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Plancharel and Parseval in CT Fourier Series

Module by: Richard Baraniuk. E-mail the author

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Summary: An introduction to to Plancharel and Parseval in CT Fourier series.

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With an unnormalized basis

ft=∑n∈ℤ c n ⅇⅈ ω 0 nt f t n n c n ω 0 n t (1)
c n =1T∫0Tftⅇ-ⅈ ω 0 ntdt c n 1 T t 0 T f t ω 0 n t (2)
The normalized coefficients are
c n ′ =T c n c n ′ T c n (3)
Parseval:
∥f∥2=∑| c n ′ |2 f 2 n c n ′ 2 (4)

Finite Signals

L20T L 0 T 2 (Figure 1).

Figure 1
Figure 1 (fig1.png)

Energy

Parseval

∥f∥2=T∑n∈ℤ| c n |2=∫0T|ft|2dt f 2 T n n c n 2 t 0 T f t 2 (5)
which is unnormalized FS!

Inner Product

ft=∑ c n ⅇⅈ ω 0 nt f t n c n ω 0 n t (6)
gt=∑ d n ⅇⅈ ω 0 nt g t n d n ω 0 n t (7)
which are both unnormalized.

Plancharel

<f,g>=T∑n∈ℤ c n d n ¯=∫0Tftgt¯dt f g T n n c n d n t 0 T f t g t (8)
which is unnormalized FS!

Parseval's Theorem says: "CTFS maps the signal ( xt∈L20T x t L 0 T 2 ) to the Fourier Series coefficients ( c n ∈l2ℤ c n l 2 )."

Periodic Signals

With each period in L20T L 0 T 2 (Figure 2).

Figure 2
Figure 2 (fig2.png)

Energy

∥f∥2=∫-∞∞|ft|2dt f 2 t f t 2 (9)
where Equation 9 also equals ∞ now.

Let's compute the power.

Power

P f =limT→∞ Energy in [ 0 , T ) T=limT→∞T∑n∈ℤ| c n |2T=limT→∞∑n∈ℤ| c n |2=∑n∈ℤ| c n |2 P f T Energy in [ 0 , T ) T T T n n c n 2 T T n n c n 2 n n c n 2 (10)
which is unnormalized FS!

Example 1: FS of Square Pulse III: Energy

Figure 3's Fourier coefficients are

c n =12sinπ2nπ2n c n 1 2 2 n 2 n (11)
Figure 3
Figure 3 (fig3.png)
From Figure 3, the energy is
∥f∥2=∫0T|ft|2dt=T2 f 2 t 0 T f t 2 T 2 (12)
From the Fourier coefficients, the energy is
T∑| c n |2=T4∑sinπ2nπ2n2=T44π2∑sin2π2nn2=T44π2∑ n odd 1n2+∑ n even 0+∑n=0π24=T44π2∑n≠0∧ n odd 1n2+π24=T44π22∑n>0∧ n odd 1n2+π24=T44π22∑n>0∧n∈ℤ1n2−∑n>0∧ n even 1n2+π24=T44π22∑n>0∧n∈ℤ1n2−∑n>0∧n∈ℤ12n2+π24=T44π22π26−14π26+π24=T44π22π28+π24=T44π2π24+π24=T44π2π22=T2 T n c n 2 T 4 n 2 n 2 n 2 T 4 4 2 n 2 n 2 n 2 T 4 4 2 n n odd 1 n 2 n n even 0 n n 0 2 4 T 4 4 2 n n 0 n odd 1 n 2 2 4 T 4 4 2 2 n n 0 n odd 1 n 2 2 4 T 4 4 2 2 n n 0 n 1 n 2 n n 0 n even 1 n 2 2 4 T 4 4 2 2 n n 0 n 1 n 2 n n 0 n 1 2 n 2 2 4 T 4 4 2 2 2 6 1 4 2 6 2 4 T 4 4 2 2 2 8 2 4 T 4 4 2 2 4 2 4 T 4 4 2 2 2 T 2 (13)
The signal and the Fourier coefficients yield the same result!

Example 2: FS of Square Pulse IV: Power in Periodic Extension

Figure 4's Fourier coefficients are Equation 11.

Figure 4
Figure 4 (fig4.png)
From Figure 4, the power is
Power= Energy in [ 0 , T ) T=T2T=12 Power Energy in [ 0 , T ) T T 2 T 1 2 (14)
From the Fourier coefficients, the power is
Power=∑| c n |2=12 Power n c n 2 1 2 (15)
which follows from Equation 13. Once again the signal and the Fourier coefficients yield the same result!

Example 3: Cauchy Schwarz "Matched Filter" Detector in FS Domain

The goal is to look for a pattern like gt g t in ft f t . CSI says take ft f t , then calculate <f,g>=∫0Tftgt¯dt f g t 0 T f t g t . If this yields a small value then the pattern is not present, if it yields a large value then the pattern is present. Here, when we say small or large, we mean compared to ∥f∥∥g∥ f g .

We can also expand ff and gg in FS where ft f t yields c n c n and gt g t yields d n d n . We can take c n c n , then calculate ∑ c n d n ¯ n c n d n . Again, if this yields a small value then the pattern is not present, if it yields a large value then the pattern is present. Here, when we say small or large, we again mean compared to ∥f∥∥g∥=∑| c n |2∑| d n |2 f g n c n 2 n d n 2 . This system gives identical output to the system discussed above.

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