Sampling at the Nyquist Rate

August 1, 2019

In Example 1, the time waveform x(t) was sampled at π–₯s = 4 Γ— 𝖿h. The resulting DTFT spectral images were spread out in the DTFT frequency domain.

By examining these results, it is apparent that the minimum sample rate is π–₯s = 2 Γ— 𝖿h. Anything less would result in the overlap of spectral images.

The π–₯s = 2Β  Γ— 𝖿h rate is called the Nyquist rate. Example 5 is similar to Example 4, but the sample rate π–₯s is reduced to the Nyquist rate.

Frequency Content of a Waveform Sampled at Nyquist Rate (Ex. 5)

Given the time waveform illustrated in Figure 2.1:

(1)   \begin{equation*} x(t)=\frac{1}{\sqrt{2\pi}}\left[\frac{\sin(t/2)}{t/2}\right]^2 \triangleq \frac{1}{\sqrt{2\pi}}sinc^2\left(\frac{t}{2}\right) \end{equation*}

The Fourier transform 𝖷̃(πœ”) is the triangle function illustrated in Figure 2.3 with the highest frequency component of 𝖿h = 1/2πœ‹ Hz.

The waveform x(t) sampled at π–₯s β‰œ 2 Γ— 𝖿h = 4 Γ— 1/2πœ‹ = 1/πœ‹ gives a sample period of 𝜏 = 1/π–₯s = πœ‹ and is illustrated in Figure 2.5.

Sample sequence

Figure 2.5. Sample sequence.

Using this sample rate and Theorem 1, a DTFT 𝖸̃(πœ”) of y(n) is expressed as:

(2)   \begin{equation*} \tilde{Y}(\omega)=\sqrt{2\pi}F_{s}\sum_{n\in\mathbb{Z}}\tilde{X}(F_{s}[\omega-2\pi n]) \end{equation*}

(3)   \begin{equation*} =\frac{\sqrt{2\pi}}{\pi}\sum_{n\in\mathbb{Z}}\tilde{X}\left(\frac{1}{\pi}[\omega-2\pi n]\right) \end{equation*}


  • 𝖷̃(πœ”) repeats every 2πœ‹ (this is true of all DTFTs)
  • 𝖷̃(πœ”) reaches 0 when (1/πœ‹) πœ” = 1 or πœ” = πœ‹
  • The height of each triangle is \frac{\sqrt{2\pi}}{\pi} or about 0.798 (see Figures 2.6 and 2.7)
Fourier transform

Figure 2.6. Fourier transform.


Figure 2.7. DTFT.