Ferkans — Interactive Telecom Tutor

Historical Note: Nyquist, Shannon, and the Sampling Theorem

Harry Nyquist (1928) and Claude Shannon (1949) independently established that a band-limited signal can be perfectly reconstructed from its samples — provided the sampling rate exceeds twice the bandwidth. This theorem is the bridge between continuous-time theory and digital implementation, underpinning every A/D converter, DSP chip, and software-defined radio in existence.

Definition:
Ideal Sampling

Ideal sampling of a continuous-time signal $x(t)$ at rate $f_s = 1/T_s$ produces the discrete-time sequence

$x[n] = x(nT_s), \qquad n \in \mathbb{Z}.$

Equivalently, the sampled signal can be written as

$x_s(t) = x(t) \sum_{n=-\infty}^{\infty} \delta(t - nT_s) = \sum_{n=-\infty}^{\infty} x(nT_s)\,\delta(t - nT_s).$

Taking the Fourier transform of $x_s(t)$ :

$X_s(f) = f_s \sum_{k=-\infty}^{\infty} X(f - k f_s).$

The spectrum of the sampled signal is a periodised version of $X(f)$ , repeated every $f_s$ Hz.

Theorem: Nyquist–Shannon Sampling Theorem

Let $x(t)$ be band-limited with bandwidth $W$ , i.e., $X(f) = 0$ for $|f| > W$ . Then $x(t)$ can be perfectly reconstructed from its samples $\{x(nT_s)\}$ if and only if

$f_s > 2W.$

The minimum rate $f_{\text{Nyquist}} = 2W$ is the Nyquist rate. Reconstruction is achieved by ideal low-pass filtering:

$x(t) = \sum_{n=-\infty}^{\infty} x(nT_s)\,\mathrm{sinc}\!\left(\frac{t - nT_s}{T_s}\right).$

If $f_s > 2W$ , the periodised copies of $X(f)$ do not overlap, so the original spectrum can be isolated with a low-pass filter. If $f_s \le 2W$ , copies overlap (aliasing) and information is irreversibly lost.

Proof

Spectrum of sampled signal

From the Poisson summation formula, $X_s(f) = f_s \sum_k X(f - kf_s)$ . If $X(f) = 0$ for $|f| > W$ and $f_s > 2W$ , the replicas $X(f - kf_s)$ for $k \ne 0$ do not overlap with $X(f)$ in $[-f_s/2, f_s/2]$ .

Reconstruction

Multiply $X_s(f)$ by the ideal LPF $H(f) = T_s\,\mathrm{rect}(f/f_s)$ to isolate $X(f)$ : $X(f) = H(f) X_s(f)$ . In the time domain this becomes convolution with $\mathrm{sinc}(t/T_s)$ , yielding the interpolation formula. $\blacksquare$

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Definition:
Aliasing

When a signal is sampled at rate $f_s \leq 2W$ (below the Nyquist rate), the periodic spectral copies overlap. This spectral folding is called aliasing: high-frequency components masquerade as lower frequencies and cannot be separated.

The maximum unambiguous frequency is the folding frequency $f_s / 2$ . Any frequency component $f_0 > f_s/2$ aliases to $f_0 \bmod f_s$ (folded into the first Nyquist zone).

To prevent aliasing in practice, an anti-aliasing filter (analog low-pass) is applied before sampling to remove energy above $f_s/2$ .

Sampling Theorem: From Alias-Free to Aliasing

Watch the spectrum of a sampled signal as the sampling rate decreases. When

f_s > 2W

, the spectral copies are separated; when

f_s < 2W

, they overlap and information is lost.

Spectral copies of a band-limited signal (

W = 2

Hz) as

f_s

sweeps from 8 Hz down to 3 Hz. Overlap indicates aliasing.

Sampling and Aliasing Demo

Adjust the sampling rate $f_s$ and observe how the spectrum changes. When $f_s < 2W$ , spectral copies overlap (aliasing) and the reconstructed signal differs from the original.

Parameters

Signal frequency

f_0

(Hz)5

Sampling rate

f_s

(Hz)15

Common Mistake: Real Signals Are Never Truly Band-Limited

Mistake:

Assuming that choosing $f_s > 2W$ eliminates aliasing for any real-world signal.

Correction:

Mathematically band-limited signals have infinite time duration (a consequence of the uncertainty principle). Real signals are time-limited and therefore have infinite bandwidth — they are never truly band-limited. In practice, an anti-aliasing filter attenuates out-of-band energy to an acceptable level (e.g., below the noise floor), and the residual aliasing is tolerable.

Definition:
Discrete-Time Fourier Transform (DTFT)

For a discrete-time signal $x[n]$ , the DTFT is

$X(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\Omega n}$

where $\Omega = 2\pi f / f_s$ is the normalised angular frequency (radians/sample). The DTFT is periodic with period $2\pi$ .

The inverse DTFT is

$x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\Omega})\,e^{j\Omega n}\,d\Omega.$

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Definition:
Discrete Fourier Transform (DFT)

For a finite-length sequence $x[n]$ , $n = 0, 1, \ldots, N-1$ , the $N$ -point DFT is

$X[k] = \sum_{n=0}^{N-1} x[n]\,e^{-j2\pi kn/N}, \qquad k = 0, 1, \ldots, N-1$

with inverse

$x[n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k]\,e^{j2\pi kn/N}.$

The DFT samples the DTFT at $N$ equally spaced points: $X[k] = X(e^{j2\pi k/N})$ .

The Fast Fourier Transform (FFT) computes the DFT in $O(N \log N)$ operations instead of $O(N^2)$ , making real-time spectral analysis practical.

Example: DFT Frequency Resolution

A signal is sampled at $f_s = 1000$ Hz and we compute an $N = 256$ -point DFT. What is the frequency resolution? If the signal contains two tones at 100 Hz and 104 Hz, can the DFT resolve them?

Solution

Frequency resolution

The frequency spacing between DFT bins is

$\Delta f = \frac{f_s}{N} = \frac{1000}{256} \approx 3.91 \text{ Hz}.$

Resolving the two tones

The tone separation is $104 - 100 = 4$ Hz $> \Delta f \approx 3.91$ Hz. So yes, the tones fall in different DFT bins and can be resolved.

If we had used $N = 128$ ( $\Delta f = 7.81$ Hz), the tones would fall in the same bin and could not be distinguished. $\blacksquare$

Common Mistake: Spectral Leakage in the DFT

Mistake:

Expecting the DFT to produce sharp spectral lines for all frequencies, regardless of the signal length.

Correction:

The DFT assumes the $N$ -point signal is one period of an $N$ -periodic sequence. If the signal frequency is not an exact multiple of $\Delta f = f_s / N$ , the energy "leaks" into adjacent bins. Windowing (Hann, Hamming, Blackman) reduces leakage at the cost of wider main lobes (reduced resolution). This is the DFT analogue of the time–frequency uncertainty principle.

Definition:
Z-Transform

The Z-transform of a discrete-time signal $x[n]$ is

$X(z) = \sum_{n=-\infty}^{\infty} x[n]\,z^{-n}$

where $z \in \mathbb{C}$ . The DTFT is the Z-transform evaluated on the unit circle: $X(e^{j\Omega}) = X(z)\big|_{z = e^{j\Omega}}$ .

Key properties (analogous to CTFT):

Delay: $x[n - k] \leftrightarrow z^{-k} X(z)$
Convolution: $(x * h)[n] \leftrightarrow X(z) H(z)$
ROC: The Z-transform converges in a region of the $z$ -plane (region of convergence). Causal sequences have ROC outside a circle; stable systems have ROC including the unit circle.

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Example: Z-Transform of a Causal Exponential

Find the Z-transform of $x[n] = a^n u[n]$ , where $|a| < 1$ . Determine the ROC and verify BIBO stability.

Solution

Z-transform

$X(z) = \sum_{n=0}^{\infty} a^n z^{-n} = \sum_{n=0}^{\infty} (a/z)^n = \frac{1}{1 - az^{-1}} = \frac{z}{z - a}$ $for$ |z| > |a| $. The ROC is$ |z| > |a|$.

Stability check

Since $|a| < 1$ , the ROC $|z| > |a|$ includes the unit circle $|z| = 1$ . Therefore the system is BIBO stable.

The DTFT exists: $X(e^{j\Omega}) = \frac{1}{1 - a e^{-j\Omega}}$ . $\blacksquare$

Quick Check

A signal has bandwidth $W = 4$ kHz. What is the minimum sampling rate to avoid aliasing?

4 kHz

8 kHz

16 kHz

2 kHz

Correction:

8 kHz

Correct. The Nyquist rate is $f_{\text{Nyquist}} = 2W = 8$ kHz. Any sampling rate $f_s > 8$ kHz avoids aliasing.

Quick Check

An $N$ -point DFT is computed with sampling rate $f_s$ . What frequency does bin $k$ correspond to?

$k \cdot f_s$

$k \cdot f_s / N$

$k / f_s$

$k \cdot N / f_s$

Correction:

k \cdot f_s / N

Correct. DFT bin $k$ corresponds to frequency $f_k = k f_s / N$ for $k = 0, 1, \ldots, N-1$ .

Sampling and Reconstruction Pipeline — The complete sampling pipeline: anti-aliasing LPF $\to$ sampler $\to$ DSP $\to$ reconstruction LPF. The anti-aliasing filter ensures $X(f) \approx 0$ for $|f| > f_s/2$ ; the reconstruction filter isolates the baseband replica from the periodised spectrum.

⚠️Engineering Note

ADC Resolution and Quantisation Noise

In practice, the analog-to-digital converter (ADC) has finite resolution: a $b$ -bit ADC maps each sample to one of $2^b$ levels. This quantisation introduces additive noise with approximate SNR:

$\text{SQNR} \approx 6.02\,b + 1.76 \;\text{dB}.$

At 5G NR mmWave frequencies, wideband signals (e.g., 400 MHz at 28 GHz) require ADCs operating at $\geq 800$ MSa/s. High-resolution ADCs (12+ bits) at these rates consume prohibitive power ( $\sim$ 1 W per ADC at 10-bit, 1 GSa/s). This motivates research into low-resolution ADCs (1–4 bits) and mixed-ADC architectures, where the quantisation noise is no longer negligible and must be modelled explicitly.

Practical Constraints

•
Power consumption scales exponentially with ADC bit width at GHz sampling rates
•
5G NR FR2 (mmWave) uses up to 400 MHz bandwidth, requiring $f_s \geq 800$ MHz
•
Typical SQNR for 10-bit ADC: approximately 62 dB

🔧Engineering Note

FFT Computation in Real-Time Systems

The FFT reduces the DFT from $O(N^2)$ to $O(N \log_2 N)$ operations. For an OFDM system with $N = 4096$ subcarriers (5G NR, 30 kHz SCS), one FFT requires approximately $4096 \times 12 = 49{,}152$ complex multiply-accumulate (MAC) operations. At 14 OFDM symbols per slot and 2000 slots/second (30 kHz SCS), the receiver must compute roughly $49{,}152 \times 14 \times 2000 \approx 1.38 \times 10^9$ MACs/s for the FFT alone — well within modern FPGA and ASIC capability, but a significant fraction of the baseband processing budget.

The choice of FFT size balances spectral efficiency (more subcarriers $\to$ smaller guard band overhead) against latency and peak-to-average power ratio (PAPR).

Practical Constraints

•
5G NR FFT sizes: 128, 256, 512, 1024, 2048, 4096 (3GPP TS 38.211)
•
Real-time FFT at 4096 points requires dedicated hardware (FPGA/ASIC), not software
•
Larger FFT reduces subcarrier spacing, increasing sensitivity to Doppler

📋 Ref: 3GPP TS 38.211, §5.3

Nyquist Rate

The minimum sampling rate for alias-free reconstruction of a band-limited signal: $f_{\text{Nyquist}} = 2W$ .

Related: Aliasing, Sampling Theorem

Aliasing

Spectral overlap caused by sampling below the Nyquist rate. High-frequency components fold into lower frequencies and cannot be recovered.

Discrete Fourier Transform (DFT)

$X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2\pi kn/N}$ . The frequency-domain representation of a finite-length sequence. Efficiently computed by the FFT.

Z-Transform

$X(z) = \sum_{n} x[n] z^{-n}$ . The discrete-time analogue of the Laplace transform. The DTFT is its restriction to $|z|=1$ .

Sampling Theorem and Discrete-Time Signals

Historical Note: Nyquist, Shannon, and the Sampling Theorem

Definition: Ideal Sampling

Theorem: Nyquist–Shannon Sampling Theorem

Spectrum of sampled signal

Reconstruction

Definition: Aliasing

Sampling Theorem: From Alias-Free to Aliasing

Sampling and Aliasing Demo

Parameters

Common Mistake: Real Signals Are Never Truly Band-Limited

Definition: Discrete-Time Fourier Transform (DTFT)

Definition: Discrete Fourier Transform (DFT)

Example: DFT Frequency Resolution

Frequency resolution

Resolving the two tones

Common Mistake: Spectral Leakage in the DFT

Definition: Z-Transform

Example: Z-Transform of a Causal Exponential

Z-transform

Stability check

Quick Check

Quick Check

Sampling and Reconstruction Pipeline

ADC Resolution and Quantisation Noise

FFT Computation in Real-Time Systems

Nyquist Rate

Aliasing

Discrete Fourier Transform (DFT)

Z-Transform

Definition:
Ideal Sampling

Definition:
Aliasing

Definition:
Discrete-Time Fourier Transform (DTFT)

Definition:
Discrete Fourier Transform (DFT)

Definition:
Z-Transform