The carrier frequency will be at 2 1 2 w and the upper

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: at the lower edge of the passband of H (ω − ω2 ), then only the upper sideband will be passed. To make this happen, set ω2 − ω1 = ω2 − π W , or ω1 = π W (f1 = W/2). The other alternative would be to place the center of M (ω − ω2 + ω1 ) at the upper edge of the passband of H (ω − ω2 ) - but this requires that ω2 − ω1 = ω2 + π W , which cannot be done (ω1 is a positive number). Thus, provided that ω1 = π W (f1 = W/2), choosing the upper sign will give USB. The carrier frequency will be at ω2 − ω1 = ω2 − π W and the upper edge of the transmitted spectrum will be at ω2 − ω1 + 2π W = ω2 + π W . S (ω ) = 1c - In part b, we determined that choosing the + sign yields USB. If the - sign is chosen then S (ω ) = 1 1 M (ω + ω2 + ω1 )H (ω + ω...
View Full Document

This homework help was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online