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1. Multiple choice: Suppose we sample the continuous-time signal f(t) = cos(2740t) at a rate of F, = 60 samples/second and let F(t) denote the continuous-time impulse train carrying the sample values. We then pass F(t) through an ideal sinc interpolator with cutoff frequency 30 Hz. f(t) F(t) sampling ↑ Fs What is the resulting continuous-time signal y(t)? ideal sinc interpolation y(t)

1. Multiple choice: Suppose we sample the continuous-time signal f(t) = cos(2740t) at a rate of F, = 60 samples/second and let F(t) denote the continuous-time impulse train carrying the sample values. We then pass F(t) through an ideal sinc interpolator with cutoff frequency 30 Hz. f(t) F(t) sampling ↑ Fs What is the resulting continuous-time signal y(t)? ideal sinc interpolation y(t)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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### Digital Signal Processing Problem **Question:** Suppose we sample the continuous-time signal \( f(t) = \cos(2\pi 40t) \) at a rate of \( F_s = 60 \) samples/second and let \( \bar{f}(t) \) denote the continuous-time impulse train carrying the sample values. We then pass \( \bar{f}(t) \) through an ideal sinc interpolator with cutoff frequency 30 Hz. *What is the resulting continuous-time signal \( y(t) \)?* **Options:** (a) \( y(t) = \cos(2\pi 20t) \) (b) \( y(t) = \cos(2\pi 30t) \) (c) \( y(t) = \cos(2\pi 40t) \) (d) \( y(t) = \cos(2\pi 60t) \) **Diagram Explanation:** A block diagram illustrates the process: - **Input:** \( f(t) \) is the original continuous-time signal. - **Sampling Block:** Converts \( f(t) \) into \( \bar{f}(t) \) by sampling at a rate \( F_s = 60 \) samples/second. - **Interpolation Block:** \( \bar{f}(t) \) is passed through an ideal sinc interpolator to produce \( y(t) \). The cutoff frequency here is 30 Hz. The question revolves around understanding the effects of sampling and interpolation on this specific signal.
Transcribed Image Text:### Digital Signal Processing Problem **Question:** Suppose we sample the continuous-time signal \( f(t) = \cos(2\pi 40t) \) at a rate of \( F_s = 60 \) samples/second and let \( \bar{f}(t) \) denote the continuous-time impulse train carrying the sample values. We then pass \( \bar{f}(t) \) through an ideal sinc interpolator with cutoff frequency 30 Hz. *What is the resulting continuous-time signal \( y(t) \)?* **Options:** (a) \( y(t) = \cos(2\pi 20t) \) (b) \( y(t) = \cos(2\pi 30t) \) (c) \( y(t) = \cos(2\pi 40t) \) (d) \( y(t) = \cos(2\pi 60t) \) **Diagram Explanation:** A block diagram illustrates the process: - **Input:** \( f(t) \) is the original continuous-time signal. - **Sampling Block:** Converts \( f(t) \) into \( \bar{f}(t) \) by sampling at a rate \( F_s = 60 \) samples/second. - **Interpolation Block:** \( \bar{f}(t) \) is passed through an ideal sinc interpolator to produce \( y(t) \). The cutoff frequency here is 30 Hz. The question revolves around understanding the effects of sampling and interpolation on this specific signal.
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