The Shannon sampling theorem states that in order to perfectly reconstruct a band-limited signal, it must be sampled at a rate at least twice the bandwidth of the signal. This can be achieved by multiplying the signal with an ideal sampling function, which is represented by a Dirac comb.

The Dirac comb is defined as:

&#x03B4;T(t)=&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nT)" role="presentation" style="text-align: center; position: relative;">δT(t)=∑n=−∞∞δ(t−nT)δT(t)=∑n=−∞∞δ(t−nT)\delta\_T(t) = \sum\_{n=-\infty}^{\infty} \delta(t-nT)

where &#x03B4;(t)" role="presentation" style="position: relative;">δ(t)δ(t)\delta(t) is the Dirac delta function, T" role="presentation" style="position: relative;">TTT is the sampling period (equal to 12B" role="presentation" style="position: relative;">12B12B\frac{1}{2B}, where B" role="presentation" style="position: relative;">BBB is the bandwidth of the signal), and n" role="presentation" style="position: relative;">nnn is an integer.

Multiplying a band-limited signal x(t)" role="presentation" style="position: relative;">x(t)x(t)x(t) with the Dirac comb yields:

xs(t)=x(t)&#x03B4;T(t)=&#x2211;n=&#x2212;&#x221E;&#x221E;x(nT)&#x03B4;(t&#x2212;nT)" role="presentation" style="text-align: center; position: relative;">xs(t)=x(t)δT(t)=∑n=−∞∞x(nT)δ(t−nT)xs(t)=x(t)δT(t)=∑n=−∞∞x(nT)δ(t−nT)x\_s(t) = x(t) \delta\_T(t) = \sum\_{n=-\infty}^{\infty} x(nT) \delta(t-nT)

Taking the Fourier transform of xs(t)" role="presentation" style="position: relative;">xs(t)xs(t)x\_s(t) gives:

Xs(f)=&#x222B;&#x2212;&#x221E;&#x221E;xs(t)e&#x2212;j2&#x03C0;ftdt" role="presentation" style="text-align: center; position: relative;">Xs(f)=∫∞−∞xs(t)e−j2πftdtXs(f)=∫−∞∞xs(t)e−j2πftdtX\_s(f) = \int\_{-\infty}^{\infty} x\_s(t) e^{-j2\pi ft} dt

Substituting the expression for xs(t)" role="presentation" style="position: relative;">xs(t)xs(t)x\_s(t) and using the sifting property of the Dirac delta function, we get:

Xs(f)=&#x2211;n=&#x2212;&#x221E;&#x221E;x(nT)&#x222B;&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nT)e&#x2212;j2&#x03C0;ftdt" role="presentation" style="text-align: center; position: relative;">Xs(f)=∑n=−∞∞x(nT)∫∞−∞δ(t−nT)e−j2πftdtXs(f)=∑n=−∞∞x(nT)∫−∞∞δ(t−nT)e−j2πftdtX\_s(f) = \sum\_{n=-\infty}^{\infty} x(nT) \int\_{-\infty}^{\infty} \delta(t-nT) e^{-j2\pi ft} dt

The integral evaluates to:

&#x222B;&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nT)e&#x2212;j2&#x03C0;ftdt=e&#x2212;j2&#x03C0;fnT" role="presentation" style="text-align: center; position: relative;">∫∞−∞δ(t−nT)e−j2πftdt=e−j2πfnT∫−∞∞δ(t−nT)e−j2πftdt=e−j2πfnT\int\_{-\infty}^{\infty} \delta(t-nT) e^{-j2\pi ft} dt = e^{-j2\pi fnT}

Substituting this back into the expression for Xs(f)" role="presentation" style="position: relative;">Xs(f)Xs(f)X\_s(f), we get:

Xs(f)=&#x2211;n=&#x2212;&#x221E;&#x221E;x(nT)e&#x2212;j2&#x03C0;fnT" role="presentation" style="text-align: center; position: relative;">Xs(f)=∑n=−∞∞x(nT)e−j2πfnTXs(f)=∑n=−∞∞x(nT)e−j2πfnTX\_s(f) = \sum\_{n=-\infty}^{\infty} x(nT) e^{-j2\pi fnT}

This is the spectrum of the sampled signal. It consists of an infinite sum of replicas of the original spectrum centered at multiples of the sampling frequency 1T" role="presentation" style="position: relative;">1T1T\frac{1}{T}. The amplitude of each replica is scaled by the value of the original signal at the corresponding sampling instant.

To avoid aliasing, the signal must be band-limited to less than half the sampling frequency (i.e., the Nyquist frequency), which is equal to 12T" role="presentation" style="position: relative;">12T12T\frac{1}{2T}.
섀넌 샘플링 정리에 따르면 대역이 제한된 신호를 완벽하게 재구성하려면 신호 대역폭의 최소 두 배 이상의 속도로 샘플링해야 합니다. 이는 신호에 디락 빗으로 표현되는 이상적인 샘플링 함수를 곱함으로써 달성할 수 있습니다.

디락 빗은 다음과 같이 정의됩니다:

&#x03B4;T(t)=&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nT)" role="presentation" style="text-align: center; position: 상대;">δT(t)=∑n=-∞∞δ(t-nT)δT(t)=∑n=-∞∞δ(t-nT)\delta\_T(t) = \sum\_{n=-\infty}^{\infty} \델타(t-nT)

여기서 &#x03B4;(t)" role="presentation" style="위치: 상대적;">δ(t)δ(t)\delta(t)는 디락 델타 함수, T" role="presentation" style="위치: 상대적;">TTT는 샘플링 주기(12B와 같음." role="presentation" style="position: 상대적;">12B12B\frac{1}{2B}, 여기서 B" role="presentation" style="위치: 상대적;">BBB는 신호의 대역폭)이고, n" role="presentation" style="위치: 상대적;">nnn은 정수입니다.

대역이 제한된 신호 x(t)" role="presentation" style="위치: 상대적;">x(t)x(t)x(t)에 디랙 빗을 곱하면 다음과 같은 결과가 나옵니다:

xs(t)=x(t)&#x03B4;T(t)=&#x2211;n=&#x2212;&#x221E;&#x221E;x(nT)&#x03B4;(t&#x2212;nT)" role="presentation" style="text-align: center; position: 상대적;">xs(t)=x(t)δT(t)=∑n=-∞∞x(nT)δ(t-nT)xs(t)=x(t)δT(t)=∑n=-∞∞x(nT)δ(t-nT)x\_s(t) = x(t) \delta\_T(t) = \sum\_{n=-\infty}^{\infty} x(nT) \delta(t-nT)

xs(t)의 푸리에 변환"role="presentation" style="위치: 상대적;">xs(t)xs(t)x\_s(t)를 구하면 다음과 같습니다:

Xs(f)=&#x222B;&#x2212;&#x221E;&#x221E;xs(t)e&#x2212;j2&#x03C0;ftdt" role="presentation" style="text-align: center; position: 상대적;">Xs(f)=∫∞-∞xs(t)e-j2πftdtXs(f)=∫-∞∞xs(t)e-j2πftdtX\_s(f) = \int\_{-\인프티}^{\인프티} x\_s(t) e^{-j2\pi ft} dt

이 식을 xs(t)"role="presentation" style="위치: 상대적;">xs(t)xs(t)x\_s(t)에 대입하고 디랙 델타 함수의 시프팅 속성을 사용하면 다음과 같이 구할 수 있습니다:

Xs(f)=&#x2211;n=&#x2212;&#x221E;&#x221E;x(nT)&#x222B;&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nT)e&#x2212;j2&#x03C0;ftdt" role="presentation" style="text-align: center; position: 상대적;">Xs(f)=∑n=-∞∞x(nT)∫∞-∞δ(t-nT)e-j2πftdtXs(f)=∑n=-∞∞x(nT)∫-∞∞δ(t-nT)e-j2πftdtX\_s(f) = \sum\_{n=-\infty}^{\infty} x(nT) \int\_{-\infty}^{\infty} \델타(t-nT) e^{-j2\pi ft} dt

적분은 다음과 같이 평가됩니다:

&#x222B;&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nT)e&#x2212;j2&#x03C0;ftdt=e&#x2212;j2&#x03C0;fnT" role="presentation" style="text-align: center; 위치: 상대적;">∫∞-∞δ(t-nT)e-j2πftdt=e-j2πfnT∫-∞∞δ(t-nT)e-j2πftdt=e-j2πfnT\int\_{-\인프티}^{\인프티} \델타(t-nT) e^{-j2\pi ft} dt = e^{-j2\pi fnT}

이를 다시 Xs(f)" role="presentation" style="위치: 상대적;">Xs(f)Xs(f)X\_s(f)에 대한 식에 대입하면 다음과 같습니다:

Xs(f)=&#x2211;n=&#x2212;&#x221E;&#x221E;x(nT)e&#x2212;j2&#x03C0;fnT" role="presentation" style="text-align: center; position: 상대적;">Xs(f)=∑n=-∞∞x(nT)e-j2πfnTXs(f)=∑n=-∞∞x(nT)e-j2πfnTX\_s(f) = \sum\_{n=-\infty}^{\infty} x(nT) e^{-j2\pi fnT}

이것은 샘플링된 신호의 스펙트럼입니다. 샘플링 주파수 1T의 배수를 중심으로 한 원본 스펙트럼의 복제본의 무한 합으로 구성됩니다." role="presentation" style="위치: 상대적;">1T1T\frac{1}{T}. 각 복제본의 진폭은 해당 샘플링 시점의 원본 신호 값에 따라 스케일링됩니다.

에일리어싱을 방지하려면 신호의 대역이 샘플링 주파수(즉, 나이퀴스트 주파수)의 절반 미만으로 제한되어야 하며, 이는 12T와 같습니다." role="presentation" style="위치: 상대적;">12T12T\frac{1}{2T}.