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Plancharel and Parseval's Theorems

By Justin Romberg

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
Figure 1

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 2 2 inner product of their expansion coefficients.

Let b i b i be an orthonormal basis for a Hilbert Space H H. xH x H , yH 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 ft f t to c n c n and gt g t to d n d n 0Tftgt¯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 ij 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 2 2 .

All H with ONB are somehow equivalent to 2 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.

Parseval's Theorem: a different approach

Theorem 2: Parseval's Theorem

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

Let xH x H , b i b i ONB

x= i α i b i x i α i b i Then xH2= i | α i |2 H x 2 i α i 2

Proof

Directly from Plancharel xH2= 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 1Tei w 0 nt 1 T w 0 n t ft=1T n c n 1Tei w 0 nt f t 1 T n c n 1 T w 0 n t 0T|ft|2d t = n =| c n |2 t 0 T f t 2 n c n 2