単位ステップシーケンス

教科書から、 DTFTは $u[n]$ 、

\begin{matrix} (1) & U (ω) = π δ (ω) + \frac{1}{1 - e^{- j ω}}, - π \leq ω < π \end{matrix}

$U(\omega)=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}},\qquad -\pi\le\omega <\pi\tag{1}$

ただし、多少の派生を与えるふりをするDSPの教科書を見たことはありません $(1)$ 。

Proakisは[1]の右側の右半分の導出 $(1)$ 設定することにより、 $z=e^{j\omega}$ で $\mathcal{Z}$ の-transform $u[n]$ 、及びそれが以外有効であることを言う $\omega=2\pi k$ （もちろん正しいです）。次に、 $\mathcal{Z}$ 変換の極で、面積がデルタインパルスを追加する必要 $\pi$ があると述べていますが、それは私にとって、他の何よりもレシピのように見えます。

オッペンハイムとシェーファー[2]この文脈での言及

完全に簡単に示すことはできませんが、このシーケンスは次のフーリエ変換で表すことができます。

相当する式が続き $(1)$ ます。残念ながら、彼らは「完全に単純ではない」証明を私たちに示すのに苦労しませんでした。

私が実際に知らなかったが、証明を探しているときに見つけた本 $(1)$ は、BA Shenoiによるデジタル信号処理とフィルター設計の紹介です。上のページ138そこの「派生」だ $(1)$ が、残念ながらそれは間違っています。「DSPパズル」の質問をして、その証明の何が悪いのかを人々に見せてもらいました。]

だから私の質問は：

数学的な傾向のあるエンジニアがアクセスできる一方で、証明/導出 $(1)$ は健全であるか、さらには厳密である可能性がありますか？本からコピーしただけでもかまいません。とにかくこのサイトに載せておくといいと思います。

math.SEであっても、関連するものはほとんどないことに注意してください。この質問には回答がありません。1つは2つの回答を持ち、1つは間違っており（Shenoiの主張と同じ）、もう1つは「蓄積プロパティ」を使用します。、私は満足していますが、そのプロパティを証明する必要があります。これにより、最初に戻ります（両方の証明は基本的に同じことを証明するため）。

最後のメモとして、私は証明のようなものを思いつきました（まあ、私はエンジニアです）、また数日後に回答として投稿しますが、他の公開または非公開の証明を収集させていただきますシンプルでエレガントであり、最も重要なのは、DSPエンジニアがアクセスできることです。

PS：妥当性を疑うことはありません $(1)$ 。1つまたはいくつかの比較的単純な証明が欲しいだけです。

[1] Proakis、JGおよびDG Manolakis、デジタル信号処理：原則、アルゴリズム、およびアプリケーション、第3版、セクション4.2.8

[2] Oppenheim、AVおよびRW Schafer、離散時間信号処理、第2版、p。54。

MarcusMüllerのコメントに触発されて、式

で与えられた

を示したいと思います。

要件を満たしている

U (ω)

$U(\omega)$

(1)

$(1)$

u [n] = u^{2} [n] \to U (ω) = \frac{1}{2 π} (U ⋆ U) (ω)

$u[n]=u^2[n]\rightarrow U(\omega)=\frac{1}{2\pi}(U\star U)(\omega)$

場合のDTFTある、次いで $U(\omega)$ $u[n]$

V (ω) = \frac{1}{1 - e^{- j ω}}

$V(\omega)=\frac{1}{1-e^{-j\omega}}$

のDTFTでなければなりません

v [n] = \frac{1}{2} sign [n]

$v[n]=\frac12\text{sign}[n]$

（ここで、を定義します $\text{sign}[0]=1$ ）、なぜなら

V (ω) = U (ω) - π δ (ω) ⟺ u [n] - \frac{1}{2} = \frac{1}{2} sign [n]

$V(\omega)=U(\omega)-\pi\delta(\omega)\Longleftrightarrow u[n]-\frac12=\frac12\text{sign}[n]$

だから私たちは

\frac{1}{2 π} (V ⋆ V) (ω) ⟺ {(\frac{1}{2} sign [n])}^{2} = \frac{1}{4}

$\frac{1}{2\pi}(V\star V)(\omega)\Longleftrightarrow \left(\frac12\text{sign}[n]\right)^2=\frac14$

それから続く

\frac{1}{2 π} (V ⋆ V) (ω) = DTFT {\frac{1}{4}} = \frac{π}{2} δ (ω)

$\frac{1}{2\pi}(V\star V)(\omega)=\text{DTFT}\left\{\frac14\right\}=\frac{\pi}{2}\delta(\omega)$

これで私たちは

\begin{aligned} \frac{1}{2 π} (U ⋆ U) (ω) & = \frac{1}{2 π} [(π δ (ω) + V (ω)) ⋆ (π δ (ω) + V (ω))] \\ = \frac{1}{2 π} [π^{2} δ (ω) + 2 π V (ω) + (V ⋆ V) (ω)] \\ = \frac{π}{2} δ (ω) + V (ω) + \frac{π}{2} δ (ω) \\ = U (ω) q.e.d. \end{aligned}

$\begin{align}\frac{1}{2\pi}(U\star U)(\omega)&=\frac{1}{2\pi}\left[(\pi\delta(\omega)+V(\omega))\star (\pi\delta(\omega)+V(\omega))\right]\\&=\frac{1}{2\pi}\left[\pi^2\delta(\omega)+2\pi V(\omega)+(V\star V)(\omega)\right]\\&=\frac{\pi}{2}\delta(\omega)+V(\omega)+\frac{\pi}{2}\delta(\omega)\\&=U(\omega)\qquad\text{q.e.d.}\end{align}$

— マットL.
ソース

u^{2} (t) = u (t)

$u^2(t) = u(t)$ , and hence (with a cont. FT definition prefactor depending constant

c

$c$ ),

\begin{aligned} DTFT (u^{2}) (ω) & = c \cdot U (ω) * U (ω) \\ = c π U (ω) + \frac{c}{1 + e^{- j ω}} * U (ω) \\ = c π (π δ (ω) + \frac{1}{1 - e^{- j ω}}) + \frac{c π}{1 + e^{- j ω}} + c \frac{1}{1 + e^{- j ω}} * \frac{1}{1 + e^{- j ω}} \\ = c π^{2} δ (ω) + \frac{2 c π}{1 + e^{- j ω}} + c \frac{1}{1 + e^{- j ω}} * \frac{1}{1 + e^{- j ω}} \\ \overset{magic?}{=} U \end{aligned}

$\begin{align}\text{DTFT}(u^2)(\omega) &=c\,\cdot\, U(\omega)*U(\omega) \\&=c\pi U(\omega)+ \frac{c}{1+e^{-j\omega}}*U(\omega)\\&=c\pi \left(\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\right)+\frac{c\pi}{1+e^{-j\omega}}+c\frac{1}{1+e^{-j\omega}}*\frac{1}{1+e^{-j\omega}}\\&=c\pi^2\delta(\omega)+\frac{2c\pi}{1+e^{-j\omega}} + c\frac{1}{1+e^{-j\omega}}*\frac{1}{1+e^{-j\omega}}\\&\overset{\text{magic?}}=U\end{align}$

u^{2} [n] = u [n]

$u^2[n]=u[n]$ works out for the given DTFT, no problem.

— Matt L.

I consider myself very simple-minded, and that means I worry when things don't feel "safe" when I can't see how they are derived.

— Marcus Müller

I see that what you're after is not to prove whether the equation is correct or not, but rather it's to rigorously and directly derive

U (w)

$U(w)$ from first principles and definition of DTFT. Then whenever one wants to make a rigorous proof involving impulses then I guess one should better refer to the cited books from generalized function theory: Lighthill-1958 is cited in Opp&Schafer for a discussion of impulse function and its use in Fourier transforms. All other proofs will inevitably rely on the proofs made on those references and will be insufficient to replace a rigorous proof.

— Fat32

@Fat32: That's a valid viewpoint. I think, however, that a reasonably sound derivation is possible if we accept basic transforms such as

DTFT {1} = 2 π δ (ω)

$\text{DTFT}\{1\}=2\pi\delta(\omega)$ , and if we're content to define integrals by their Cauchy principal value.

— Matt L.

回答:

Cedron Dawg posted an interesting initial point in this answer. It begins with these steps:

\begin{aligned} U (ω) & = \sum_{n = 0}^{+ \infty} e^{- j ω n} \\ = lim_{N \to \infty} \sum_{n = 0}^{N - 1} e^{- j ω n} \\ = lim_{N \to \infty} [\frac{1 - e^{- j ω N}}{1 - e^{- j ω}}] \\ = \frac{1}{1 - e^{- j ω}} - lim_{N \to \infty} [\frac{e^{- j ω N}}{1 - e^{- j ω}}] \end{aligned}

$\begin{align} U(\omega) &= \sum\limits_{n=0}^{+\infty} e^{-j \omega n} \\ &= \lim_{ N \to \infty } \sum\limits_{n=0}^{N-1} e^{-j \omega n}\\ &= \lim_{ N \to \infty } \left[ \frac{ 1 - e^{-j \omega N} }{ 1 - e^{-j \omega } } \right] \\ &= \frac{ 1 }{ 1 - e^{-j \omega } } - \lim_{ N \to \infty } \left[ \frac{ e^{-j \omega N} }{ 1 - e^{-j \omega } } \right] \end{align}$

It turns out the term inside the limit can be expanded as follows:

\begin{aligned} \frac{e^{- j ω N}}{1 - e^{- j ω}} = & \frac{1}{\sin^{2} (ω) + (1 - \cos (ω))^{2}} \cdot \\ [- \cos (ω) \cos (N ω) + \cos (N ω) - \sin (ω) \sin (N ω) + j (- \sin (ω) \cos (N ω) + \cos (ω) \sin (ω) - \sin (N ω))] \end{aligned}

$\begin{aligned} \frac{ e^{-j \omega N} }{ 1 - e^{-j \omega } } ={} &\frac{1}{\sin^2(\omega)+(1-\cos(\omega))^2} \cdot \\ &\left[-\cos(\omega)\cos(N\omega)+\cos(N\omega)-\sin(\omega)\sin(N\omega)+ \\ j(-\sin(\omega)\cos(N\omega)+\cos(\omega)\sin(\omega)-\sin(N\omega))\right] \end{aligned}$

The common factor outside the brackets can be expressed as:

\frac{1}{\sin^{2} (ω) + (1 - \cos (ω))^{2}} = \frac{1}{4 \sin^{2} (ω / 2)}

$\frac{1}{\sin^2(\omega)+(1-\cos(\omega))^2} = \frac{1}{4\sin^2(\omega/2)}$

The real part inside the brackets also equals:

- \cos (ω) \cos (N ω) + \cos (N ω) - \sin (ω) \sin (N ω) = 2 \sin (ω / 2) \sin [ω (- N + 1 / 2)]

$-\cos(\omega)\cos(N\omega)+\cos(N\omega)-\sin(\omega)\sin(N\omega)= 2\sin(\omega/2)\sin[\omega(-N+1/2)]$

On the other hand, the imaginary part can be rewritten as:

- \sin (ω) \cos (N ω) + \cos (ω) \sin (ω) - \sin (N ω) = - 2 \sin (ω / 2) \cos [ω (- N + 1 / 2)]

$-\sin(\omega)\cos(N\omega)+\cos(\omega)\sin(\omega)-\sin(N\omega)=-2\sin(\omega/2)\cos[\omega(-N+1/2)]$

Rewritting the original term we get that:

\begin{aligned} \frac{e^{- j ω N}}{1 - e^{- j ω}} & = \frac{2 \sin (\frac{ω}{2})}{4 \sin^{2} (\frac{ω}{2})} (\sin [ω (- N + 1 / 2)] - j \cos [ω (- N + 1 / 2)]) \\ = - \frac{\sin [ω (M + 1 / 2)]}{2 \sin (\frac{ω}{2})} - j \frac{\cos [ω (M + 1 / 2)]}{2 \sin (\frac{ω}{2})} \end{aligned}

$\begin{align} \frac{ e^{-j \omega N} }{ 1 - e^{-j \omega } } &=\frac{2\sin\left(\frac \omega 2\right)}{4\sin^2\left(\frac \omega 2\right) } \left( \sin[\omega(-N+1/2)] - j\cos[\omega(-N+1/2)]\right) \\ &=-\frac{\sin[\omega(M+1/2)]}{2\sin\left(\frac \omega 2\right)} - j\frac{\cos[\omega(M+1/2)]}{2\sin\left(\frac \omega 2\right)} \end{align}$

where I used $M=N-1$ and the limit stays unaffected as $M\to \infty$ as well.

According to the 7th definition in this site:

lim_{M \to \infty} - \frac{1}{2 \sin (ω / 2)} \sin [ω (M + 1 / 2)] = - π δ (ω)

$\lim_{M\to \infty} -\frac{1}{2\sin(\omega/2)}\sin[\omega(M+1/2)] = -\pi \delta(\omega)$

So far we have that:

lim_{M \to \infty} \frac{e^{- j ω (M + 1)}}{1 - e^{- j ω}} = - π δ (ω) - j lim_{M \to \infty} \frac{\cos [ω (M + 1 / 2)]}{2 \sin (ω / 2)}

$\lim_{ M \to \infty } \frac{ e^{-j \omega (M+1)} }{ 1 - e^{-j \omega } } =-\pi\delta(\omega) -j\lim_{M\to \infty} \frac{\cos[\omega(M+1/2)]}{2\sin(\omega/2)}$

If we could prove that the second term on the right of the equality is $0$ in some sense, then we are done. I asked it at math.SE and, indeed, that sequence of functions tends to the zero distribution. So, we have that:

\begin{aligned} U (ω) & = \frac{1}{1 - e^{- j ω}} - lim_{N \to \infty} [\frac{e^{- j ω N}}{1 - e^{- j ω}}] \\ = \frac{1}{1 - e^{- j ω}} + π δ (ω) + j lim_{M \to \infty} \frac{\cos [ω (M + 1 / 2)]}{2 \sin (ω / 2)} \\ = \frac{1}{1 - e^{- j ω}} + π δ (ω) \end{aligned}

$\begin{align} U(\omega) &= \frac{ 1 }{ 1 - e^{-j \omega } } - \lim_{ N \to \infty } \left[ \frac{ e^{-j \omega N} }{ 1 - e^{-j \omega } } \right] \\ &=\frac{ 1 }{ 1 - e^{-j \omega } }+\pi\delta(\omega) +j\lim_{M\to \infty} \frac{\cos[\omega(M+1/2)]}{2\sin(\omega/2)} \\ & =\frac{ 1 }{ 1 - e^{-j \omega } }+\pi\delta(\omega) \end{align}$

— Tendero
ソース

This is very nice! I checked it and everything seems to be correct, so that imaginary part must tend to zero in some sense. I'll think about it for a bit.

— Matt L.

@MattL. Let me know if you are able to make any progress!

— Tendero

@MattL. The proof is finally complete!

— Tendero

Good work! I had figured out that the cosine term would tend to zero due to the Riemann-Lebesgue lemma, but my problem was the case

ω = 0

$\omega=0$ . Because the very first formula is based on the geometric sum, which is only valid for

ω \neq 0

$\omega\neq 0$ . It all somehow works out after all, but that's still a minor flaw. I have another derivation that does not split out the term

1 / (1 - e^{- j ω})

$1/(1-e^{-j\omega})$ , in which the case

ω = 0

$\omega=0$ is handled with a bit more care, but it's still an "engineer's proof". I might post it when I have more time.

— Matt L.

I'll provide two relatively simple proofs that do not require any knowledge of distribution theory. For a proof that computes the DTFT by a limit process using results from distribution theory, see this answer by Tendero.

I will only mention (and not elaborate on) the first proof here, because I've posted it as an answer to this question, the purpose of which was to show that a certain published proof is faulty.

The other proof goes as follows. Let's first write down the even part of the unit step sequence $u[n]$ :

\begin{matrix} (1) & u_{e} [n] = \frac{1}{2} (u [n] + u [- n]) = \frac{1}{2} + \frac{1}{2} δ [n] \end{matrix}

$u_e[n]=\frac12\left(u[n]+u[-n]\right)=\frac12+\frac12\delta[n]\tag{1}$

The DTFT of $(1)$ is

\begin{matrix} (2) & DTFT {u_{e} [n]} = π δ (ω) + \frac{1}{2} \end{matrix}

$\text{DTFT}\{u_e[n]\}=\pi\delta(\omega)+\frac12\tag{2}$

which equals the real part of the DTFT of $u[n]$ :

\begin{matrix} (3) & U_{R} (ω) = Re {U (ω)} = π δ (ω) + \frac{1}{2} \end{matrix}

$U_R(\omega)=\text{Re}\{U(\omega)\}=\pi\delta(\omega)+\frac12\tag{3}$

Since $u[n]$ is a real-valued sequence we're done because the real and imaginary parts of $U(\omega)$ are related via the Hilbert transform, and, consequently, $U_R(\omega)$ uniquely determines $U(\omega)$ . However, in most DSP texts, these Hilbert transform relations are derived from the equation $h[n]=h[n]u[n]$ (which is valid for any causal sequence $h[n]$ ), from which it follows that $H(\omega)=\frac{1}{2\pi}(H\star U)(\omega)$ . So in order to show the Hilbert transform relation between the real and imaginary parts of the DTFT we need the DTFT of $u[n]$ , which we actually want to derive here. So the proof becomes circular. That's why we'll choose a different way to derive the imaginary part of $U(\omega)$ .

For deriving $U_I(\omega)=\text{Im}\{U(\omega)\}$ we write the odd part of $u[n]$ as follows:

\begin{matrix} (4) & u_{o} [n] = \frac{1}{2} (u [n] - u [- n]) = u [n - 1] - \frac{1}{2} + \frac{1}{2} δ [n] \end{matrix}

$u_o[n]=\frac12\left(u[n]-u[-n]\right)=u[n-1]-\frac12+\frac12\delta[n]\tag{4}$

Taking the DTFT of $(4)$ gives

\begin{aligned} j U_{I} (ω) & = e^{- j ω} U (ω) - π δ (ω) + \frac{1}{2} \\ = e^{- j ω} (U_{R} (ω) + j U_{I} (ω)) - π δ (ω) + \frac{1}{2} \\ = e^{- j ω} (π δ (ω) + \frac{1}{2}) + e^{- j ω} j U_{I} (ω) - π δ (ω) + \frac{1}{2} \\ (5) & = \frac{1}{2} (1 + e^{- j ω}) + e^{- j ω} j U_{I} (ω) \end{aligned}

$\begin{align}jU_I(\omega)&=e^{-j\omega}U(\omega)-\pi\delta(\omega)+\frac12\\&=e^{-j\omega}(U_R(\omega)+jU_I(\omega))-\pi\delta(\omega)+\frac12\\&=e^{-j\omega}\left(\pi\delta(\omega)+\frac12\right)+e^{-j\omega}jU_I(\omega)-\pi\delta(\omega)+\frac12\\&=\frac12(1+e^{-j\omega})+e^{-j\omega}jU_I(\omega)\tag{5}\end{align}$

where I've used $(3)$ . Eq. $(5)$ can be written as

\begin{matrix} (6) & j U_{I} (ω) (1 - e^{- j ω}) = \frac{1}{2} (1 + e^{- j ω}) \end{matrix}

$jU_I(\omega)(1-e^{-j\omega})=\frac12(1+e^{-j\omega})\tag{6}$

The correct conclusion from $(6)$ is (see this answer for more details)

\begin{matrix} (7) & j U_{I} (ω) = \frac{1}{2} \frac{1 + e^{- j ω}}{1 - e^{- j ω}} + c δ (ω) \end{matrix}

$jU_I(\omega)=\frac12\frac{1+e^{-j\omega}}{1-e^{-j\omega}}+c\delta(\omega)\tag{7}$

But since we know that $U_I(\omega)$ must be an odd function of $\omega$ (because $u[n]$ is real-valued), we can immediately conclude that $c=0$ . Hence, from $(3)$ and $(7)$ we finally get

\begin{aligned} U (ω) & = U_{R} (ω) + j U_{I} (ω) \\ = π δ (ω) + \frac{1}{2} + \frac{1}{2} \frac{1 + e^{- j ω}}{1 - e^{- j ω}} \\ = π δ (ω) + \frac{1}{2} (1 + \frac{1 + e^{- j ω}}{1 - e^{- j ω}}) \\ (8) & = π δ (ω) + \frac{1}{1 - e^{- j ω}} \end{aligned}

$\begin{align}U(\omega)&=U_R(\omega)+jU_I(\omega)\\&=\pi\delta(\omega)+\frac12+\frac12\frac{1+e^{-j\omega}}{1-e^{-j\omega}}\\&=\pi\delta(\omega)+\frac12\left( 1+\frac{1+e^{-j\omega}}{1-e^{-j\omega}}\right)\\&=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\tag{8}\end{align}$

— Matt L.
ソース