# 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 π seconds, yielding the sequence y(n) = x(nπ). 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.

### Theorem 1

Defining the DTFT of a sampled sequence in terms of the Fourier transform of the sampled waveform.

Let π(π) β π(ππ) be the sample sequence of a waveform π(π‘). π₯s = 1/π is the sample rate. Let πΈΜ(π) be the DTFT of π(π) and π·Μ (π) be the Fourier transform of π(π‘). Then:

(1)

#### Proof

This theorem is difficult to prove without resorting to an βengineering hackβ 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).

 (2) by definition of DTFT (3) by definition of y(n) (4) because Ο/Ο = 1 (5) by IPSF (6) and absolute summability

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

1. Repeated every 2π radians (in discrete world frequency, which is every 2 Γ π₯s Hz in real-world frequency)
2. Dilated (made narrower) by a factor of π₯s
3. Scaled (made taller) by a factor of β2π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 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:

(7)

The Fourier transform π·Μ (π) is the triangle function illustrated in Figure 2.3. The highest frequency component is πh = 1 radian/second or πΏh = 1/2π Hz. Therefore, the minimum sample rate is π₯s β₯ 2 Γ πΏh = 1/π and the maximum sample period is π β€ π₯s = π.

The waveform π(π‘) sampled at π₯s β 4 Γ πΏh = 4 Γ 1/2π = 2/π (giving a sample period of π = 1/π₯s = π/2) is illustrated in Figure 2.2.

With this sample rate and Theorem 1, the DTFT πΈΜ(π) of π(π) is a repeated and scaled version of π·Μ (π). It is expressed as:

(8)

(9)

### Conclusion

• π·Μ (π) repeats every 2π (this is true of all DTFTs by definition)
• π·Μ (π) reaches zero when (2/π) π = 1 or π = π/2; the higher the sample rate π₯s, the narrower the triangle illustrated in Figure 2.3
• The height of each triangle isΒ  or about 1.596. The higher the sample rate π₯s, the higher the triangle illustrated in Figure 2.3