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Plancharel and Parseval's Theorems

Module by: Justin Romberg. E-mail the author

Summary: This module contains the definition of the Plancharel theorem and Parseval's theorem along with proofs and examples.

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
(1)

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 )
(2)

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)
(3)

Example 1: Fourier Series of square pulse III -- Compute the Energy

Figure 1
Figure 1 (ex1.png)

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

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

Plancharels Theorem Demonstration

Figure 2: Interact (when online) with a Mathematica CDF demonstrating Plancharels Theorem visually. To Download, right-click and save target as .cdf.
PlancharelsTheoremDemo

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 2

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

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