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# Parseval's Theorem

Continuous Time Fourier Series preserves signal energy

i.e.:

$∫ 0 T | f ( t ) | 2 d t = T ∑ n = - ∞ ∞ | C n | 2 with unnormalized basis e j 2 π T n t ∫ 0 T | f ( t ) | 2 d t = ∑ n = - ∞ ∞ | C n ' | 2 with unnormalized basis e j 2 π T n t T | | f | | 2 2 ︸ L 2 [ 0 , T ) e n e r g y = | | C n ' | | 2 2 ︸ l 2 ( Z ) e n e r g y ∫ 0 T | f ( t ) | 2 d t = T ∑ n = - ∞ ∞ | C n | 2 with unnormalized basis e j 2 π T n t ∫ 0 T | f ( t ) | 2 d t = ∑ n = - ∞ ∞ | C n ' | 2 with unnormalized basis e j 2 π T n t T | | f | | 2 2 ︸ L 2 [ 0 , T ) e n e r g y = | | C n ' | | 2 2 ︸ l 2 ( Z ) e n e r g y$
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# Prove: Plancherel theorem

$Given f ( t ) → C T F S c n g ( t ) → C T F S d n Then ∫ 0 T f ( t ) g * ( t ) d t = T ∑ n = - ∞ ∞ c n d n * with unnormalized basis e j 2 π T n t ∫ 0 T f ( t ) g * ( t ) d t = ∑ n = - ∞ ∞ c n ' ( d n ' ) * with normalized basis e j 2 π T n t T 〈 f , g 〉 L 2 ( 0 , T ] = 〈 c , d 〉 l 2 ( Z ) Given f ( t ) → C T F S c n g ( t ) → C T F S d n Then ∫ 0 T f ( t ) g * ( t ) d t = T ∑ n = - ∞ ∞ c n d n * with unnormalized basis e j 2 π T n t ∫ 0 T f ( t ) g * ( t ) d t = ∑ n = - ∞ ∞ c n ' ( d n ' ) * with normalized basis e j 2 π T n t T 〈 f , g 〉 L 2 ( 0 , T ] = 〈 c , d 〉 l 2 ( Z )$
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# Periodic Signals Power

$Energy = | | f | | 2 = ∫ - ∞ ∞ | f ( t ) | 2 d t = ∞ Power = lim T → ∞ Energy in [ 0 , T ) T = lim T → ∞ T ∑ n | c n | 2 T = ∑ n ∈ Z | c n | 2 (unnormalized FS) Energy = | | f | | 2 = ∫ - ∞ ∞ | f ( t ) | 2 d t = ∞ Power = lim T → ∞ Energy in [ 0 , T ) T = lim T → ∞ T ∑ n | c n | 2 T = ∑ n ∈ Z | c n | 2 (unnormalized FS)$
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Example 1: Fourier Series of square pulse III -- Compute the Energy

$f ( t ) = ∑ n = - ∞ ∞ c n e j 2 π T n t → FS c n = 1 2 sin π 2 n π 2 n energy in time domain: | | f | | 2 2 = ∫ 0 T | f ( t ) | 2 d t = T 2 apply Parseval's Theorem: T ∑ n | c n | 2 = T 4 ∑ n sin π 2 n π 2 n 2 = T 4 4 π 2 ∑ n sin π 2 n 2 n 2 = T π 2 π 2 4 + ∑ n odd 1 n 2 ︸ π 2 4 = T 2 □ f ( t ) = ∑ n = - ∞ ∞ c n e j 2 π T n t → FS c n = 1 2 sin π 2 n π 2 n energy in time domain: | | f | | 2 2 = ∫ 0 T | f ( t ) | 2 d t = T 2 apply Parseval's Theorem: T ∑ n | c n | 2 = T 4 ∑ n sin π 2 n π 2 n 2 = T 4 4 π 2 ∑ n sin π 2 n 2 n 2 = T π 2 π 2 4 + ∑ n odd 1 n 2 ︸ π 2 4 = T 2 □$

# Plancharel Theorem

Theorem 1: Plancharel Theorem

The inner product of two vectors/signals is the same as the $ℓ2 ℓ 2$ inner product of their expansion coefficients.

Let $b i b i$ be an orthonormal basis for a Hilbert Space $H H$. $x∈H x H$, $y∈H y H$ $x=∑ i α i ⁢ b i x i α i b i$ $y=∑ i β i ⁢ b i y i β i b i$ then $⟨x,y⟩ H =∑ i α i ⁢ β i ¯ x y H i α i β i$

Example 2

Applying the Fourier Series, we can go from $f⁢t f t$ to $c n c n$ and $g⁢t g t$ to $d n d n$ $∫0Tf⁢t⁢g⁢t¯d t =∑ n =−∞∞ c n ⁢ d n ¯ t 0 T f t g t n c n d n$ inner product in time-domain = inner product of Fourier coefficients.

Proof

$x=∑ i α i ⁢ b i x i α i b i$ $y=∑ j β j ⁢ b j y j β j b j$ $⟨x,y⟩ H =⟨∑ i α i ⁢ b i ,∑ j β j ⁢ b j ⟩=∑ i α i ⁢⟨( b i ,∑ j β j ⁢ b j )⟩=∑ i α i ⁢∑ j β j ¯⁢⟨( b i , b j )⟩=∑ i α i ⁢ β i ¯ x y H i α i b i j β j b j i α i b i j β j b j i α i j β j b i b j i α i β i$ by using inner product rules

Note:
$⟨ b i , b j ⟩=0 b i b j 0$ when $i≠j i j$ and $⟨ b i , b j ⟩=1 b i b j 1$ when $i=j i j$

If Hilbert space H has a ONB, then inner products are equivalent to inner products in $ℓ2 ℓ 2$.

All H with ONB are somehow equivalent to $ℓ2 ℓ 2$.

Point of interest:
square-summable sequences are important

# Parseval's Theorem: a different approach

Theorem 2: Parseval's Theorem

Energy of a signal = sum of squares of its expansion coefficients

Let $x∈H x H$, $b i b i$ ONB

$x=∑ i α i ⁢ b i x i α i b i$ Then $∥x∥H2=∑ i | α i |2 H x 2 i α i 2$

Proof

Directly from Plancharel $∥x∥H2= ⟨x,x⟩ H =∑ i α i ⁢ α i ¯=∑ i | α i |2 H x 2 x x H i α i α i i α i 2$

Example 3

Fourier Series $1T⁢ei⁢ w 0 ⁢n⁢t 1 T w 0 n t$ $f⁢t=1T⁢∑ n c n ⁢1T⁢ei⁢ w 0 ⁢n⁢t f t 1 T n c n 1 T w 0 n t$ $∫0T|f⁢t|2d t =∑ n =−∞∞| c n |2 t 0 T f t 2 n c n 2$