# HW9S.pdf - HW9 Fourier Transform Q1(10pts For each waveform...

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HW9: Fourier Transform Q1 (10pts.): For each waveform shown below find the DC average and period. Figure 1: T=1s, Average=0 Figure 2: T=1s, Average=.5 Figure 3: T=.5s, Average=.5 Figure 4: T=1s, average=.25 Figure 5: T=1s, Average=1.25
Figure 6: T=1s, Average=1.875 Q2 (10pts.): Find the ? ? and ? ? constants of the Fourier series for: 𝐴 × ?𝑖?(2𝜋??) 2 Which parts of the spectrum would you want to keep if you were using the above signal for a DC power supply? ? 0 , the DC average Which parts of the spectrum would you use if you wanted a sinusoid that is 4𝜋? ?
Which parts of the spectrum would you want to keep if you were using the above signal for a DC power supply? Which parts of the spectrum would you use if you wanted a sinusoid that is 4𝜋? To better see what is going on plot out the functions: ? 1 (?) = sin(2𝜋?) ?
? 2 (?) = sin(2𝜋?) 2 You can see that the frequency doubles to 2 Hz! It looks like an inverted cosine wave with a DC offset. ?(?) = ? 0 + ∑[? ? cos(?𝜔 0 ?) + ? ? sin(?𝜔 0 ?)] ?=1 ? 0 𝑖? ?ℎ? ?? ???????, which can be approximated by a box as seen below. Figure 7: Approximate with a box. ? 0 = .5 A triangle can also approximate it.
𝐴??? 1 = 1 2 ? × ℎ = 1 2 𝑇 0 × 1 2 × 1 = 𝑇 0 4 𝐴??? 2 = 𝐴??? 1 𝐴??? 𝑡?𝑡?? = 𝑇 0 2 ? 0 = ?? ??????? = 𝐴??? 𝑡?𝑡?? 𝑇 0 = 1 2 To find the a or b constants we look at the signal with the .5 subtracted. There is one frequency at 2Hz, and by inspection, we see the amplitude is .5. It does not start at zero at t=0, so it is not a sine wave. It looks like −.5cos(4𝜋?) ? 0 = .5 ?