Related Notes: 4.02 Spectral Replications & Aliasing Dynamics | 4.03 Ideal, Natural & Flat-Top Sampling Techniques | 4.04 Multi-Rate Signal Processing: Decimation & Interpolation

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4.01 The Shannon-Nyquist Sampling Theorem

Core Idea

The Shannon-Nyquist Sampling Theorem is the mathematical bridge that allows us to convert a continuous-time analog signal into a discrete-time sequence without any loss of information. It asserts that if a band-limited continuous-time signal has no frequency components higher than $f_{\text{max}}$ Hz, it can be uniquely and perfectly reconstructed if the sampling frequency $f_s$ satisfies $f_s\ge 2f_{\text{max}}$.

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1. Mathematical Definitions and Concepts

To understand sampling, we must mathematically define the transition from continuous-time $t$ to discrete-time indices $n$ under a uniform sampling period $T_s$ {the time interval between successive samples}.

1.1 The Sampling Equations

Let $x_a(t)$ represent a continuous-time analog input signal. Uniform sampling is modeled as taking snapshots of the signal every $T_s$ seconds: $x[n]=x_a(n{T}_s)\text{for }n\in \mathbb{Z}$

Here:

• $T_s$ = Sampling Period [seconds]
• $f_s=\frac{1}{T_s}$ = Sampling Frequency [Hz or samples/second]
• ${\Omega }_s=\frac{2\pi }{T_s}=2\pi f_s$ = Sampling Angular Frequency [rad/s]

1.2 The Nyquist Conditions

1. Nyquist Rate ($f_{\text{Nyquist}}$): The absolute minimum theoretical sampling rate that avoids overlapping spectral replicas. $f_{\text{Nyquist}}=2f_{\text{max}}$
2. Nyquist Interval ($T_{\text{Nyquist}}$): The maximum allowable time spacing between adjacent samples: $T_{\text{Nyquist}}=\frac{1}{f_{\text{Nyquist}}}=\frac{1}{2f_{\text{max}}}$

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2. Rigorous Proof of the Sampling Theorem [PYQ 2019, 2017]

The proof of the sampling theorem relies on modeling the physical sampling switch as a continuous-time multiplication of the analog signal $x_a(t)$ by a periodic impulse train $p(t)$:

2.1 Time-Domain Formulation

Let $p(t)$ be an impulse train of unit area spaced by $T_s$: $p(t)={\sum }_{n=-\infty }^{\infty }\delta (t-n{T}_s)$

The sampled signal $x_s(t)$ is the product: $x_s(t)=x_a(t)\cdot p(t)=x_a(t){\sum }_{n=-\infty }^{\infty }\delta (t-n{T}_s)$

Using the sampling property of the impulse ($x(t)\delta (t-t_0)=x(t_0)\delta (t-t_0)$): $x_s(t)={\sum }_{n=-\infty }^{\infty }{x}_a(n{T}_s)\delta (t-n{T}_s)$

2.2 Frequency-Domain Formulation

Since multiplication in the time domain corresponds to convolution in the frequency domain (scaled by $1/2\pi$): $X_s(j\Omega )=\frac{1}{2\pi }[X_a(j\Omega )\ast P(j\Omega )]$

Because $p(t)$ is periodic with period $T_s$, we can express it as a complex exponential Fourier Series: $p(t)={\sum }_{k=-\infty }^{\infty }{c}_k{e}^{jk{\Omega }_{s}t}\text{where }{\Omega }_s=\frac{2\pi }{T_s}$

The Fourier coefficients $c_k$ are: $c_k=\frac{1}{T_s}{\int }_{-T_s/2}^{T_s/2}p(t)e^{-jk{\Omega }_{s}t}dt=\frac{1}{T_s}{\int }_{-T_s/2}^{T_s/2}\delta (t)e^{-jk{\Omega }_{s}t}dt=\frac{1}{T_s}$

Thus: $p(t)=\frac{1}{T_s}{\sum }_{k=-\infty }^{\infty }{e}^{jk{\Omega }_{s}t}$

Taking the Fourier transform of this exponential series yields a train of impulses in the frequency domain: $P(j\Omega )=\frac{2\pi }{T_s}{\sum }_{k=-\infty }^{\infty }\delta (\Omega -k{\Omega }_s)$

Now, convolve $X_a(j\Omega )$ with $P(j\Omega )$: $X_s(j\Omega )=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{X}_a(j\theta )P(j(\Omega -\theta ))d\theta$ $X_s(j\Omega )=\frac{1}{2\pi }\cdot \frac{2\pi }{T_s}{\sum }_{k=-\infty }^{\infty }{X}_a(j\Omega )\ast \delta (\Omega -k{\Omega }_s)$

Since convolving any function with a shifted impulse simply shifts the function itself: $X_s(j\Omega )=\frac{1}{T_s}{\sum }_{k=-\infty }^{\infty }{X}_a(j(\Omega -k{\Omega }_s))$

Physical Meaning of the Result

This equation proves that the spectrum of the sampled signal $X_s(j\Omega )$ consists of an infinite series of replicated copies of the original analog spectrum $X_a(j\Omega )$, spaced at integer multiples of the sampling frequency ${\Omega }_s$ and scaled in amplitude by $\frac{1}{T_s}$.

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3. High-Yield Solved Numericals

3.1 Example 1: Multi-Tone Composite Waveform [PYQ 2016, 2015]

Question: Calculate the Nyquist Rate and Nyquist Interval for the analog signal: $x(t)=5\cos (200\pi t)+10\sin (500\pi t)$

Step-by-Step Solution:

1. Extract individual frequencies:

   • For Term 1 ($5\cos (200\pi t)$): ${\omega }_1=200\pi \text{ rad/s}⟹f_1=\frac{{\omega }_1}{2\pi }=100\text{ Hz}$
   • For Term 2 ($10\sin (500\pi t)$): ${\omega }_2=500\pi \text{ rad/s}⟹f_2=\frac{{\omega }_2}{2\pi }=250\text{ Hz}$

2. Identify maximum frequency ($f_{\text{max}}$): $f_{\text{max}}=\max (f_1,f_2)=250\text{ Hz}$

3. Evaluate Nyquist parameters:

   • Nyquist Rate: $f_{\text{Nyquist}}=2f_{\text{max}}=2\times 250\text{ Hz}=500\text{ Hz}$
   • Nyquist Interval: $T_{\text{Nyquist}}=\frac{1}{f_{\text{Nyquist}}}=\frac{1}{500}=0.002\text{ seconds}(2\text{ ms})$

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3.2 Example 2: Sinc and Sinc-Squared Functions

Question: Calculate the Nyquist Rate for the signal: $x(t)={\text{sinc}}^2(100t)$

Step-by-Step Solution:

To solve this, we must check both standard engineering definitions of the Sinc function since different instructors use different standards.

• Interpretation A: Normalized Sinc (Rao Standard: $\text{sinc}(u)=\frac{\sin (\pi u)}{\pi u}$)

  1. Express the base term: $\text{sinc}(100t)=\frac{\sin (100\pi t)}{100\pi t}$
  2. Find the base maximum frequency: ${\omega }_{\text{base}}=100\pi \text{ rad/s}⟹f_{\text{base}}=50\text{ Hz}$
  3. Because squaring in time corresponds to frequency-domain convolution, the band limit doubles: $f_{\text{max}}=2\times f_{\text{base}}=2\times 50=100\text{ Hz}$
  4. Calculate the Nyquist Rate: $f_{\text{Nyquist}}=2f_{\text{max}}=2\times 100\text{ Hz}=200\text{ Hz}$

• Interpretation B: Unnormalized Sinc ($\text{sinc}(u)=\frac{\sin (u)}{u}$)

  1. Express the base term: $\text{sinc}(100t)=\frac{\sin (100t)}{100t}$
  2. Find base maximum frequency: ${\omega }_{\text{base}}=100\text{ rad/s}⟹f_{\text{base}}=\frac{50}{\pi }\text{ Hz}$
  3. Doubling the band limit due to multiplication: ${\omega }_{\text{max}}=2\times 100=200\text{ rad/s}⟹f_{\text{max}}=\frac{100}{\pi }\text{ Hz}$
  4. Calculate the Nyquist Rate: $f_{\text{Nyquist}}=2f_{\text{max}}=\frac{200}{\pi }\approx 63.66\text{ Hz}$

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4. Common Mistakes That Cost Marks

The Sinc frequency factor error

When squaring a signal (like $x(t)=g^2(t)$ or convolving in frequency), the highest frequency component doubles ($f_{\text{max}}=2f_{\text{base}}$). Many students forget to double the base frequency and evaluate the Nyquist rate incorrectly, losing major marks.

The Angular vs. Cyclic frequency trap ($\Omega$ vs. $f$)

Always verify if the question asks for the Nyquist rate in Hz or rad/s.

• Nyquist rate in Hz: $f_s=2f_{\text{max}}$
• Nyquist rate in rad/s: ${\Omega }_s=2{\Omega }_{\text{max}}$ Confusing the two factors in the proof or numericals leads to scaling errors by $2\pi$.

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5. PYQ Bank — Verbatim Questions & Answer Plans

5.1 PYQ 2019/2017 [10 Marks]

Question: State and prove the Shannon-Nyquist sampling theorem.

• Answer Plan:
  1. State the theorem formally (Section 1).
  2. Model uniform impulse train sampling in the time domain (Section 2.1).
  3. Compute the Fourier series representation of the impulse train $p(t)$ (Section 2.2).
  4. Carry out the frequency-domain convolution integration to derive $X_s(j\Omega )=\frac{1}{T_s}\sum X_a(j(\Omega -k{\Omega }_s))$ (Section 2.2).
  5. Explain the reconstruction criteria (over-sampling, Nyquist, under-sampling).

5.2 PYQ 2016/2015 [5 Marks]

Question: Determine the Nyquist Rate and Nyquist Interval of the signal $x(t)=5\cos (200\pi t)+10\sin (500\pi t)$.

• Answer Plan: Follow the step-by-step extraction and numerical solution laid out in Section 3.1 to secure all 5 marks.

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6. Self-Check Before Moving On

• Can you state the Shannon-Nyquist theorem verbatim without skipping the “band-limited” constraint?
• Can you derive the Fourier Transform of an impulse train $P(j\Omega )=\frac{2\pi }{T_s}\sum \delta (\Omega -k{\Omega }_s)$?
• Can you prove that multiplication by $p(t)$ in time replicates the spectrum in frequency?
• Do you know how to calculate the maximum frequency for composite signals and sinc-squared signals?

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 6 & 7), 1.05 The Sampling Theorem Bridge & Multi-Rate Processing.md.