DTFT of a Sampled Sequence

July 24, 2019

Suppose the following: There is a signal x(t) with a Fourier transform of \tilde{X}(\omega). The signal is sampled every 𝜏 seconds yielding the sequence y(n) = x(n𝜏). From the sampled sequence, a calculation yields the DTFT \tilde{Y}(\omega) of y(n).

Now, there is a Fourier transform \tilde{X}(\omega) and a DTFT \tilde{Y}(\omega). They come from two different transform definitions but originated from the same signal, x(t).

This prompts the question: do \tilde{X}(\omega) and \tilde{Y}(\omega) look anything alike? The answer is yes; in fact, \tilde{Y}(\omega) looks like a scaled and repeated version of \tilde{X}(\omega). Below, Theorem 1 and Example 1 offer a more precise description as to why this is.

Theorem 1

Defining the DTFT of a sampled sequence in terms of the FT of the sampled waveform

Let 𝗒(𝑛) β‰œ 𝗑(π‘›πœ) be the sample sequence of a waveform 𝗑(𝑑) and π–₯s = 1/𝜏 be the sample rate. Let 𝖸̆(πœ”) be the DTFT of 𝗒(𝑛) and 𝖷̃ (πœ”) be the Fourier transform of 𝗑(𝑑). Then:

(1)   \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*}


This theorem is difficult to prove without resorting to β€œengineering hack” methodology such as β€œmultiply the waveform by a train of impulse functions.” However, the proof becomes quite manageable by leveraging the somewhat obscure yet extremely powerful inverse Poisson summation formula (IPSF).

(2)   \begin{equation*} \tilde{Y}(\omega)\triangleq\sum_{n\in\mathbb{Z}}y(n)e^{-i\omega n}\text{ by definition of DTFT} \end{equation*}

(3)   \begin{equation*} =\sum_{n\in\mathbb{Z}}x(n\tau)e^{-i\omega n}\text{ by definition of }y(n) \end{equation*}

(4)   \begin{equation*} =\sum_{n\in\mathbb{Z}}x(n\tau)e^{-i\frac{\omega}{\tau}n\tau}\text{ because }\frac{\tau}{\tau}=1 \end{equation*}

(5)   \begin{equation*} =\frac{\sqrt{2\pi}}{\tau}\sum_{n\in\mathbb{Z}}\tilde{X}\left(\frac{\omega}{\tau}+\frac{2\pi}{\tau}n\right)\text{ by IPSF} \end{equation*}

(6)   \begin{equation*} =\sqrt{2\pi}\text{F}_{s}\sum_{n\in\mathbb{Z}}\tilde{X}(\text{F}_{s}[\omega-2\pi n])\text{ by }\text{F}_{s}=1/\tau\text{ and absolute summability} \end{equation*}

The DTFT of the sample sequence is the Fourier transform of the original waveform. It is:

  1. Repeated every 2πœ‹ radians (in discrete world frequency, which is every 2 Γ— π–₯s Hertz in real-world frequency)
  2. Dilated (made β€œnarrower”) by a factor of π–₯s
  3. Scaled (made β€œtaller”) by a factor of √2πœ‹π–₯s

Figures 2.1-2.4 illustrate the extraction of the frequency content of a sampled waveform with a sample rate of FsΒ = 4 x fh.

Figure 2.1. Time waveform x(t).

Figure 2.2. Sample sequence.

Figure 2.3. Fourier transform.

Figure 2.4. DTFT of y(n).

Examples 4 through 7 (in this and the following lessons) illustrate the importance of Theorem 1.

Example 4

Frequency Content of a Sampled Waveform at Fs = 4 x fh

Given the time waveform illustrated in Figure 2.1:

(7)   \begin{equation*} x(t)=\frac{1}{\sqrt2\pi}\left[\frac{sin(t/2)}{t/2}\right]^2 \triangleq \frac{1}{\sqrt2\pi}\text{sinc}^2\left(\frac{t}{2}\right) \end{equation*}

The Fourier transform 𝖷̃ (πœ”) is the triangle function illustrated in Figure 2.3. The highest frequency component is πœ”h = 1 radian/second or 𝖿h = 1/2πœ‹ Hz. Therefore, the minimum sample rate is π–₯s β‰₯ 2 Γ— 𝖿h = 1/πœ‹ and the maximum sample period is 𝜏 ≀ π–₯s = πœ‹.

The waveform 𝗑(𝑑) sampled at π–₯s β‰œ 4 Γ— 𝖿h = 4 Γ— 1/2πœ‹ = 2/πœ‹ (giving a sample period of 𝜏 = 1/π–₯s = πœ‹/2) is illustrated in Figure 2.2.

With this sample rate and Theorem 1, the DTFT 𝖸̃(πœ”) of 𝗒(𝑛) is a repeated and scaled version of 𝖷̃ (πœ”). It is expressed as:

(8)   \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*}

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

In conclusion:

  • 𝖷̃ (πœ”) repeats every 2πœ‹ (this is true of all DTFTs from the definition of DTFT)
  • 𝖷̃ (πœ”) reaches zero when (2/πœ‹) πœ” = 1 or πœ” = πœ‹/2; the higher the sample rate π–₯s, the narrower the triangle illustrated in Figure 2.3
  • The height of each triangle isΒ 2\sqrt{2\pi}/\pi or about 1.596. The higher the sample rate π–₯s, the higher the triangle illustrated in Figure 2.3