Lesson_310 - Lesson 31 Lesson 31 Challenge 30 Lesson 31 –...

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Unformatted text preview: Lesson 31 Lesson 31 Challenge 30 Lesson 31 – Continuous-time Fourier transform Challenge 31 Lesson 31 Challenge 30 A full-wave rectified signal is shown below. ϖ =2 π /T T T /2 Lesson 31 Challenge 30 The Fourier transform is: What is the value of a 0 (DC level)? ( 29 ( 29 ( 29 ∑ ∞ ≠ =-- = 2 2 1 cos 4 i even i i i t i V a t v ϖ π ( 29 ( 29 π ϖ ϖ π π ϖ 2 / T /2 T T 2 cos 2 / T 2 ) sin( 2 / T 1 x(t)dt 2 / T 1 / 2 2 V t V dt t V a T f =- = = = = = ∫ ∫ Lesson 31 Lesson 31 The continuous-time Fourier transform (CTFT) is defined in terms of the analysis equation : and the synthesis equation ∫ ∞ ∞-- = dt x(t)e ) X(j t j ϖ ϖ ∫ ∞ ∞- Ω = d )e X(j x(t) j t ϖ ϖ Lesson 31 Lesson 31 The Fourier transform of x(t) exists if x(t) is absolutely integrable or square integrable (finite energy), plus some exceptions. The so-called Dirichlet conditions state that: The inverse Fourier transform of X(j ϖ ) equals x(t) at points where x(t) is continuous. The inverse Fourier transform of X(j ϖ ) converges to the mid-point of x(t) at points where x(t) is discontinuous. ∫ ∞ ∞-- = dt x(t)e ) X(j t j ϖ ϖ Lesson 31 Lesson 31 Example: Single sideband ϖ ϖ ( 29 ( 29 ( 29 t j t j t j e d e d e 2 1 2 1 2 1 1 ϖ ϖ ϖ π ϖ π ϖ ϖ ϖ δ π ϖ ϖ δ = =- =- ℑ ∫ ∫ ∞ ∞- ∞ ∞-- ( 29 ( 29 2 / 1 ϖ ϖ δ π ϖ- → ← ℑ t j e ( 29 ( 29 2 ϖ ϖ δ π ϖ- → ← ℑ t j e Lesson 31 Lesson 31 Example: x(t)=sgn(t)={1 if t>0, -1 if t<0, 0 if t=0} ( 29 [ ] ) ( ) ( sgn →--- = a at at t u e t u e t ( 29 ( 29 [ ] ( 29 ϖ ϖ ϖ ϖ ϖ j a j j a j a t u e t u e t a a a at at 2 2 1 1 ) ( ) ( sgn 2 2 = +- = -- + =-- ℑ = ℑ → → →-...
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Lesson_310 - Lesson 31 Lesson 31 Challenge 30 Lesson 31 –...

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