I want part D only **Analysis of Analog Signal and Nyquist Rate**Consider an analog signal \( x_a(t) = \sin(250\pi t) + \sin(400\pi t) + \sin(750\pi t) \).**(a) What is the Nyquist rate for \( x_a(t) \)?**The Nyquist rate is the minimum sampling rate required to avoid aliasing and accurately reconstruct the signal. It is twice the maximum frequency present in the signal.**(b) What range of sampling rates allows exact reconstruction of this signal from its samples?**To accurately reconstruct the signal, the sampling rate should be greater than the Nyquist rate, which is twice the highest frequency component in the signal.**(c) Suppose we sample \( x_a(t) \) at the sampling rate of 600 Hz. What is the resulting discrete-time signal \( x(n) \)? What are the discrete-time frequencies in \( x(n) \)?**Sampling at 600 Hz allows us to capture the discrete-time version of the signal. The discrete-time frequencies can be derived based on the sampling theorem.**(d) If \( x(n) \) in part (c) is passed through an ideal D/A converter, what is the reconstructed signal \( y_a(t) \)? Is \( x_a(t) = y_a(t) \)?**An ideal D/A converter would reconstruct the continuous-time signal from the discrete-time samples. Check if the output \( y_a(t) \) is equivalent to the original \( x_a(t) \).**Detailed Explanation:**The given signal is a summation of three sine waves with frequencies 125 Hz, 200 Hz, and 375 Hz (since frequency = 1/2 of the coefficient of \(\pi t\)). The Nyquist rate is 750 Hz, as it is double the highest frequency (375 Hz) present in the signal.For exact reconstruction, the sampling rate must be above 750 Hz. At 600 Hz, certain frequencies might lead to aliasing, impacting the reconstruction. Discrete-time frequencies are calculated by mapping continuous-time frequencies to the normalized frequency (considering sampling rate). An ideal D/A conversion is expected to restore the signal, assuming no aliasing occurs.

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Answered: (d) If x(n) in part (c) is passed…

(d) If x(n) in part (c) is passed through an ideal D/A converter, what is the reconstructed signal ya(t)? Is xa(t) = ya(t)?

Introductory Circuit Analysis (13th Edition)

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Publisher:Robert L. Boylestad

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Question

I want part D only

**Analysis of Analog Signal and Nyquist Rate**

Consider an analog signal \( x_a(t) = \sin(250\pi t) + \sin(400\pi t) + \sin(750\pi t) \).

**(a) What is the Nyquist rate for \( x_a(t) \)?**

The Nyquist rate is the minimum sampling rate required to avoid aliasing and accurately reconstruct the signal. It is twice the maximum frequency present in the signal.

**(b) What range of sampling rates allows exact reconstruction of this signal from its samples?**

To accurately reconstruct the signal, the sampling rate should be greater than the Nyquist rate, which is twice the highest frequency component in the signal.

**(c) Suppose we sample \( x_a(t) \) at the sampling rate of 600 Hz. What is the resulting discrete-time signal \( x(n) \)? What are the discrete-time frequencies in \( x(n) \)?**

Sampling at 600 Hz allows us to capture the discrete-time version of the signal. The discrete-time frequencies can be derived based on the sampling theorem.

**(d) If \( x(n) \) in part (c) is passed through an ideal D/A converter, what is the reconstructed signal \( y_a(t) \)? Is \( x_a(t) = y_a(t) \)?**

An ideal D/A converter would reconstruct the continuous-time signal from the discrete-time samples. Check if the output \( y_a(t) \) is equivalent to the original \( x_a(t) \).

**Detailed Explanation:**

The given signal is a summation of three sine waves with frequencies 125 Hz, 200 Hz, and 375 Hz (since frequency = 1/2 of the coefficient of \(\pi t\)). The Nyquist rate is 750 Hz, as it is double the highest frequency (375 Hz) present in the signal.

For exact reconstruction, the sampling rate must be above 750 Hz. At 600 Hz, certain frequencies might lead to aliasing, impacting the reconstruction. Discrete-time frequencies are calculated by mapping continuous-time frequencies to the normalized frequency (considering sampling rate). An ideal D/A conversion is expected to restore the signal, assuming no aliasing occurs.

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