trigger at 2V positive slope and average in the time domain, I get a good estimate of the sine wave, cleaning it from noise 4 Auto covariance (3 punti) Is another name for the auto correlation function, so they are the same Auto covariance at zero lag is the standard deviation squared Autocovariance and autocorrelation have nothing to share Auto covariance is defined in the frequency domain while auto correlation is defined in the time domain Auto Covariance is evaluated including the mean values in the calculations 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 4/12Auto covariance at zero lag is the RMS squared 5 Given the two records on the left, which is the right cross correlation among the six given? (the first is fixed and the second is sliding) (3 punti) f) c) b) a) e) d) 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 5/126 Which is the right convolution of signals 1) and 2)? (3 punti) a) d) b) c) 7 Spectra (3 punti) The imaginary part of a spectrum is always even only for a sine wave In spectral analysis the corresponding components at positive and negative frequencies are complex conjugate 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 6/12The real part of a spectrum is always even Theoretically speaking if a signal is not periodic, the Fourier series cannot be applied Spectra cannot be defined at negative frequencies e^(-j*omega) is a rotating vector Both real and imaginary parts are not even nor odd The imaginary part of a spectrum is always even 8 I have a function made up by two harmonics at 4 Hz and 8 Hz, sampled for 1s, with a bias of 1V, which of the following apply? (4 punti) I do have leakage for both harmonics Amplitude at frequency=0 is equal to 0 The first line made available by the Fourier algorithms is 4 Hz I do not have leakage for the first harmonic, but I have for the second The first line made available by the Fourier algorithms is 8 Hz The first line made available by the Fourier algorithms is 1 Hz The first line different from 0 is the fourth (not counting f=0) 9 I have the record given in picture, with some spikes; the standard deviation of the record is 0.00047 mm/s^2. The second picture gives a zoom in the surroundings of the main peaks; the legend gives the coordinates of the maximum and minimum, together with its value (named Y in the cursor palette) Please evaluate the Crest Factor also writing the formula you plan to use (2 punti) 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 7/12CF= |x_peak| / x_RMS RMS = mean^2+stddev^2, here mean=0 so is 0.00047 so CF=0.00442047/0.00047=9.4058 10 I have the record given in picture, with some spikes; the standard deviation of the record is 0.47 mm/s^2 (3 punti) 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 8/12The RMS is 0.47 mm/s2 In this case the crest factor is better than the skewness to get the presence of spikes To get the variance I need to sum the RMS squared and the mean squared In this case the kurtosis is the best choice to get spikes in the given signal The RMS does not change at all due to the presence of the spikes The skewness is expected to be higher than 3 The skewness is expected to be around 0 11 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQV … 9/12In the picture you have an acceleration record on top, a perfect random signal in the middle and the histograms of the two signals in the third plot. In the histogram the empty bars refer to the acceleration signal, the red ones to the perfect gaussian process. (3 punti) The acceleration signal is expected to have an excess kurtosis higher than 0 〖������������������〗^2=〖��������������〗^2+〖���������������������������� 〗^2 The skewness is different as the two distributions are different The excess kurtosis is the same in both cases I cannot evaluate the kurtosis as I have a mean different from 0 The Gaussian signal is expected to have an excess kurtosis lower than 0 12 The delta function (2 punti) Is capable of exciting all frequencies up to infinity It is defined only at positive lags Has unit amplitude Its amplitude tends to infinity It can only be defined for time = 0 Its duration can be finite It can only excite frequencies below the sampling frequency 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQ … 10/1213 I have the couple of signals on the left column and then the couple on the right column. For both cases (left and right column) I want to estimate the cross correlation among these two couples of signals. Which of the following apply? (3 punti) The cross-correlation function has only a single peak in both cases I get the first maximum in the cross correlation function by shifting the second record to the left (Delta t0) The position of the peaks in the cross-correlation function is not affected by the presence of noise The cross correlation function has one peak only but just in the case with noise 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQ … 11/1214 I convolve a sine function (������=2sin�������)) with a delta function ������(�). The output is (2 punti) -2 -1 0 The value is the same if I convolve ������=2 cos�������) with the delta function 2 1 15 The two images show two acceleration records from a stadium grandstand during the night. (3 punti) The process is complex periodic 8/7/2021 Meas 2021 July 18 Part1 https://forms.office.com/Pages/ResponsePage.aspx?id=K3EXCvNtXUKAjjCd8ope60_E5LGpsJVDo_abJAuvijNUNDBETjQ5MzcyOUtVRUFJQ … 12/12 Questo contenuto è creato dal proprietario del modulo. I dati inoltrati verranno inviati al proprietario del modulo. Microsoft non è responsabile per la privacy o le procedure di sicurezza dei propri clienti, incluse quelle del proprietario di questo modulo. Non fornire mai la password. Con tecnologia Microsoft Forms | Privacy e cookie | Condizioni per l'utilizzoInvia The process is a sequence of transients The process is not stationary but not random as the signals have no mean The process is stationary and not ergodic The process is ergodic but not stationary The process is deterministic The process is stationary and ergodic The process is quasi periodic