Relating Continuous and Discrete Frequencies

Hello! Welcome back to our journey from the analog to the digital realm.

In our last lesson, we moved from the idealized theory of sampling to the world of practical engineering. We established the necessity of anti-aliasing and reconstruction filters, the analog "gatekeepers" that manage the transition between continuous and discrete signals. We saw how the spectrum of a sampled continuous-time signal, $x_{s} (t)$ , consists of replicas of the original signal's spectrum, centered at multiples of the sampling frequency, $Ω_{s}$ .

Today, we will formalize the crucial link between the two worlds of frequency analysis. Our goal is to derive the precise mathematical relationship between the continuous-time frequency, $Ω$ (measured in radians per second), and the discrete-time frequency, $ω$ (measured in radians per sample). This relationship is the cornerstone of all digital signal processing, as it allows us to interpret the spectrum of a discrete sequence of numbers in terms of the physical frequencies of the original analog signal.

1. Two Worlds, Two Frequencies

Before we can connect them, we must clearly define our two frequency domains.

The Continuous-Time Perspective: Spectral Replicas

As a brief recap, when we sample a continuous-time signal $x_{c} (t)$ , we can model this process as multiplying it by an impulse train. In the frequency domain, this results in the original signal's spectrum, $X_{c} (j Ω)$ , being replicated at every integer multiple of the sampling frequency, $Ω_{s} = 2 π / T_{s}$ .

The following video segment provides a quick visual reminder of this phenomenon.

Continuous Time and Discrete Time Fourier Transforms

Let's begin with a review of what sampling does to a signal in the continuous-time frequency domain. The video 'Continuous Time and Discrete Time Fourier Transforms' by Iain Explains Signals, Systems, and Digital Comms offers a clear illustration.

Please watch from the beginning to 02:04. Focus on the distinction between the original signal's spectrum, X(j\Omega), and the spectrum of the sampled signal, X_s(j\Omega), which contains repeating copies.

This spectrum, $X_{s} (j Ω)$ , still uses the continuous-time frequency variable $Ω$ in radians per second.

The Discrete-Time Perspective: A Normalized, Periodic World

Once we have our sequence of samples, $x [n] = x_{c} (n T_{s})$ , we can treat it as a purely mathematical, discrete-time signal. We analyze its frequency content using the Discrete-Time Fourier Transform (DTFT), which gives us $X (e^{j ω})$ .

A fundamental property of the DTFT is that it is always periodic with a period of $2 π$ . The frequency variable $ω$ is a normalized angle, typically considered over the range $[- π, π]$ or $[0, 2 π]$ . Its units are radians per sample.

But why is it periodic? The reason is that discrete-time complex exponentials, $e^{j ω n}$ , are only unique over a $2 π$ interval of $ω$ . A frequency of $ω_{0}$ is indistinguishable from a frequency of $ω_{0} + 2 π k$ for any integer $k$ .

The same video explains this essential property both mathematically and visually. Given your background, the mathematical derivation will be particularly satisfying.

Continuous Time and Discrete Time Fourier Transforms

Now, let's understand the nature of the discrete-time frequency domain. The same video provides an excellent explanation for the 2\pi-periodicity of the DTFT.

Please watch from 02:04 to 06:03. First (02:04 - 04:47), follow the proof showing that cos(ω₀n + 2πn) is identical to cos(ω₀n) for integer n. This is the mathematical heart of periodicity. Then (04:47 - 06:03), observe the visual demonstration. It shows how two different continuous-time sinusoids can produce the exact same set of samples, illustrating why high frequencies 'alias' to lower frequencies in the discrete domain.

So, we have two representations for the frequency content of our sampled signal:

Continuous: $X_{s} (j Ω)$ , which is periodic with period $Ω_{s}$ .
Discrete: $X (e^{j ω})$ , which is periodic with period $2 π$ .

Since both describe the same information, there must be a direct relationship between them.

2. Deriving the Relationship: The Formal Bridge

The derivation relies on a direct comparison of the defining equations for the CTFT of the impulse-sampled signal and the DTFT of the resulting sequence.

The resource below provides a concise mathematical derivation that formalizes this comparison.

1. Working Between Continuous and Discrete Time

The document 'Working Between Continuous and Discrete Time' from Carnegie Mellon University provides a direct and elegant derivation of the frequency relationship. We will walk through its logic.

Please read the short section 'Relating continuous-time and discrete-time frequency' (equations 1.8 through 1.11). The core of the argument is comparing the final expression for the DTFT with the final expression for the CTFT of the sampled signal.

Let's break down the key steps from that derivation:

Start with the DTFT: The DTFT of the discrete sequence $x [n]$ is defined as:
$X (e^{j ω}) = n = - \infty \sum \infty x [n] e^{- j ω n}$
Since $x [n]$ is obtained by sampling the continuous signal $x_{c} (t)$ at intervals of $T_{s}$ , we have $x [n] = x_{c} (n T_{s})$ . Substituting this gives:
$X (e^{j ω}) = n = - \infty \sum \infty x_{c} (n T_{s}) e^{- j ω n} (*_{1})$
Move to the CTFT: The impulse-sampled continuous signal is $x_{s} (t) = \sum_{n = - \infty}^{\infty} x_{c} (n T_{s}) δ (t - n T_{s})$ . Its CTFT is defined as:
$X_{s} (j Ω) = \int_{- \infty}^{\infty} (n = - \infty \sum \infty x_{c} (n T_{s}) δ (t - n T_{s})) e^{- j Ω t} d t$
By swapping the integral and summation and using the sifting property of the delta function ( $\int f (t) δ (t - a) d t = f (a)$ ), we get:
$X_{s} (j Ω) = n = - \infty \sum \infty x_{c} (n T_{s}) e^{- j Ω n T_{s}} (*_{2})$
Compare the Results: Now, look closely at equations $(*_{1})$ and $(*_{2})$ . They describe the same spectral shape but are expressed in terms of different frequency variables, $ω$ and $Ω$ .
- Equation $(*_{1})$ : $X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x_{c} (n T_{s}) e^{- j ω n}$
- Equation $(*_{2})$ : $X_{s} (j Ω) = \sum_{n = - \infty}^{\infty} x_{c} (n T_{s}) e^{- j Ω T_{s} n}$
The two expressions become identical if we set the exponents equal to each other:
$- j ω n = - j Ω T_{s} n$
This simplifies to the fundamental relationship:
$ω = Ω T_{s}$

3. Interpreting the Frequency Scaling

The equation $ω = Ω T_{s}$ is more than just a formula; it's a scaling factor that maps the entire continuous frequency axis onto the normalized discrete frequency axis.

Let's check the units, as this often clarifies the meaning:

ω [\frac{radians}{sample}] = Ω [\frac{radians}{second}] \times T_{s} [\frac{seconds}{sample}]

The units match perfectly, confirming that $ω$ is a measure of phase change per sample.

The video you watched earlier concludes with an excellent intuitive summary of this scaling relationship.

Continuous Time and Discrete Time Fourier Transforms

Let's revisit Iain's explanation to see an intuitive visualization of this scaling.

Please watch from 06:03 to 09:08. The key insight here is that the period in the continuous frequency domain, \Omega_s, must map to the period in the discrete frequency domain, 2\pi, and the relationship \omega = \Omega T is the scaling factor that makes this happen.

Mapping Key Frequencies

Let's see how this scaling maps critical points on the frequency axis:

DC (Direct Current): For $Ω = 0$ rad/s, the discrete frequency is $ω = 0 \times T_{s} = 0$ . As expected, zero frequency maps to zero frequency.
Nyquist Frequency: The highest unambiguous frequency in the continuous domain is the Nyquist frequency, $Ω_{N y q} = Ω_{s} /2 = (\frac{2 π}{T _{s}}) /2 = \frac{π}{T _{s}}$ . Let's see where it maps: $ω = Ω_{N y q} \times T_{s} = (\frac{π}{T _{s}}) \times T_{s} = π$ This is a profound result: the edge of the baseband in the continuous-time spectrum maps precisely to $π$ in the discrete-time spectrum. The entire unique frequency range of the sampled signal, $[- Ω_{N y q}, Ω_{N y q}]$ , is mapped to the discrete frequency range $[- π, π]$ .
Sampling Frequency: The continuous sampling frequency is $Ω_{s} = \frac{2 π}{T _{s}}$ . This maps to: $ω = Ω_{s} \times T_{s} = (\frac{2 π}{T _{s}}) \times T_{s} = 2 π$ This confirms our intuition: the replication interval in the $Ω$ domain ( $Ω_{s}$ ) becomes the fundamental period in the $ω$ domain ( $2 π$ ).

This diagram illustrates the mapping from the continuous frequency axis ( $Ω$ ) to the discrete frequency axis ( $ω$ ). The primary interval $[- Ω_{s} /2, Ω_{s} /2]$ in the continuous domain, representing the unique spectral content, is scaled to fit the interval $[- π, π]$ in the discrete domain.

Test your understanding!

A signal containing a pure sinusoidal tone at a frequency of 4 kHz is sampled at a rate of 48 kHz.

What is the continuous-time frequency $Ω$ of the tone in rad/s?
What is the sampling period $T_{s}$ ?
What is the resulting discrete-time frequency $ω$ of the tone in rad/sample?
Another continuous-time tone at 52 kHz is present in the signal. What discrete-time frequency $ω^{'}$ does it map to? How does this relate to your answer for question 3?

Show answer

The continuous-time frequency in Hertz is $f = 4000$ Hz. The frequency in rad/s is $Ω = 2 π f = 2 π (4000) = 8000 π$ rad/s.
The sampling period is the inverse of the sampling rate: $T_{s} = 1/ f_{s} = 1/48000$ s.
Using our derived relationship: $ω = Ω T_{s} = (8000 π) \times (\frac{1}{48000}) = \frac{8000 π}{48000} = \frac{π}{6} rad/sample$
For the 52 kHz tone, $Ω^{'} = 2 π (52000) = 104000 π$ rad/s. Its discrete-time frequency is: $ω^{'} = Ω^{'} T_{s} = (104000 π) \times (\frac{1}{48000}) = \frac{104 π}{48} = \frac{13 π}{6} rad/sample$ Because the DTFT is $2 π$ -periodic, the frequency $\frac{13 π}{6}$ is indistinguishable from $\frac{13 π}{6} - 2 π = \frac{13 π}{6} - \frac{12 π}{6} = \frac{π}{6}$ . This is a perfect example of aliasing. The 52 kHz tone, which is above the Nyquist frequency of 24 kHz, has aliased down and appears as the same discrete frequency as the 4 kHz tone.

Conclusion

In this lesson, we have formally established the fundamental bridge between the continuous and discrete frequency domains. By comparing the Fourier transforms of a sampled signal from both perspectives, we derived the simple but powerful scaling relationship $ω = Ω T_{s}$ .

Key Takeaways:

The spectrum of a continuous-time signal sampled with an impulse train, $X_{s} (j Ω)$ , is periodic with period $Ω_{s} = 2 π / T_{s}$ .
The Discrete-Time Fourier Transform (DTFT) of the resulting sequence, $X (e^{j ω})$ , is always periodic with period $2 π$ .
The relationship between the two frequency variables is a linear scaling: $ω = Ω T_{s}$ .
This relationship maps the unique continuous-time frequency interval $[- π / T_{s}, π / T_{s}]$ onto the discrete-time frequency interval $[- π, π]$ .

Preview of the Next Lesson:

The sampling theorem we've discussed so far assumes we are interested in low-pass signals, where the frequency content starts at DC. But what if we want to sample a signal that is already band-limited but centered around a high carrier frequency, like an AM radio signal? Sampling at twice the highest frequency would be incredibly inefficient. In our next lesson, we will explore a powerful extension of sampling theory by deriving and applying the bandpass sampling theorem, which allows for efficient sampling of such signals.

Recommended resources

1. Two Worlds, Two Frequencies

The Continuous-Time Perspective: Spectral Replicas

The Discrete-Time Perspective: A Normalized, Periodic World

2. Deriving the Relationship: The Formal Bridge

3. Interpreting the Frequency Scaling

Mapping Key Frequencies

Conclusion