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

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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 ...
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