2. Consider the sampling expansion, u(t) = Σk u(kT)sinc ( – k), of u(t) (Figure 1). u(t) 1 1/2 3 O 6 t Figure 1: u(t). (a) Write u(t)= Σku(kT)sinc ( + − k) for T = 1. (b) Find u5(t) = Σ½–0 u(kT)sinc († – k) for T = 1. k=0 (c) Compute the energy of the reconstruction error, e(t) = u(t) — u5(t).

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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2.
Consider the sampling expansion, u(t) = Σk u(kT)sinc († — k), of u(t) (Figure 1).
u(t)
1
1/2
3
Figure 1: u(t).
(a) Write u(t) = Σk u(kT)sinc († — k) for T = 1.
(b) _Find u5(t) = Σ%-o u(kT)sinc († − k) for T = 1.
-
k=0
(c) Compute the energy of the reconstruction error, e(t) = u(t) — u5(t).
Transcribed Image Text:2. Consider the sampling expansion, u(t) = Σk u(kT)sinc († — k), of u(t) (Figure 1). u(t) 1 1/2 3 Figure 1: u(t). (a) Write u(t) = Σk u(kT)sinc († — k) for T = 1. (b) _Find u5(t) = Σ%-o u(kT)sinc († − k) for T = 1. - k=0 (c) Compute the energy of the reconstruction error, e(t) = u(t) — u5(t).
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