Shannon's sampling theorem states that a continuous-time signal that is band-limited to a maximum frequency of fmax" role="presentation" style="position: relative;">fmaxfmaxf\_{max} can be reconstructed from its uniformly spaced samples, as long as the sampling rate is at least 2fmax" role="presentation" style="position: relative;">2fmax2fmax2f\_{max} samples per second.

In the case of sampling a band-limited signal with an ideal sampling function (Dirac comb) by multiplying the signal with the Dirac comb, the resulting sampled signal is given by:

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

where x(t)" role="presentation" style="position: relative;">x(t)x(t)x(t) is the continuous-time signal, Ts" role="presentation" style="position: relative;">TsTsT\_s is the sampling period, and &#x03B4;(t)" role="presentation" style="position: relative;">δ(t)δ(t)\delta(t) is the Dirac delta function. The resulting spectrum of the sampled signal can be obtained by taking the Fourier transform of the sampled signal:

Xs(f)=Fxs(t)=Fx(t)&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)" role="presentation" style="text-align: center; position: relative;">Xs(f)=Fxs(t)=Fx(t)∑n=−∞∞δ(t−nTs)Xs(f)=Fxs(t)=Fx(t)∑n=−∞∞δ(t−nTs) X\_s(f) = \mathcal{F}{x\_s(t)} = \mathcal{F}{x(t) \sum\_{n=-\infty}^{\infty} \delta(t - nT\_s)} 

Using the property of the Fourier transform that the Fourier transform of a product of two functions is the convolution of their Fourier transforms, we have:

Xs(f)=Fx(t)&#x2217;F&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)" role="presentation" style="text-align: center; position: relative;">Xs(f)=Fx(t)∗F∑n=−∞∞δ(t−nTs)Xs(f)=Fx(t)∗F∑n=−∞∞δ(t−nTs) X\_s(f) = \mathcal{F}{x(t)} \* \mathcal{F}{\sum\_{n=-\infty}^{\infty} \delta(t - nT\_s)} 

The Fourier transform of the Dirac comb is given by:

F&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)=1Ts&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(f&#x2212;n/Ts)" role="presentation" style="text-align: center; position: relative;">F∑n=−∞∞δ(t−nTs)=1Ts∑n=−∞∞δ(f−n/Ts)F∑n=−∞∞δ(t−nTs)=1Ts∑n=−∞∞δ(f−n/Ts) \mathcal{F}{\sum\_{n=-\infty}^{\infty} \delta(t - nT\_s)} = \frac{1}{T\_s} \sum\_{n=-\infty}^{\infty} \delta(f - n/T\_s) 

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

Xs(f)=1Ts&#x2211;n=&#x2212;&#x221E;&#x221E;X(f&#x2212;n/Ts)" role="presentation" style="text-align: center; position: relative;">Xs(f)=1Ts∑n=−∞∞X(f−n/Ts)Xs(f)=1Ts∑n=−∞∞X(f−n/Ts) X\_s(f) = \frac{1}{T\_s} \sum\_{n=-\infty}^{\infty} X(f - n/T\_s) 

where X(f)" role="presentation" style="position: relative;">X(f)X(f)X(f) is the Fourier transform of the continuous-time signal x(t)" role="presentation" style="position: relative;">x(t)x(t)x(t). This is known as the Fourier series of the sampled signal.

The spectrum of the sampled signal consists of replicas of the spectrum of the continuous-time signal, spaced at intervals of 1/Ts" role="presentation" style="position: relative;">1/Ts1/Ts1/T\_s in frequency. If the continuous-time signal is band-limited to a maximum frequency of fmax" role="presentation" style="position: relative;">fmaxfmaxf\_{max}, then the spectrum of the sampled signal will be band-limited to the range [&#x2212;fmax,fmax]" role="presentation" style="position: relative;">[−fmax,fmax][−fmax,fmax][-f\_{max}, f\_{max}], with the first replica centered at f=0" role="presentation" style="position: relative;">f=0f=0f=0 and subsequent replicas centered at multiples of 1/Ts" role="presentation" style="position: relative;">1/Ts1/Ts1/T\_s.

In the case where the sampling rate is exactly twice the bandwidth of the signal (i.e., Ts=1/2fmax" role="presentation" style="position: relative;">Ts=1/2fmaxTs=1/2fmaxT\_s = 1/2f\_{max}), the first replica of the spectrum will be centered at f=fmax" role="presentation" style="position: relative;">f=fmaxf=fmaxf=f\_{max} and will exactly overlap with the original spectrum of the continuous-time signal, without any distortion or aliasing. This is known as the Nyquist rate. If the sampling rate is lower than the Nyquist rate, then the replicas will overlap and cause aliasing, resulting in distortion of the reconstructed signal.
한논의 샘플링 정리에 따르면 최대 주파수 fmax" role="presentation" style="위치: 상대적;">fmaxfmaxf\_{max}로 대역이 제한된 연속 시간 신호는 샘플링 속도가 최소 2fmax" role="presentation" style="위치: 상대적;">2fmax2fmax2f\_{max} 초당 샘플로부터 재구성될 수 있습니다.이상적인 샘플링 함수(디락 콤)를 사용하여 대역이 제한된 신호를 샘플링하는 경우 신호에 디락 콤을 곱하여 샘플링된 결과 신호는 다음과 같이 주어집니다:xs(t)=x(t)&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)" role="presentation" style="text-align: center; position: 상대;">xs(t)=x(t)∑n=-∞∞δ(t-nTs)xs(t)=x(t)∑n=-∞∞δ(t-nTs) x\_s(t) = x(t) \sum\_{n=-\infty}^{\infty} \델타(t - nT\_s) 여기서 x(t)" role="presentation" style="위치: 상대적;">x(t)x(t)x(t)는 연속 시간 신호, Ts" role="presentation" style="위치: 상대적;">TsTsT\_s는 샘플링 주기, &#x03B4;(t)" role="presentation" style="위치: 상대적;">δ(t)δ(t)\delta(t)는 Dirac 델타 함수입니다. 샘플링된 신호의 결과 스펙트럼은 샘플링된 신호의 푸리에 변환을 취하여 얻을 수 있습니다:Xs(f)=Fxs(t)=Fx(t)&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)" role="presentation" style="text-align: center; position: 상대적;">Xs(f)=Fxs(t)=Fx(t)∑n=-∞∞δ(t-nTs)Xs(f)=Fxs(t)=Fx(t)∑n=-∞∞δ(t-nTs) X\_s(f) = \mathcal{F}{x\_s(t)} = \mathcal{F}{x(t) \sum\_{n=-\infty}^{\infty} \델타(t - nT\_s)}. 두 함수의 곱의 푸리에 변환은 두 함수의 푸리에 변환의 컨볼루션이라는 푸리에 변환의 속성을 사용하여 다음과 같은 식을 구할 수 있습니다:Xs(f)=Fx(t)&#x2217;F&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)" role="presentation" style="text-align: center; position: 상대적;">Xs(f)=Fx(t)∗F∑n=-∞∞δ(t-nTs)Xs(f)=Fx(t)∗F∑n=-∞∞δ(t-nTs) X\_s(f) = \mathcal{F}{x(t)} \* \mathcal{F}{\sum\_{n=-\인프티}^{\인프티} \델타(t - nT\_s)}. 디락 빗의 푸리에 변환은 다음과 같이 주어집니다:F&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(t&#x2212;nTs)=1Ts&#x2211;n=&#x2212;&#x221E;&#x221E;&#x03B4;(f&#x2212;n/Ts)" role="presentation" style="text-align: center; position: 상대;">F∑n=-∞∞δ(t-nTs)=1Ts∑n=-∞∞δ(f-n/Ts)F∑n=-∞∞δ(t-nTs)=1Ts∑n=-∞∞δ(f-n/Ts) \mathcal{F}{\sum\_{n=-\infty}^{\infty} \델타(t - nT\_s)} = \frac{1}{T\_s} \sum\_{n=-\infty}^{\infty} \델타(f - n/T\_s) 이를 Xs(f)"role="presentation" style="위치: 상대적;">Xs(f)Xs(f)X\_s(f)에 대한 식에 대입하면 다음과 같이 됩니다:Xs(f)=1Ts&#x2211;n=&#x2212;&#x221E;&#x221E;X(f&#x2212;n/Ts)" role="presentation" style="text-align: 가운데; 위치: 상대적;">Xs(f)=1Ts∑n=-∞∞X(f-n/Ts)Xs(f)=1Ts∑n=-∞∞X(f-n/Ts) X\_s(f) = \frac{1}{T\_s} \sum\_{n=-\infty}^{\infty} X(f - n/T\_s) 여기서 X(f)" role="presentation" style="위치: 상대적;">X(f)는 연속 시간 신호 x(t)의 푸리에 변환입니다." role="presentation" style="위치: 상대적;">x(t)x(t)x(t)x(t). 이를 샘플링된 신호의 푸리에 급수라고 합니다.샘플링된 신호의 스펙트럼은 연속 시간 신호 스펙트럼의 복제본으로 구성되며, 주파수는 1/Ts" role="presentation" style="위치: 상대적;">1/Ts1/Ts1/T\_s 간격으로 배치됩니다. 연속 시간 신호의 대역이 최대 주파수 fmax로 제한되어 있는 경우" role="presentation" style="위치: 상대적;">fmaxfmaxf\_{max}, 샘플링된 신호의 스펙트럼은 [&#x2212;fmax,fmax] 범위로 대역이 제한됩니다." role="presentation" style="위치: 상대적;">[-fmax,fmax][-fmax,fmax][-f\_{최대}, f\_{최대}], 첫 번째 복제본은 f=0에 중심" role="presentation" style="위치: 상대적;">f=0f=0f=0 및 후속 복제본은 1/Ts의 배수에 중심" role="presentation" style="위치: 상대적;"> 1/Ts1/Ts1/T\_s.샘플링 속도가 신호 대역폭의 정확히 두 배인 경우(즉,, Ts=1/2fmax" role="presentation" style="위치: 상대적;">Ts=1/2fmaxTs=1/2fmaxT\_s = 1/2f\_{max}), 스펙트럼의 첫 번째 복제본은 f=fmax" role="presentation" style="위치: 상대적;">f=fmaxf=fmaxf=f\_{max}의 중심에 위치하며 왜곡이나 앨리어싱 없이 연속 시간 신호의 원래 스펙트럼과 정확히 겹치게 됩니다. 이를 나이퀴스트 속도라고 합니다. 샘플링 속도가 나이퀴스트 속도보다 낮으면 복제본이 겹쳐서 에일리어싱이 발생하여 재구성된 신호가 왜곡됩니다.