DTFT Derivation and Sampled Signal Relationship

Hello! Welcome back to our exploration of discrete-time signals and systems.

In the last lesson, we traced the historical development of the z-transform, understanding it as a powerful tool for solving difference equations that is analogous to the Laplace transform. We also established the crucial link between the two worlds: the mapping $z = e^{s T_{s}}$ , which connects the continuous s-plane to the discrete z-plane.

This lesson builds directly on that foundation. Our goal is to derive the Discrete-Time Fourier Transform (DTFT) and establish its fundamental relationship to the Continuous-Time Fourier Transform (CTFT) of a sampled signal. Just as the CTFT is the Laplace transform evaluated on the imaginary axis ( $s = j Ω$ ), we will see that the DTFT is the z-transform evaluated on the unit circle ( $z = e^{j ω}$ ). Most importantly, you will see how the DTFT spectrum is formed directly from the original continuous-time signal's spectrum, providing a deep, mathematical bridge between the analog and digital frequency domains.

1. Revisiting the Spectrum of a Sampled Signal

To understand the frequency content of a discrete-time sequence, we must first understand what happens to a continuous-time signal's spectrum when it is sampled.

The standard model for ideal sampling, which you've encountered before, represents the process as the multiplication of the continuous-time signal, $x (t)$ , with a periodic impulse train, $p (t)$ .
$p (t) = \sum_{n = - \infty}^{\infty} δ (t - n T_{s})$
The resulting sampled signal, $x_{s} (t)$ , is a train of impulses where the weight of each impulse corresponds to the value of $x (t)$ at that instant.
$x_{s} (t) = x (t) p (t) = \sum_{n = - \infty}^{\infty} x (n T_{s}) δ (t - n T_{s})$

From your knowledge of Fourier properties, you know that multiplication in the time domain corresponds to convolution in the frequency domain. To find the spectrum of $x_{s} (t)$ , we must convolve the spectrum of $x (t)$ with the spectrum of the impulse train $p (t)$ .

The following video provides a clear, step-by-step derivation of this process.

Sampling Theorem

The video 'Sampling Theorem' from Neso Academy excellently derives the Fourier transform of an ideally sampled signal. It shows precisely how the original spectrum is replicated at integer multiples of the sampling frequency.

Please watch from 04:43 to 13:28. Focus on these key steps: The mathematical representation of the sampling process as a multiplication of the message signal m(t) and an impulse train c(t). The derivation of the Fourier transform of the impulse train. The use of the convolution property to derive the spectrum of the sampled signal, S(\Omega). The final expression showing that the sampled spectrum consists of scaled copies of the original spectrum, repeated periodically.

As the video derives, the Fourier transform of the impulse train $p (t)$ is another impulse train in the frequency domain:
$P (j Ω) = \frac{2 π}{T _{s}} \sum_{k = - \infty}^{\infty} δ (Ω - k Ω_{s})$
where $Ω_{s} = 2 π / T_{s}$ is the sampling frequency in radians per second.

Convolving the original signal's spectrum, $X (j Ω)$ , with $P (j Ω)$ (and including the $1/2 π$ scaling factor for convolution) yields the spectrum of the sampled signal, $X_{s} (j Ω)$ :

X_{s} (j Ω) = \frac{1}{2 π} [X (j Ω) * P (j Ω)] = \frac{1}{T _{s}} k = - \infty \sum \infty X (j (Ω - k Ω_{s}))

This is a critical intermediate result. It tells us that sampling in the time domain causes the original signal's spectrum to be replicated periodically in the frequency domain at integer multiples of the sampling frequency $Ω_{s}$ .

2. Deriving the Discrete-Time Fourier Transform (DTFT)

The expression for $X_{s} (j Ω)$ is the CTFT of the impulse-sampled continuous-time signal. However, our goal is to find a frequency representation for the discrete-time sequence $x [n] = x (n T_{s})$ itself. This is the Discrete-Time Fourier Transform (DTFT).

The DTFT is defined as:

X (e^{j ω}) = n = - \infty \sum \infty x [n] e^{- j ω n}

Here, $ω$ is the normalized discrete-time frequency, with units of radians per sample. Notice the notation $X (e^{j ω})$ ; this emphasizes its relationship to the z-transform, $X (z)$ , evaluated at $z = e^{j ω}$ (i.e., on the unit circle).

Now, let's establish the connection between the CTFT of the sampled signal, $X_{s} (j Ω)$ , and the DTFT of the sequence, $X (e^{j ω})$ .

Let's re-derive $X_{s} (j Ω)$ from its definition, but this time keeping the discrete samples $x [n]$ explicit:

X_{s} (j Ω) = F {x_{s} (t)} = \int_{- \infty}^{\infty} (n = - \infty \sum \infty x (n T_{s}) δ (t - n T_{s})) e^{- j Ω t} d t

Swapping the integral and summation:

X_{s} (j Ω) = n = - \infty \sum \infty x (n T_{s}) \int_{- \infty}^{\infty} δ (t - n T_{s}) e^{- j Ω t} d t

Using the sifting property of the Dirac delta function, the integral evaluates to $e^{- j Ω n T_{s}}$ . Substituting $x [n]$ for $x (n T_{s})$ , we get:

X_{s} (j Ω) = n = - \infty \sum \infty x [n] e^{- j Ω n T_{s}}

Now, compare this result with the definition of the DTFT:

CTFT of sampled signal: $X_{s} (j Ω) = \sum_{n = - \infty}^{\infty} x [n] e^{- j (Ω T_{s}) n}$
DTFT of sequence: $X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n}$

By direct comparison, we see that the two expressions are identical if we make the substitution:

ω = Ω T_{s}

This simple equation is the bridge between the continuous and discrete frequency domains. It shows that the normalized discrete frequency $ω$ is simply the continuous frequency $Ω$ scaled by the sampling period $T_{s}$ . This scaling relationship is key to understanding everything that follows.

Continuous Time and Discrete Time Fourier Transforms

The video 'Continuous Time and Discrete Time Fourier Transforms' provides an excellent visual and conceptual explanation of this scaling relationship.

Please watch the segment from 06:03 to 09:08. The speaker directly addresses the relationship between the two transforms. Focus on how the sampling frequency in the continuous domain, \Omega_s = 2\pi/T_s, scales to become the period of the DTFT, 2\pi, and how any continuous frequency \Omega_C scales to \omega_c = \Omega_C T_s in the discrete domain.

3. The Full Relationship: A Mathematical Derivation

We now have all the pieces to state the complete relationship between the original signal's CTFT and the DTFT of its samples. This derivation provides the rigorous mathematical link you appreciate.

We have two expressions for the spectrum of the sampled signal:

From periodic replication: $X_{s} (j Ω) = \frac{1}{T _{s}} \sum_{k = - \infty}^{\infty} X (j (Ω - k Ω_{s}))$
From the DTFT relationship: $X_{s} (j Ω) = X (e^{j Ω T_{s}})$

Equating these gives:

X (e^{j Ω T_{s}}) = \frac{1}{T _{s}} k = - \infty \sum \infty X (j (Ω - k Ω_{s}))

Substituting $ω = Ω T_{s}$ (so $Ω = ω / T_{s}$ ) and $Ω_{s} = 2 π / T_{s}$ , we arrive at the final, fundamental equation:

X (e^{j ω}) = \frac{1}{T _{s}} k = - \infty \sum \infty X (j (\frac{ω}{T _{s}} - k \frac{2 π}{T _{s}})) = \frac{1}{T _{s}} k = - \infty \sum \infty X (\frac{j ( ω - 2 π k )}{T _{s}})

This equation is the cornerstone of sampling theory. It explicitly shows that the DTFT spectrum is a summation of scaled and frequency-shifted versions of the original CTFT spectrum.

For a rigorous, step-by-step presentation of this derivation, please refer to the following text.

Sampling and Reconstruction

The textbook chapter 'Sampling and Reconstruction' by Lee and Varaiya provides a formal derivation of this relationship. It uses a slightly different notation but follows the exact same logic we have just outlined.

Please read the section titled '11.3 The Nyquist-Shannon sampling theorem', focusing on pages 388-389. Pay close attention to: Step 1: The formula relating the DTFT of the samples (Y(\omega) in their notation) to the CTFT of the original signal (X(\omega)). This is exactly the formula we just derived. Figure 11.10: This diagram perfectly illustrates the relationship for a band-limited signal where the spectral copies do not overlap. Figure 11.11: This shows the case where aliasing distortion occurs because the spectral copies overlap.

4. The Periodicity of the DTFT

The formula we just derived provides a direct mathematical proof of a core property of the DTFT: it is always periodic with a period of $2 π$ .

Let's prove this by replacing $ω$ with $ω + 2 π$ in our final equation:

X (e^{j (ω + 2 π)}) = \frac{1}{T _{s}} k = - \infty \sum \infty X (\frac{j ( ω + 2 π - 2 π k )}{T _{s}})

= \frac{1}{T _{s}} k = - \infty \sum \infty X (\frac{j ( ω - 2 π ( k - 1 ))}{T _{s}})

Now, let's introduce a new summation index $m = k - 1$ . As $k$ ranges over all integers from $- \infty$ to $\infty$ , so does $m$ . The sum becomes:

X (e^{j (ω + 2 π)}) = \frac{1}{T _{s}} m = - \infty \sum \infty X (\frac{j ( ω - 2 π m )}{T _{s}})

This expression is identical to the original expression for $X (e^{j ω})$ . Therefore, we have proven that $X (e^{j (ω + 2 π)}) = X (e^{j ω})$ .

This periodicity is not just a mathematical curiosity; it is the frequency-domain manifestation of aliasing. Because the spectrum repeats every $2 π$ radians/sample, we cannot distinguish between a frequency $ω$ and a frequency $ω + 2 π k$ for any integer $k$ just by looking at the samples. This is the same phenomenon we saw in the previous module, now explained from the perspective of the DTFT.

Test your understanding!

A continuous-time audio signal has a strong sinusoidal component at a frequency of $f = 4$ kHz. The signal is sampled at a rate of $f_{s} = 48$ kHz.

What is the continuous-time frequency of the sinusoid in radians/second ( $Ω$ )?
What is the sampling period $T_{s}$ ?
What is the corresponding discrete-time frequency in radians/sample ( $ω$ )?
If another continuous-time sinusoid at $f^{'} = 52$ kHz was also present in the original signal, what would its discrete-time frequency $ω^{'}$ be? What does this tell you?

Show answer

$Ω = 2 π f = 2 π (4000) = 8000 π$ rad/s.
$T_{s} = 1/ f_{s} = 1/48000$ s.
$ω = Ω T_{s} = (8000 π) \times (1/48000) = π /6$ rad/sample.
For the $f^{'} = 52$ kHz tone: $Ω^{'} = 2 π f^{'} = 104000 π$ rad/s.
$ω^{'} = Ω^{'} T_{s} = (104000 π) \times (1/48000) = \frac{13}{6} π = 2 π + \frac{π}{6}$ rad/sample.
Since the DTFT is periodic with $2 π$ , the frequency $ω^{'} = 2 π + π /6$ is indistinguishable from $ω = π /6$ . This shows aliasing in action: after sampling, the 52 kHz tone appears at the exact same discrete-time frequency as the 4 kHz tone.

Conclusion

In this lesson, we have formally derived the DTFT and, more importantly, established its precise mathematical relationship with the CTFT of the original analog signal. This connection is the theoretical backbone for all of digital signal processing.

Key Takeaways:

The process of ideal sampling in the time domain results in a periodic replication of the signal's spectrum in the continuous-frequency domain.
The DTFT, $X (e^{j ω})$ , is the Fourier transform for a discrete-time sequence $x [n]$ .
The DTFT of a sequence is mathematically equivalent to the CTFT of the corresponding impulse-sampled signal, with the frequency axes related by the scaling $ω = Ω T_{s}$ .
The full relationship is given by the formula: $X (e^{j ω}) = \frac{1}{T _{s}} \sum_{k = - \infty}^{\infty} X (\frac{j ( ω - 2 π k )}{T _{s}})$ .
This formula mathematically proves that the DTFT is always periodic with a period of $2 π$ , which is the frequency-domain explanation for aliasing.

Preview of the Next Lesson:

Now that we have derived the DTFT and established its fundamental periodicity, we will analyze the implications of this periodicity in greater detail. We will revisit the concept of aliasing and the Nyquist-Shannon sampling theorem from this new perspective and explore why the DTFT spectrum contains all the necessary information, provided the signal was sampled correctly.

Recommended resources

1. Revisiting the Spectrum of a Sampled Signal

2. Deriving the Discrete-Time Fourier Transform (DTFT)

3. The Full Relationship: A Mathematical Derivation

4. The Periodicity of the DTFT

Conclusion