DTFT of a Sampled Sequence
July 24, 2019
Back to: Sampling & Reconstruction
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(ππ) be the sample sequence of a waveform π(π‘). π₯s = 1/π is the sample rate. Let πΈΜ(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 βengineering 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:
- Repeated every 2π radians (in discrete world frequency, which is every 2 Γ π₯s Hz in real-world frequency)
- Dilated (made narrower) by a factor of π₯s
- Scaled (made taller) 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 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 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 πΏh = 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:
(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