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A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"?
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A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The sig...
Why Bandpass Sampling Cannot Be Used?
Introduction:
In the given scenario, a voice signal m(t) is amplitude-modulated to generate an AM signal ƒ(t) = A(1 - m(t)) cos 2πƒct, where ƒc = 600 kHz. The objective is to digitize and archive the AM signal ƒ(t). However, it is stated that the signal should be sampled at 1.2 times the Nyquist frequency. In this context, it is necessary to understand why bandpass sampling cannot be used.
Explanation:
1. Nyquist Sampling Theorem:
The Nyquist Sampling Theorem states that in order to accurately reconstruct a continuous-time signal, it must be sampled at a rate greater than or equal to twice the highest frequency component in the signal. This is known as the Nyquist rate or Nyquist frequency.
2. Nyquist Frequency for AM Signal:
The AM signal ƒ(t) is generated by modulating the voice signal m(t). The frequency range of m(t) is mentioned as 5 kHz to 15 kHz. Therefore, the highest frequency component of the AM signal is the sum of the highest frequency component of m(t) and the carrier frequency ƒc. Hence, the highest frequency component of ƒ(t) is 15 kHz + 600 kHz = 615 kHz.
3. Nyquist Sampling Rate:
According to the Nyquist theorem, the AM signal ƒ(t) should be sampled at a rate greater than or equal to twice the highest frequency component. Therefore, the Nyquist sampling rate for ƒ(t) is 2 * 615 kHz = 1.23 MHz.
4. Sampling at 1.2 times the Nyquist Frequency:
In the given scenario, it is specified that the AM signal ƒ(t) should be sampled at 1.2 times the Nyquist frequency. This means that the sampling rate is 1.2 * 1.23 MHz = 1.476 MHz.
5. Bandpass Sampling Concept:
Bandpass sampling is a technique used for signals with a narrow bandwidth centered around a carrier frequency. It allows the sampling rate to be reduced to twice the bandwidth instead of the Nyquist rate. However, in this scenario, the AM signal ƒ(t) has a bandwidth of 615 kHz, which is significantly larger than the desired sampling rate of 1.476 MHz.
Conclusion:
Bandpass sampling cannot be used in this scenario because the AM signal ƒ(t) has a bandwidth that exceeds the desired sampling rate. The Nyquist sampling theorem requires a sampling rate of at least twice the highest frequency component in the signal, and in this case, the Nyquist rate is higher than the specified sampling rate. Therefore, the signal should be sampled at 1.2 times the Nyquist frequency to ensure accurate reconstruction.
Introduction:
In the given scenario, a voice signal m(t) is amplitude-modulated to generate an AM signal ƒ(t) = A(1 - m(t)) cos 2πƒct, where ƒc = 600 kHz. The objective is to digitize and archive the AM signal ƒ(t). However, it is stated that the signal should be sampled at 1.2 times the Nyquist frequency. In this context, it is necessary to understand why bandpass sampling cannot be used.
Explanation:
1. Nyquist Sampling Theorem:
The Nyquist Sampling Theorem states that in order to accurately reconstruct a continuous-time signal, it must be sampled at a rate greater than or equal to twice the highest frequency component in the signal. This is known as the Nyquist rate or Nyquist frequency.
2. Nyquist Frequency for AM Signal:
The AM signal ƒ(t) is generated by modulating the voice signal m(t). The frequency range of m(t) is mentioned as 5 kHz to 15 kHz. Therefore, the highest frequency component of the AM signal is the sum of the highest frequency component of m(t) and the carrier frequency ƒc. Hence, the highest frequency component of ƒ(t) is 15 kHz + 600 kHz = 615 kHz.
3. Nyquist Sampling Rate:
According to the Nyquist theorem, the AM signal ƒ(t) should be sampled at a rate greater than or equal to twice the highest frequency component. Therefore, the Nyquist sampling rate for ƒ(t) is 2 * 615 kHz = 1.23 MHz.
4. Sampling at 1.2 times the Nyquist Frequency:
In the given scenario, it is specified that the AM signal ƒ(t) should be sampled at 1.2 times the Nyquist frequency. This means that the sampling rate is 1.2 * 1.23 MHz = 1.476 MHz.
5. Bandpass Sampling Concept:
Bandpass sampling is a technique used for signals with a narrow bandwidth centered around a carrier frequency. It allows the sampling rate to be reduced to twice the bandwidth instead of the Nyquist rate. However, in this scenario, the AM signal ƒ(t) has a bandwidth of 615 kHz, which is significantly larger than the desired sampling rate of 1.476 MHz.
Conclusion:
Bandpass sampling cannot be used in this scenario because the AM signal ƒ(t) has a bandwidth that exceeds the desired sampling rate. The Nyquist sampling theorem requires a sampling rate of at least twice the highest frequency component in the signal, and in this case, the Nyquist rate is higher than the specified sampling rate. Therefore, the signal should be sampled at 1.2 times the Nyquist frequency to ensure accurate reconstruction.
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A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"?
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A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"?.
A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"?.
Solutions for A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal ƒ(t) = A(1+ m(t)) cos 2πƒct, where ƒc = 600 kHz. The AM signal ƒ(t) is to be digitized and archived. This is done by first sampling ƒ(t) at1.2times the Nyquist frequency My question is "why can't we use bandpass sampling concept here"? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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