Question

4. (20 pts) Using a delicate temperature sensor, two students are continuously measuring temperature which fluctuates with a frequency of 50 Hz. 4-1) (5 pts) What sampling frequency is required so that the 50 Hz temperature signal will not be aliased? 4-2) (5 pts) Student A decided to sample the temperature signal at 200 Hz. What is the time between samples? If the sampling continues for 20 seconds, how many samples will be taken? 4-3) (5 pts) Student B sampled the temperature signal at 1000 Hz, with 1024 samples, and then performed an FFT routine to create a spectrum of the signal. What is the increment between frequencies in the FFT and what is the maximum useful frequency returned by the FFT?

Answer 1

satuc fluctuates with toeguency ) DHZ Temperatus sampling frequency required so that 50Hz will not be aliased. By Nyquist coitesia, sampling > 2x Maximum foco uency date in the signal since the signal has only 50 Hz toequency. Sampling frequency (or gate) z 2x50 :. Sam foco > 100 HZ 81 100 samples/sec 21 student decided to sample at 200 HZ' Ssampling frequency = 200 HZ" am - = 0.005 sec Time between signals, T - SF2DD 200 HZ = 200 samples in ISCC 2005 -> seconds) 0 I sam - zo seconds No of samples for 20 SLC = 20x200 = 4000 Samples in 20 seconds Student 8 sampled at 1000 Hz with 1024 samples In coement between toeguency is also called frequency resolution - sampling frequency = 1000 = 0.9765 1024 No of Samples .. Increment between frequency = 0.9765 nue Manimum useful focquency octurned by FFT: = 1000 - 500 HZ

+) spectal lеаkіnge occnic d in FT → To select lowest sampling frequency which will not cause leakage and aliaging problem → Spectdal leakage occurs when the signal(under consideration is Sampied non-integer number of cycles. considco the 1000 HZ sampling of 50 Hz Signal for 1024 samples. 50 Hz Signal mean 50 cycles) sec. Its time period = 0.02ste A 50 Hz Signal repeats itself after 0.02 sec. 1024 Time of sampling : no of sampics Sampling tregi 1024 1000 is sampled for 1.024 seconds. = No of cycles of signal sampled = 1.024 = 51.2 cycles 0.02 The 50 Hz Signal is sampled too 51.2 cycles, a non-integer number. 50 spectral leakage occured To avoid it integer number of cycles should be recorded. no of samples sampling frequency 0 = integer (+ve). Signal toequency Signal focquency = 50, no of samples = 1024 To avoid leakage, sampling foco z 2150 > 100

1024 sampling trea =D 50 Samp foea z 100. 1024 x 50 Sampling foco For sampling toeq to be least & in to be integer. » Sampling toege should be factor of 1024 | Samp 2.100] factors of 1024 above 100 are 128, 256, 512. 128. For sampling trea to be least & above 100 is Sampling toequency = 128 [Hz] No of cycles recorded when sampling is done at 128 HZ 1024 24 = 400 cycles (Integer number] 128
I hope you understand. Comment if you have any doubt.