DTFT of a Sampled Sequence
July 24, 2019
Suppose there is a signal x(t) with a Fourier transform of . The signal is sampled every 𝜏 seconds, yielding the sequence y(n) = x(n𝜏). From the sampled sequence, a calculation yields the discrete-time Fourier transform (DTFT) of y(n).
Both a Fourier transform and a DTFT have been identified. They come from two different transform definitions but originated from the same signal, x(t).
This prompts the question: do and look anything alike? The answer is yes; in fact, looks like a scaled and repeated version of . Below, Theorem 1 and Example 1 offer a more precise description as to why this is.
Defining the DTFT of a sampled sequence in terms of the Fourier transform of the sampled waveform.
Let 𝗒(𝑛) ≜ 𝗑(𝑛𝜏) be the sample sequence of a waveform 𝗑(𝑡). 𝖥s = 1/𝜏 is the sample rate. Let 𝖸̆(𝜔) be the DTFT of 𝗒(𝑛) and 𝖷̃ (𝜔) be the Fourier transform of 𝗑(𝑡). Then:
This theorem is difficult to prove without resorting to an “engineering hack” methodology such as multiplying the waveform by a train of impulse functions. However, the proof is manageable when we leverage the somewhat obscure yet powerful inverse Poisson summation formula (IPSF).
|by definition of DTFT|
|by definition of y(n)|
|because τ/τ = 1|
|and absolute summability|
The DTFT of the sample sequence is the Fourier transform of the original waveform. It is:
- Repeated every 2𝜋 radians (in discrete world frequency, which is every 2 × 𝖥s Hz in real-world frequency)
- Dilated (made narrower) by a factor of 𝖥s
- Scaled (made taller) by a factor of √2𝜋𝖥s
Figures 2.1-2.4 illustrate the extraction of the frequency content of a sampled waveform with a sample rate of Fs = 4 x fh.
Examples 4 through 7 in this lesson and those that follow illustrate the importance of Theorem 1.
Frequency Content of a Sampled Waveform (Ex. 4)
Fs = 4 x fh
Given the time waveform illustrated in Figure 2.1:
The Fourier transform 𝖷̃ (𝜔) is the triangle function illustrated in Figure 2.3. The highest frequency component is 𝜔h = 1 radian/second or 𝖿h = 1/2𝜋 Hz. Therefore, the minimum sample rate is 𝖥s ≥ 2 × 𝖿h = 1/𝜋 and the maximum sample period is 𝜏 ≤ 𝖥s = 𝜋.
The waveform 𝗑(𝑡) sampled at 𝖥s ≜ 4 × 𝖿h = 4 × 1/2𝜋 = 2/𝜋 (giving a sample period of 𝜏 = 1/𝖥s = 𝜋/2) is illustrated in Figure 2.2.
With this sample rate and Theorem 1, the DTFT 𝖸̃(𝜔) of 𝗒(𝑛) is a repeated and scaled version of 𝖷̃ (𝜔). It is expressed as:
- 𝖷̃ (𝜔) repeats every 2𝜋 (this is true of all DTFTs by definition)
- 𝖷̃ (𝜔) reaches zero when (2/𝜋) 𝜔 = 1 or 𝜔 = 𝜋/2; the higher the sample rate 𝖥s, the narrower the triangle illustrated in Figure 2.3
- The height of each triangle is or about 1.596. The higher the sample rate 𝖥s, the higher the triangle illustrated in Figure 2.3