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Course Level Date Clip# Line Interlocutor Content
Oral Assessments Graduate 2023-08-01 3 113 S01 occurs this many times throughout the sample↘ (1.7)
Oral Assessments Graduate 2023-08-01 3 114 S01 so (0.7) why we needed xxx xxx↘ FFT it's not the (1.4)
Oral Assessments Graduate 2023-08-01 3 115 S01 the (1.9) usual f- uh fourier transform which
Oral Assessments Graduate 2023-08-01 3 116 S01 we are using↘ (0.5)the fourier transform (0.4)
Oral Assessments Graduate 2023-08-01 3 117 S01 as in the term (0.8) is described as (5.0) °minus one°
Oral Assessments Graduate 2023-08-01 3 118 S01 (0.3) °x of n (0.7) e bar minus j two pi (1.7) k n°
Oral Assessments Graduate 2023-08-01 3 119 S01 (1.2) °divide n° so (4.3) this is the frequency
Oral Assessments Graduate 2023-08-01 3 120 S01 representation of a signal↘ so what the frequency
Oral Assessments Graduate 2023-08-01 3 121 S01 is pointing to n parts when uh where k is from
Oral Assessments Graduate 2023-08-01 3 122 S01 uh k goes to n minus one↗ (0.6) and we multiply
Oral Assessments Graduate 2023-08-01 3 123 S01 each term↗ with a complex exponent↘ (0.7) uhm
Oral Assessments Graduate 2023-08-01 3 124 S01 (1.2) so
Oral Assessments Graduate 2023-08-01 3 125 T02 um so how is frequency represented in that graph↘
Oral Assessments Graduate 2023-08-01 3 126 S01 (1.5) so this term is the frequency component
Oral Assessments Graduate 2023-08-01 3 127 T02 uh huh
Oral Assessments Graduate 2023-08-01 3 128 S01 so two pi by n we are diving a circle
Oral Assessments Graduate 2023-08-01 3 129 T02 mm hmm
Oral Assessments Graduate 2023-08-01 3 130 S01 into n parts↘
Oral Assessments Graduate 2023-08-01 3 131 T02 (0.4) mm hmm
Oral Assessments Graduate 2023-08-01 3 132 S01 and we are taking each part differently↘ so (0.5)
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