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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...
The signal x(t) = cos(6t) is a cosine wave with a frequency of 6 radians per unit time. The "cos" function represents the cosine waveform, and the argument inside the cosine function, 6t, determines the frequency of the waveform.
The frequency of 6 radians per unit time means that the waveform completes 6 full cycles in one unit of time. The time period of one cycle is given by T = 2π/ω, where ω is the angular frequency. In this case, ω = 6, so the time period is T = 2π/6 = π/3.
The waveform starts at t = 0 with a value of x(0) = cos(6(0)) = cos(0) = 1. As time progresses, the waveform oscillates between -1 and 1, completing 6 full cycles in one unit of time.
The waveform x(t) = cos(6t) is a periodic signal with a frequency of 6 radians per unit time and an amplitude of 1. It can be used to model various physical phenomena, such as oscillations in mechanical systems, electrical signals, or sound waves.
The frequency of 6 radians per unit time means that the waveform completes 6 full cycles in one unit of time. The time period of one cycle is given by T = 2π/ω, where ω is the angular frequency. In this case, ω = 6, so the time period is T = 2π/6 = π/3.
The waveform starts at t = 0 with a value of x(0) = cos(6(0)) = cos(0) = 1. As time progresses, the waveform oscillates between -1 and 1, completing 6 full cycles in one unit of time.
The waveform x(t) = cos(6t) is a periodic signal with a frequency of 6 radians per unit time and an amplitude of 1. It can be used to model various physical phenomena, such as oscillations in mechanical systems, electrical signals, or sound waves.
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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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