--- title: "Sampling 3cos200πt + 2sin300πt − 3cos500πt" subject: "Signals and Systems" chapter: 4 key: "semester_4/signals_and_systems:sampling-reconstruction-analog" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-reconstruction-analog site: BCT Engineering --- # Sampling 3cos200πt + 2sin300πt − 3cos500πt Signals and Systems · Chapter 4: Sampling - Asked in: 2080 Asoj, 8 marks - Video: [Sampling Frequency Problem Example 1](https://www.youtube.com/watch?v=dDv-vSJ6JS8) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-reconstruction-analog - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4 ## Question Consider an analog signal x(t) = 3cos 200πt + 2sin 300πt − 3cos 500πt. i) Determine the minimum sampling rate required to avoid aliasing. ii) Suppose that the signal is to be sampled at the rate of Fs = 400 Hz. What is the discrete time signal obtained after sampling? iii) Find reconstructed signal xᵣ(t) from sampled signal calculated above in (ii). ## Short answer Minimum sampling rate is twice the highest frequency 250 Hz, so 500 samples/s. At Fs = 400 Hz, x[n] = 3cos(πn/2) + 2sin(3πn/4) − 3cos(3πn/4), because the 250 Hz term aliases to 150 Hz. Ideal reconstruction gives xr(t) = 3cos200πt + 2sin300πt − 3cos300πt. ## Answer **Frequencies in $x(t)$** ($\omega = 2\pi f$): | Term | $\omega$ (rad/s) | $f$ (Hz) | |---|---|---| | $3\cos200\pi t$ | $200\pi$ | 100 | | $2\sin300\pi t$ | $300\pi$ | 150 | | $-3\cos500\pi t$ | $500\pi$ | 250 | #### (i) Minimum sampling rate $f_m = 250$ Hz, so $$ F_{s,\min} = 2f_m = 2\times250 = 500\ \text{Hz} $$ **Answer:** 500 samples/s (Nyquist rate). #### (ii) Discrete-time signal for $F_s = 400$ Hz Put $t = nT = n/F_s = n/400$: $$ \begin{aligned} x[n] &= 3\cos\left(\frac{200\pi n}{400}\right) + 2\sin\left(\frac{300\pi n}{400}\right) - 3\cos\left(\frac{500\pi n}{400}\right) \\ &= 3\cos\left(\frac{\pi n}{2}\right) + 2\sin\left(\frac{3\pi n}{4}\right) - 3\cos\left(\frac{5\pi n}{4}\right) \end{aligned} $$ The last term has digital frequency $5\pi/4 > \pi$, so it is aliased. Since $\cos\left(\frac{5\pi n}{4}\right) = \cos\left(2\pi n - \frac{3\pi n}{4}\right) = \cos\left(\frac{3\pi n}{4}\right)$: $$ x[n] = 3\cos\left(\frac{\pi n}{2}\right) + 2\sin\left(\frac{3\pi n}{4}\right) - 3\cos\left(\frac{3\pi n}{4}\right) $$ In normalized frequency: $\frac{100}{400} = \frac14$, $\frac{150}{400} = \frac38$, and $\frac{250}{400} = \frac58$, which aliases to $1 - \frac58 = \frac38$ (i.e. 150 Hz). #### (iii) Reconstructed signal An ideal reconstruction filter passes only $|f| \le F_s/2 = 200$ Hz, so each digital frequency $\omega$ in $[0, \pi]$ maps back to $f = \frac{\omega F_s}{2\pi}$: | Digital term | $\omega$ | Analog $f$ | |---|---|---| | $3\cos(\pi n/2)$ | $\pi/2$ | 100 Hz | | $2\sin(3\pi n/4)$ | $3\pi/4$ | 150 Hz | | $-3\cos(3\pi n/4)$ | $3\pi/4$ | 150 Hz | $$ x_r(t) = 3\cos(200\pi t) + 2\sin(300\pi t) - 3\cos(300\pi t) $$ **Answer:** $x_r(t) = 3\cos 200\pi t + 2\sin 300\pi t - 3\cos 300\pi t$. Because $F_s = 400$ Hz is below the Nyquist rate (500 Hz), the 250 Hz component has been aliased to 150 Hz, so $x_r(t) \ne x(t)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Minimum sampling rate for 3cos200πt + 2sin300πt - 3cos500πt A. 250 Hz B. 1000 Hz C. 300 Hz D. 500 Hz Answer: D. fm = 250 Hz, so 2fm = 500 Hz. (Key point: Min sampling rate) 2. At Fs = 400 Hz which component is aliased? A. The 250 Hz term (digital frequency 5π/4 > π) B. The 100 Hz term C. The 150 Hz term D. None Answer: A. 250 Hz exceeds Fs/2 = 200 Hz. (Key point: Aliased term) 3. The 250 Hz tone sampled at 400 Hz becomes digital frequency A. 5π/4, i.e. 250 Hz B. 3π/4, i.e. 150 Hz C. π/4, i.e. 50 Hz D. π/2, i.e. 100 Hz Answer: B. cos(5πn/4) = cos(3πn/4). (Key point: Alias frequency) 4. The reconstructed xr(t) is A. Same as x(t) B. 3cos200πt + 2sin300πt - 3cos500πt C. 3cos200πt + 2sin300πt - 3cos300πt D. 3cos200πt + 2sin300πt Answer: C. The ideal filter passes |f| <= 200 Hz, so the aliased term appears at 150 Hz. (Key point: Reconstructed signal) ## Related in Signals and Systems - [Frequency of a signal sampled at 300 Hz](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampled-signal-frequency-300) — Signals and Systems, chapter 4 - [Aliasing; Nyquist rate of 1 + cos + sin](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-nyquist-1-cos-sin) — Signals and Systems, chapter 4, asked 4× - [Sampling theorem; aliasing cause; Nyquist](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-aliasing-cause-nyquist) — Signals and Systems, chapter 4 - [Impulse train sampling, aliasing, reconstruction](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aimpulse-train-aliasing-reconstruction) — Signals and Systems, chapter 4 - [Band-limited signals; aliasing; Nyquist rate](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Abandlimited-aliasing-nyquist) — Signals and Systems, chapter 4 - [Aliasing via impulse-train sampling](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-impulse-train-frequency) — Signals and Systems, chapter 4 ## Related in other subjects - [Aperture and aliasing; sampling 10cos(20πt)cos(200πt) at 250 samples/s](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aaperture-aliasing-10cos20pit-250) — Communication Systems, chapter 4 - [Laplace: 3x3 grid 200/300/400/500](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-3x3-200-300-400-500) — Numerical Method, chapter 7, asked 5× - [Information rate of a quantized 100 Hz signal](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ainformation-rate-100hz-four-symbols) — Communication Systems, chapter 6 - [Impulse invariance Butterworth, 200 Hz and 500 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Aimpulse-invariance-butter-200-500hz) — Digital Signal Analysis and Processing, chapter 6