# DTFT of a Sampled Sequence

July 24, 2019

Back to: Sampling & Reconstruction

Suppose the following: There isΒ a signal x(*t*) with a Fourier transform of . The signal is sampled every π seconds yielding the sequence y(*n*) = x(*n*π). From the sampled sequence, a calculation yields the DTFT of y(*n*).

Now, there is a Fourier transform and a DTFT . 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 FT of the sampled waveform

Let π(π) β π(ππ) be the sample sequence of a waveform π(π‘) and π₯_{s} = 1/π be the sample rate. Let πΈΜ(π) be the DTFT of π(π) and π·Μ (π) be the Fourier transform of π(π‘). Then:

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#### Proof

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

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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 Γ π₯
_{s}Hertz in real-world frequency) - Dilated (made βnarrowerβ) by a factor of π₯
_{s} - Scaled (made βtallerβ) by a factor of β2ππ₯
_{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.

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 F_{s} = 4 x f_{h}

Given the time waveform illustrated in Figure 2.1:

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The Fourier transform π·Μ (π) is the triangle function illustrated in Figure 2.3. The highest frequency component is πh = 1 radian/second or πΏh = 1/2π Hz. Therefore, the minimum sample rate is π₯s β₯ 2 Γ πΏh = 1/π and the maximum sample period is π β€ π₯_{s} = π.

The waveform π(π‘) sampled at π₯_{s} β 4 Γ πΏ_{h} = 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 πΈΜ(π) of π(π) is a repeated and scaled version of π·Μ (π). It is expressed as:

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### In conclusion:

- π·Μ (π) repeats every 2π (this is true of all DTFTs from the definition of DTFT)
- 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