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Dirichlet Kernel 📂Fourier Analysis

Dirichlet Kernel

Definition

Dirichlet Kernel DnD_{n} is defined as follows.

Dn(t):=12+∑k=1ncos⁡kt \begin{equation} D_{n}(t) := \dfrac{1}{2}+\sum \limits_{k=1}^{n} \cos kt \end{equation}

Explanation

The Dirichlet kernel is related to delta functions, exponential functions, etc., and appears in Fourier analysis. Here are some related theorems and their proofs.

Theorem 1

The Dirichlet Kernel satisfies the following equation.

Dn(t)=sin⁡(n+12)t2sin⁡12t D_{n}(t)=\dfrac{\sin\left(n+\frac{1}{2}\right) t}{2\sin \frac{1}{2}t}

Proof

If we express the cosine function as a complex exponential form, we get the following.

Dn(t)= 12+12∑k=1n(eikt+e−ikt)= 12[1+∑k=1n(eikt+e−ikt)]= 12∑k=−nneikt \begin{align*} D_{n}(t) =&\ \dfrac{1}{2}+\dfrac{1}{2}\sum \limits_{k=1}^n( e^{ikt}+e^{-ikt} ) \\ =&\ \dfrac{1}{2} \left[ 1+\sum \limits_{k=1}^{n} (e^{ikt}+e^{-ikt} ) \right] \\ =&\ \dfrac{1}{2} \sum \limits_{k=-n}^{n} e^{ikt} \end{align*}

In this case

Geometric Series Summation Formula

∑k=1nak=a(rn−1)r−1 \sum_{k=1}^{n} a_{k}= \dfrac{a (r^{n} -1) }{ r-1 }

Using this, with the first term being a1=e−inta_{1}=e^{-int} and the common ratio r=eitr=e^{it}, we can arrange it as follows.

Dn(t)= 12∑k=−nneikt=12∑k=12n+1ei(k−n−1)t= 12(e−int)(ei(2n+1)t−1)eit−1= 12e−intei(n+12)t−e−i(n+12)tei12t−e−i12tei(n+12)tei12t= 12ei(n+12)t−e−i(n+12)tei12t−e−i12tei(n+12)tei(n+12)t= 12sin⁡(n+12)tsin⁡12t \begin{align*} D_{n}(t) =&\ \dfrac{1}{2} \sum \limits_{k=-n}^{n} e^{ikt} = \dfrac{1}{2} \sum \limits_{k=1}^{2n+1} e^{i(k-n-1)t} \\ =&\ \dfrac{1}{2} \dfrac{ ( e^{-int} ) \left( e^{i(2n+1)t -1} \right) }{e^{it}-1} \\ =&\ \dfrac{1}{2}e^{-int}\dfrac{e^{i(n+\frac{1}{2}) t }-e^{-i(n+\frac{1}{2})t} }{e^{i\frac{1}{2}t}-e^{-i\frac{1}{2}t }} \dfrac{e^{i(n+\frac{1}{2})t}} {e^{i\frac{1}{2}t}} \\ =&\ \dfrac{1}{2}\dfrac{e^{i(n+\frac{1}{2}) t }-e^{-i(n+\frac{1}{2})t} }{e^{i\frac{1}{2}t}-e^{-i\frac{1}{2}t }} \dfrac{e^{i(n+\frac{1}{2})t}} {e^{i(n+\frac{1}{2})t}} \\ =&\ \dfrac{1}{2}\dfrac{\sin (n+\frac{1}{2})t} {\sin \frac{1}{2} t} \end{align*}

In the last equality, we used that sin⁡x=eix−e−ix2i\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}.

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Theorem 2

Let’s say the following equation is the 2L2L-periodic function f(t)f(t)’s Fourier series partial sum.

SNf(t)=12a0+∑n=1N(ancos⁡nπtL+bnsin⁡nπtL) \begin{equation} S_{N} ^{f} (t)=\dfrac{1}{2}a_{0}+\sum \limits_{n=1}^{N} \left( a_{n}\cos\dfrac{n\pi t}{L}+b_{n}\sin\frac{n\pi t}{L} \right) \end{equation}

Then, the partial sum SNf(t)S_{N}^{f}(t) can be represented as an integral including the Dirichlet kernel as follows.

SNf(t)=1L∫−LLf(x)Dn(π(x−t)L)dx S_{N}^{f} (t)=\dfrac{1}{L}\int_{-L}^{L}f(x)D_{n}\left(\dfrac{\pi (x-t)}{L}\right)dx

Proof

After calculating the Fourier coefficients a0a_{0}, ana_{n}, bnb_{n}, we get the following.

a0=1L∫−LLf(x)dxan=1L∫−LLf(x)cos⁡nπxLdxbn=1L∫−LLf(x)sin⁡nπxLdx \begin{align*} a_{0} &= \dfrac{1}{L}\int_{-L}^{L} f(x) dx \\ a_{n} &= \dfrac{1}{L}\int_{-L}^{L}f(x)\cos\dfrac{n\pi x}{L} dx \\ b_{n} &= \dfrac{1}{L}\int_{-L}^{L}f(x)\sin\dfrac{n\pi x}{L} dx \end{align*}

Then we obtain the following equation.

ancos⁡nπtL+bnsin⁡nπtL= (1L∫−LLf(x)cos⁡nπxLdx)cos⁡nπtL+(1L∫−LLf(x)sin⁡nπxLdx)sin⁡nπtL= 1L∫−LLf(x)[cos⁡nπxLcos⁡nπtL+sin⁡nπxLsin⁡nπtL]dx \begin{align*} & a_{n} \cos\dfrac{n\pi t}{L} + b_{n} \sin\dfrac{n\pi t}{L} \\ =&\ \left( \dfrac{1}{L}\int_{-L}^{L}f(x)\cos\dfrac{n\pi x}{L} dx \right) \cos\dfrac{n\pi t}{L} + \left( \dfrac{1}{L}\int_{-L}^{L}f(x)\sin\dfrac{n\pi x}{L} dx \right)\sin\dfrac{n\pi t}{L} \\ =&\ \dfrac{1}{L}\int_{-L}^{L}f(x) \left[ \cos\dfrac{n\pi x}{L} \cos\dfrac{n\pi t}{L} + \sin\dfrac{n\pi x}{L} \sin\dfrac{n\pi t}{L} \right] dx \end{align*}

Then, by the sum and difference identities for trigonometric functions, we get the following equation.

ancos⁡nπtL+bncos⁡nπtL=1L∫−LLf(x)[cos⁡nπ(x−t)L]dx a_{n}\cos\dfrac{n\pi t}{L} + b_{n}\cos\dfrac{n\pi t}{L} = \dfrac{1}{L}\int_{-L}^{L}f(x) \left[ \cos\dfrac{n\pi (x-t)}{L} \right] dx

Substituting this into (2)(2) yields the following.

SNf(t)= 12a0+∑n=1N(ancos⁡nπtL+bnsin⁡nπtL)= 121L∫−LLf(x)dx+∑n=1N(1L∫−LLf(x)[cos⁡nπ(x−t)L]dx)= 1L∫−LLf(x)[12+∑n=1Ncos⁡nπ(x−t)L]dx= 1L∫−LLf(x)Dn(π(x−t)L)dx \begin{align*} S_{N} ^{f} (t) =&\ \dfrac{1}{2}a_{0}+\sum \limits_{n=1}^{N} \left( a_{n}\cos\dfrac{n\pi t}{L}+b_{n}\sin\frac{n\pi t}{L} \right) \\ =&\ \dfrac{1}{2}\dfrac{1}{L}\int_{-L}^{L} f(x)dx+\sum\limits_{n=1}^{N}\left( \dfrac{1}{L}\int_{-L}^{L}f(x) \left[ \cos\dfrac{n\pi (x-t)}{L} \right] dx \right) \\ =&\ \dfrac{1}{L} \int_{-L}^{L} f(x) \left[ \dfrac{1}{2} + \sum\limits_{n=1}^{N} \cos \dfrac{n\pi (x-t)}{L}\right]dx \\ =&\ \dfrac{1}{L}\int_{-L}^{L}f(x)D_{n}\left(\dfrac{\pi (x-t)}{L}\right)dx \end{align*}

In the last equality, the definition of the Dirichlet kernel was used.

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Theorem 3

For any integer n∈Zn \in \mathbb{Z}, the following equation holds.

1L∫−LLDn(π(x−t)L)dx=1 \begin{equation} \dfrac{1}{L}\int_{-L}^{L}D_{n}\left( \dfrac{\pi (x-t)}{L} \right)dx=1 \end{equation}

Proof

Let’s substitute π(x−t)L=y\dfrac{\pi (x-t)}{L}=y. Then, the left side of (3)(3) is as follows.

1L∫−π−πLtπ−πLtDn(y)Lπdy= 1π∫−π−πLtπ−πLt(12+∑n=1Ncos⁡ny)dy by (1)= 12π∫−π−πLtπ−πLtdy+∑n=1N∫−π−πLtπ−πLtcos⁡nydy= 12π2π+∑n=1N∫−π−πLtπ−πLtcos⁡nydy= 1 \begin{align*} &\dfrac{1}{L}\int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} D_{n}(y) \dfrac{L}{\pi}dy \\ =&\ \dfrac{1}{\pi}\int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} \left( \dfrac{1}{2} + \sum \limits_{n=1}^{N} \cos ny \right) dy & \text{ by } (1) \\ =&\ \dfrac{1}{2\pi}\int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} dy+ \sum \limits_{n=1}^{N} \int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} \cos ny dy \\ =&\ \dfrac{1}{2\pi} 2\pi + \sum \limits_{n=1}^{N} \int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} \cos ny dy \\ =&\ 1 \end{align*}

At this point, the reason why the second term integrates to become 00 is because cos⁡0y=1\cos 0y=1 and cos⁡ny(n≠0)\cos ny (n\ne 0) are orthogonal to each other.

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