Sampling Theorem for Bandpass Signals: Minimum Sampling Rate & Allowed Ranges (Solved Examples + MCQs)

A radio signal occupying 20–25 kHz contains frequencies up to 25 kHz, so the ordinary (low-pass) sampling theorem says we need at least 50 kHz. Yet it can be sampled perfectly at only 10 kHz! This is the bandpass sampling theorem (also called undersampling or harmonic sampling). It is a regular short question in digital communication and DSP exams (BCS EEE/Telecom, BTCL, university finals) and the key idea behind modern software-defined radios.

⚡ Key takeaways

• A bandpass signal has its spectrum between f_L and f_H with bandwidth B = f_H − f_L, and nothing near DC.
• Theorem: it can be recovered from samples taken at f_s = 2f_H/k, where k = ⌊f_H/B⌋ (largest integer not exceeding f_H/B).
• The minimum rate always lies between 2B and 4B; it equals exactly 2B when f_H is an integer multiple of B.
• General allowed ranges: 2f_H/n ≤ f_s ≤ 2f_L/(n − 1), n = 1, 2, …, k.
• Not every rate above the minimum works — some higher rates cause aliasing!

📑 Contents

1. Recap: low-pass sampling theorem
2. What is a bandpass signal?
3. Statement of the bandpass sampling theorem
4. Why it works (animation)
5. Allowed sampling-rate ranges
6. Solved examples
7. Applications and practical points
8. Practice MCQs
9. বাংলায় দ্রুত রিভিশন

1. Recap: Low-Pass Sampling Theorem

For a low-pass (baseband) signal with highest frequency f_m, the sampling rate must satisfy f_s ≥ 2f_m (Nyquist rate). If it is lower, spectral copies overlap and aliasing occurs. Applying this rule blindly to a high-frequency bandpass signal wastes a huge sampling rate, because the band from 0 to f_L is empty.

2. What Is a Bandpass Signal?

A bandpass signal has significant spectral content only in a band f_L ≤ |f| ≤ f_H, where f_L > 0. Examples: an AM/FM station around its carrier, a GSM channel, a 70 MHz IF signal in a receiver. Its bandwidth B = f_H − f_L is usually much smaller than f_H.

3. Statement of the Bandpass Sampling Theorem

📜 TheoremA bandpass signal with highest frequency f_H and bandwidth B can be completely recovered from its samples if it is sampled at f_s = 2f_H / k, where k = ⌊ f_H / B ⌋ The minimum sampling rate lies in the range 2B ≤ f_s,min < 4B. If f_H/B is an integer, f_s,min = 2B exactly.

Compare: low-pass sampling of the same signal needs 2f_H. Bandpass sampling reduces this by the factor k.

4. Why It Works (Animation)

Sampling copies the spectrum (both the positive band and its negative-frequency mirror) to every multiple of f_s. For low-pass signals we need a big f_s to keep copies apart. For a bandpass signal there is a lot of empty space below f_L, so a much smaller f_s can drop the copies into the gaps without overlap. The animation sweeps f_s from 9 to 42 kHz for a 20–25 kHz signal: the green zones on the bar are the sampling rates that work.

Animation 1 – Dark blue = original band; light blue/orange = shifted copies of the positive/negative bands. Overlap means aliasing.

5. Allowed Sampling-Rate Ranges

The complete condition for no aliasing is:

2f_H / n ≤ f_s ≤ 2f_L / (n − 1), n = 1, 2, …, ⌊f_H/B⌋

(For n = 1 the upper limit is ∞, i.e. ordinary Nyquist sampling f_s ≥ 2f_H.) For f_L = 20 kHz, f_H = 25 kHz (B = 5 kHz, f_H/B = 5):

n	2fH/n (kHz)	2fL/(n−1) (kHz)	Allowed fs
1	50	∞	fs ≥ 50 kHz (Nyquist)
2	25	40	25 – 40 kHz
3	16.67	20	16.67 – 20 kHz
4	12.5	13.33	12.5 – 13.33 kHz
5	10	10	10 kHz only (= 2B, minimum)

⚠ ImportantRates between the zones — for example 15 kHz or 22 kHz — cause aliasing even though they are higher than the 10 kHz minimum. In practice, engineers choose f_s near the middle of a wide allowed zone, because a rate exactly at an edge (like 10 kHz) leaves no guard band and fails with any drift or imperfect filter.

6. Solved Examples

Example 1. A bandpass signal occupies 18 kHz to 22 kHz. Find the minimum sampling rate and compare with low-pass sampling.

1. B = 22 − 18 = 4 kHz; f_H/B = 22/4 = 5.5 ⇒ k = 5
2. f_s,min = 2f_H/k = 44/5 = 8.8 kHz (between 2B = 8 and 4B = 16 ✔)
3. Allowed range for n = 5: 8.8 ≤ f_s ≤ 2 × 18/4 = 9 kHz
4. Low-pass sampling would need 2 × 22 = 44 kHz — five times higher.

Example 2. A signal has f_L = 100 kHz and B = 20 kHz. Find f_s,min.

1. f_H = 120 kHz; f_H/B = 6 (integer) ⇒ k = 6
2. f_s,min = 240/6 = 40 kHz = 2B

Example 3. An IF signal is centred at 70 MHz with 10 MHz bandwidth (65–75 MHz). Can it be sampled at 40 MHz?

1. Check n with 2f_H/n ≤ 40 ≤ 2f_L/(n−1): n = 4 gives 150/4 = 37.5 ≤ 40 ≤ 130/3 = 43.3 ✔
2. Yes — 40 MHz lies in an allowed zone, so a 40 MSPS ADC can digitise this 75 MHz signal directly.

7. Applications and Practical Points

• Software-defined radio and digital IF receivers: the ADC samples the IF signal directly, removing a mixer stage.
• Reduced ADC speed, memory and processing power compared with Nyquist sampling at 2f_H.
• Requirements: a sharp bandpass anti-aliasing filter (to remove noise outside f_L–f_H, which would otherwise fold in), and an ADC whose analog input bandwidth exceeds f_H with low aperture jitter.
• Depending on n, the sampled spectrum may appear inverted (frequency-reversed); this is corrected digitally.
• Quadrature (I/Q) sampling is an alternative: the band is shifted to baseband and two channels (I and Q) are each sampled at a rate of at least B.

8. Practice MCQs (click to see answer)

1. A bandpass signal has f_L = 10 kHz and f_H = 12 kHz. Its bandwidth is: (a) 22 kHz (b) 12 kHz (c) 2 kHz (d) 10 kHz

✔ (c) 2 kHz.

2. In the bandpass sampling theorem, f_s = 2f_H/k where k is: (a) f_L/B (b) largest integer ≤ f_H/B (c) f_H/f_L (d) 2B

✔ (b) Largest integer not exceeding f_H/B.

3. Minimum sampling rate for a bandpass signal lies between: (a) B and 2B (b) 2B and 4B (c) 4B and 8B (d) f_L and f_H

✔ (b) 2B and 4B.

4. Signal band 20–25 kHz. Minimum sampling rate is: (a) 50 kHz (b) 25 kHz (c) 10 kHz (d) 5 kHz

✔ (c) 10 kHz (k = 5, 50/5).

5. When f_H is an integer multiple of B, the minimum sampling rate equals: (a) B (b) 2B (c) 4B (d) 2f_H

✔ (b) 2B.

6. Bandpass sampling is also known as: (a) Oversampling (b) Undersampling (c) Flat-top sampling (d) Natural sampling

✔ (b) Undersampling (or harmonic/IF sampling).

7. For a 20–25 kHz signal, which rate causes aliasing? (a) 12.8 kHz (b) 18 kHz (c) 22 kHz (d) 30 kHz

✔ (c) 22 kHz — it lies between the 16.67–20 and 25–40 kHz zones.

8. Low-pass sampling of an 18–22 kHz signal would require at least: (a) 8 kHz (b) 8.8 kHz (c) 22 kHz (d) 44 kHz

✔ (d) 44 kHz (2f_H).

9. The filter used before bandpass sampling is a: (a) Low-pass filter (b) High-pass filter (c) Bandpass filter (d) Notch filter

✔ (c) Bandpass anti-aliasing filter.

10. The main practical benefit of bandpass sampling is: (a) Higher SNR (b) Lower required ADC sampling rate (c) No need for filters (d) Infinite bandwidth

✔ (b) Lower required ADC sampling rate.

9. বাংলায় দ্রুত রিভিশন

📝 পরীক্ষার আগে এক নজরে

• ব্যান্ডপাস সংকেত: যার স্পেকট্রাম f_L থেকে f_H পর্যন্ত; ব্যান্ডউইথ B = f_H − f_L।
• উপপাদ্য: f_s = 2f_H/k, যেখানে k = f_H/B-এর পূর্ণসংখ্যা অংশ। সর্বনিম্ন হার 2B থেকে 4B-এর মধ্যে; f_H যদি B-এর পূর্ণ গুণিতক হয় তবে ঠিক 2B।
• অনুমোদিত পরিসর: 2f_H/n ≤ f_s ≤ 2f_L/(n − 1)। এর মাঝের কিছু বেশি হারেও অ্যালিয়াসিং হতে পারে।
• উদাহরণ: ২০–২৫ kHz সংকেত ১০ kHz-এ নমুনায়ন করা যায়, যেখানে সাধারণ নিয়মে ৫০ kHz লাগত।
• ব্যবহার: সফটওয়্যার-ডিফাইন্ড রেডিও, ডিজিটাল IF রিসিভার — কম গতির ADC যথেষ্ট।

Related Posts

• What Is Aliasing? Nyquist Rate & Anti-Aliasing
• Aperture Effect vs Aliasing
• Pulse Code Modulation (PCM) MATLAB Code
• Why Is Modulation Needed?

References: Simon Haykin, Communication Systems; Vaughan, Scott & White, “The Theory of Bandpass Sampling”, IEEE Trans. Signal Processing, 1991; Lyons, Understanding Digital Signal Processing. Images and animations: original work by Totalecer.