= ∞o is expected. Assume that the peak-to-peak amplitude, A = 5 V. The period is 40 ms with no DC content. Time Domain a. Pulse 1---1 r=0 1=0 d=k/T z=0 b. Square c. Triangle IMA IwWw I Frequency Domain A- 0 0f 2f 3f 4f 5f 6f 0 0 A- f 2f 3f 4f Sf 6f 0 0f 2f 3f 4f Sf 6f Figure 01 a = Ad 2.A na - 0 (d- 0.27 in this example) a=0 a = 24 NT sin (nad) a = 0 = b₁ = 0 (all even harmonies are zero) sin 4.A (nn)² b₁ = 0 (all even harmonies are zero) (20 marks for formula)
Fig: 1