Explanation Calculation: Consider that the given function. v ( t ) = V m − 2 V m 1 + e − T / 2 R C e − t / R C , 0 ≤ t ≤ T / 2 v ( t ) = V m − 2 V m ρ e − t / R C (1) Here, ρ = 1 + e − T / 2 R C The given function is half-wave symmetry. Therefore, a v = 0 , b k = 0 for k even, a k = 0 for k -even. Write the expression for Fourier co-efficient a k with the half-wave symmetry when k -odd. a k = 4 T ∫ 0 T / 2 v ( t ) cos k ω 0 t d t (2) Substitute Equation (1) in (2). a k = 4 T ∫ 0 T / 2 [ V m − 2 V m ρ e − t / R C ] cos k ω 0 t d t = 4 T ∫ 0 T / 2 V m cos k ω 0 t d t − 8 V m ρ T ∫ 0 T / 2 e − t / R C cos k ω 0 t d t = 4 V m T sin k ω 0 t k ω 0 | 0 T / 2 − 8 V m ρ T { e − t / R C ( 1 / R C ) 2 + ( k ω 0 ) 2 [ − cos k ω 0 t R C + k ω 0 sin k ω 0 t ] | 0 T / 2 } = [ 4 V m T [ sin k ω 0 ( T 2 ) k ω 0 − 0 ] − 8 V m ρ T { e − T / 2 R C ( 1 / R C ) 2 + ( k ω 0 ) 2 [ − cos k ω 0 ( T 2 ) R C + k ω 0 sin k ω 0 ( T / 2 ) ] − 1 ( 1 / R C ) 2 + ( k ω 0 ) 2 [ − 1 R C ] } ] a k = { 4 V m T [ sin k ω 0 ( 2 π 2 ω 0 ) k ω 0 ] − 8 V m ρ T { e − 2 π / 2 ω 0 R C ( 1 / R C ) 2 + ( k ω 0 ) 2 [ − cos k ω 0 ( 2 π 2 ω 0 ) R C + k ω 0 sin k ω 0 ( 2 π / 2 ω 0 ) ] + 1 ( 1 / R C ) 2 + ( k ω 0 ) 2 ( 1 R C ) } } = { 4 V m T [ sin k π k ω 0 ] − 8 V m ( 1 + e − T / 2 R C ) T { e − 2 π / 2 ω 0 R C ( 1 / R C ) 2 + ( k ω 0 ) 2 [ − cos k π R C + sin k π ] + 1 ( 1 / R C ) 2 + ( k ω 0 ) 2 ( 1 R C ) } } = { 0 − 8 V m ( 1 + e − 2 π / 2 ω 0 R C ) T { e − π / ω

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Unijunction Transistor (UJT)

The Unijunction Transistor (UJT) is a three-terminal switching semiconductor device. It contains an n-shaped silicone object bar with a terminal attached to its two ends known as base one and base two. The third terminal is connected to a highly enclose…

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Chapter 16, Problem 23P

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Derive the Fourier co-efficients ak and bk when k is odd.

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