# Tuto8_sol - EE3210 Tutorial 8(Solution Question 1 a...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE3210 Tutorial 8 (Solution) Question 1 a) Frequency Response: H ( e j ω ) e j ωn − 1 H (e jω )e jω ( n −1) = e jωn 4 4 H ( e jω ) = 4 − e − jω b) Magnitude Response: 4 4 4(4 − cos ω − j sin ω ) = = − jω 4 − cos ω + j sin ω (4 − cos ω ) 2 + sin 2 ω 4−e 4 4 H ( e jω ) = = 17 − 8 cos ω (4 − cos ω ) 2 + sin 2 ω H ( e jω ) = c) The magnitude increases when ω changes from -π to 0, and it decreases when ω changes from 0 to π. Hence, it is a low-pass filter. d) The fundamental period of the signal is 8. The fundamental frequency is π/4. Using Euler’s formula, 3πn 1 j 3πn / 4 1 − j 3πn / 4 x[n] = sin = e − e 4 2j 2j e) The Fourier series representation of the output is 2 2 y[ n] = e j 3πn / 4 − e − j 3πn / 4 . − j 3π / 4 ) j (4 − e j ( 4 − e j 3π / 4 ) Question 2 a) The Fourier transform of cos ω0t is equal to πδ (ω − ω 0 ) + πδ (ω + ω 0 ) . Hence, R ( jω ) = 2πδ (ω − 110π ) + 2πδ (ω + 110π ) + 8πδ (ω − 120π ) + 8πδ (ω + 120π ) + 2πδ (ω − 130π ) + 2πδ (ω + 130π ) b) The Fourier transform of x(t) is given by X ( jω ) = 4πδ (ω − 10π ) + 16πδ (ω ) + 4πδ (ω + 10π ) c) The Fourier transform of 1 is equal to 2πδ (ω ) . Taking inverse transform, we get x(t ) = 4 cos 10πt + 8 . You can verify that the answer is correct by applying the following formula: cos A cos B = 1 [cos( A + B) + cos( A − B)] . 2 ...
View Full Document

## This note was uploaded on 11/22/2009 for the course ELECTRONIC EE3210 taught by Professor Sungchiwan during the Spring '09 term at École Normale Supérieure.

Ask a homework question - tutors are online