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-3. Determine the Ayquist rate for each of the Co following signals assuming the dyquist. nate for act) is fs Hz. of offs Bad/sec, a) gixel) b) 4₂ (4)= c) (z (t) = d) fult) = dae(t) dz 2<² (t) xelt) cos & fs t

-3. Determine the Ayquist rate for each of the Co following signals assuming the dyquist. nate for act) is fs Hz. of offs Bad/sec, a) gixel) b) 4₂ (4)= c) (z (t) = d) fult) = dae(t) dz 2<² (t) xelt) cos & fs 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
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Title: Nyquist Rate Determination for Various Signals** **Problem Statement:** Determine the Nyquist rates for each of the following signals, assuming the Nyquist rate for \( x(t) \) is \( f_s \) Hz or \( 2\pi f_s \) Rad/sec. **Signals:** a) \( y_1(t) = 2x(t) + x(t-1) \) b) \( y_2(t) = \frac{dx(t)}{dt} \) c) \( y_3(t) = x^2(t) \) d) \( y_4(t) = x(t) \cos(4\pi f_s t) \) **Explanation:** To determine the Nyquist rate for each signal, we consider the effect of each operation (such as delay, differentiation, squaring, and modulation) on the frequency spectrum of \( x(t) \). - **Signal a:** - \( y_1(t) = 2x(t) + x(t-1) \) involves a delayed component of the original signal. The delay in the time domain does not change the bandwidth but affects the phase. The Nyquist rate remains unaffected by the constant and the delay. - **Signal b:** - \( y_2(t) = \frac{dx(t)}{dt} \) is the derivative of \( x(t) \). Differentiation in the time domain corresponds to a multiplication by a linear term in the frequency domain, potentially increasing the bandwidth. - **Signal c:** - \( y_3(t) = x^2(t) \) involves squaring the signal, which leads to frequency doubling. This operation typically increases the bandwidth, potentially doubling the original Nyquist rate. - **Signal d:** - \( y_4(t) = x(t) \cos(4\pi f_s t) \) represents a modulation with a cosine wave at frequency \( 2f_s \), which results in frequency shifting. The bandwidth can effectively double due to the new frequency components introduced. By analyzing these operations, we can conclude the necessary adjustments in the Nyquist rate for each transformed signal.
Transcribed Image Text:**Title: Nyquist Rate Determination for Various Signals** **Problem Statement:** Determine the Nyquist rates for each of the following signals, assuming the Nyquist rate for \( x(t) \) is \( f_s \) Hz or \( 2\pi f_s \) Rad/sec. **Signals:** a) \( y_1(t) = 2x(t) + x(t-1) \) b) \( y_2(t) = \frac{dx(t)}{dt} \) c) \( y_3(t) = x^2(t) \) d) \( y_4(t) = x(t) \cos(4\pi f_s t) \) **Explanation:** To determine the Nyquist rate for each signal, we consider the effect of each operation (such as delay, differentiation, squaring, and modulation) on the frequency spectrum of \( x(t) \). - **Signal a:** - \( y_1(t) = 2x(t) + x(t-1) \) involves a delayed component of the original signal. The delay in the time domain does not change the bandwidth but affects the phase. The Nyquist rate remains unaffected by the constant and the delay. - **Signal b:** - \( y_2(t) = \frac{dx(t)}{dt} \) is the derivative of \( x(t) \). Differentiation in the time domain corresponds to a multiplication by a linear term in the frequency domain, potentially increasing the bandwidth. - **Signal c:** - \( y_3(t) = x^2(t) \) involves squaring the signal, which leads to frequency doubling. This operation typically increases the bandwidth, potentially doubling the original Nyquist rate. - **Signal d:** - \( y_4(t) = x(t) \cos(4\pi f_s t) \) represents a modulation with a cosine wave at frequency \( 2f_s \), which results in frequency shifting. The bandwidth can effectively double due to the new frequency components introduced. By analyzing these operations, we can conclude the necessary adjustments in the Nyquist rate for each transformed signal.
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