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: The balanced modulator in fig. below is to be investigated for the possible generation of AM - DSB signal with m<1 each diode has the characteristic: i(t) = a¡e(t) + aze² (t) %3D e;(t) cos o (t) e(t) cos w(t) Acos om(t) e(t) i>(1) R. (a) Determine the maximum allowable value of A. (b) Determine the maximum and minimum bandwidth of an ideal BPF required on the output if one of the diodes is open - circuited.

: The balanced modulator in fig. below is to be investigated for the possible generation of AM - DSB signal with m<1 each diode has the characteristic: i(t) = a¡e(t) + aze² (t) %3D e;(t) cos o (t) e(t) cos w(t) Acos om(t) e(t) i>(1) R. (a) Determine the maximum allowable value of A. (b) Determine the maximum and minimum bandwidth of an ideal BPF required on the output if one of the diodes is open - circuited.

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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Type your question here The balanced modulator in fig. below is to be investigated for the possible generation of AM - DSB signal with m
Q8: The balanced modulator in fig. below is to be investigated for the possible generation of AM - DSB signal with m<1 each diode has the characteristic: i(t) = a,e(t) + aze? (t) e(t) cos o (t) (t) cos @(t) Acos @m(t) e:(t) i(1) (a) Determine the maximum allowable value of A. (b) Determine the maximum and minimum bandwidth of an ideal BPF required on the output if one of the diodes is open - circuited. Ans: (a) (b) Bmax = 2(fe - 2fm); Bmin = 2 fm: 2a2
Transcribed Image Text:Q8: The balanced modulator in fig. below is to be investigated for the possible generation of AM - DSB signal with m<1 each diode has the characteristic: i(t) = a,e(t) + aze? (t) e(t) cos o (t) (t) cos @(t) Acos @m(t) e:(t) i(1) (a) Determine the maximum allowable value of A. (b) Determine the maximum and minimum bandwidth of an ideal BPF required on the output if one of the diodes is open - circuited. Ans: (a) (b) Bmax = 2(fe - 2fm); Bmin = 2 fm: 2a2
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