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Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t) = x(2t + 5) is
- a)8
- b)12
- c)16
- d)32
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in se...
Πt) + 2cos(10πt)
1. What is the frequency of the first cosine term?
The frequency of the first cosine term is 6 Hz, which can be calculated as 6π/2π = 3 cycles per second.
2. What is the frequency of the second cosine term?
The frequency of the second cosine term is 10 Hz, which can be calculated as 10π/2π = 5 cycles per second.
3. What is the amplitude of the first cosine term?
The amplitude of the first cosine term is 1, as it has a coefficient of 1 in front of the cosine function.
4. What is the amplitude of the second cosine term?
The amplitude of the second cosine term is 2, as it has a coefficient of 2 in front of the cosine function.
5. What is the total amplitude of the signal x(t)?
The total amplitude of the signal x(t) is the sum of the amplitudes of each cosine term, which is 1 + 2 = 3.
1. What is the frequency of the first cosine term?
The frequency of the first cosine term is 6 Hz, which can be calculated as 6π/2π = 3 cycles per second.
2. What is the frequency of the second cosine term?
The frequency of the second cosine term is 10 Hz, which can be calculated as 10π/2π = 5 cycles per second.
3. What is the amplitude of the first cosine term?
The amplitude of the first cosine term is 1, as it has a coefficient of 1 in front of the cosine function.
4. What is the amplitude of the second cosine term?
The amplitude of the second cosine term is 2, as it has a coefficient of 2 in front of the cosine function.
5. What is the total amplitude of the signal x(t)?
The total amplitude of the signal x(t) is the sum of the amplitudes of each cosine term, which is 1 + 2 = 3.
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Community Answer
Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in se...
Given,
x(t) = cos(6πt) + sin(8πt)
The bandwidth of the signal will be:
Now, y(t) = x(2t + 5) can be written as:
Taking the Fourier transform, we get:
Thus, the frequency spectrum of X(jω) expands by 2.
This will make the highest frequency component of Y(jω) as:
2 × 4 = 8 Hz
Hence, the Nyquist rate will be:
fs = 2 × 8 = 16 samples/sec
x(t) = cos(6πt) + sin(8πt)
The bandwidth of the signal will be:
Now, y(t) = x(2t + 5) can be written as:
Taking the Fourier transform, we get:
Thus, the frequency spectrum of X(jω) expands by 2.
This will make the highest frequency component of Y(jω) as:
2 × 4 = 8 Hz
Hence, the Nyquist rate will be:
fs = 2 × 8 = 16 samples/sec
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