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Unformatted text preview: Chapters 4 and 5 Discrete Time Fourier Transform Recall that we wrote the sampled signal x s ( t ) = ∑ ∞ k =∞ x ( kT ) δ ( t kT ). We calculate its Fourier Transform. We do the following: Ex. Find the Continuous Time Fourier Transform of δ ( t kT ). Ex. Using superposition, find the CT Fourier Transform of x s ( t ). 1 Now, you just calculated that x s ( t ) ↔ ∞ X n =∞ x ( nT ) e jnωT Let x ( nT ) = x [ n ] and make a change of variables Ω = ωT (we’ll talk more about this later — it relates the discretefrequency variable Ω to the continuous frequency variable ω via the sampling period T ) and we get: DTFT : X (Ω) = ∞ X n =∞ x [ n ] e j Ω n • Discrete in time but continuous in frequency and periodic • Spectrum of discrete signal x [ n ] • Will compare the DTFT of a discrete signal x [ n ] with the Continuous Time Fourier Transform of a sampled continuous time signal x s ( t ) = x ( t ) p ( t ) Formula to calculate inverse DTFT (since X (Ω) is periodic): x [ n ] = 1 2 π Z 2 π X (Ω) e j Ω n d Ω where DTFT is periodic in frequency with period 2 π . Why? Because e j Ω is periodic with period 2 π . e j Ω = e j (Ω+2 π ) = e j Ω e j 2 π = e j Ω . Not all DTFTs converge due to the infinite sum. 2 DTFT Theorems We cover a few here and you can read about the rest in the textbook. LINEARITY ax 1 [ n ] + bx 2 [ n ] ←→ aX 1 (Ω) + bX 2 (Ω) TIME SHIFT x [ n ] ←→ X (Ω) x [ n n ] ←→ ? x [ n n ] ←→ e j Ω n X (Ω) So a shift in time causes a linear phase shift in frequency – adds a linear term to the phase of the DTFT. 3 Frequency Shift x [ n ] ←→ X (Ω) e j Ω n x [ n ] ←→ ? Modulation causes a shift in frequency. CONVOLUTION IN TIME As usual, x 1 [ n ] * x 2 [ n ] ←→ X 1 (Ω) X 2 (Ω) Ex. Given h [ n ] = a n u [ n ] ,  a  < 1. Find its inverse system h i [ n ]. 4 Modulation– MULTIPLICATION OF SIGNALS x 1 [ n ] x 2 [ n ] ←→ 1 2 π X 1 (Ω) O X 2 (Ω) where N denotes CIRCULAR CONVOLUTION: X 1 (Ω) O X 2 (Ω) = Z 2 π X 1 ( θ ) X 2 (Ω θ )) dθ y [ n ] = x 1 [ n ] x 2 [ n ] ←→ Take its DTFT : Y (Ω) = X n y [ n ] e j Ω n = X n x 1 [ n ] x 2 [ n ] e j Ω n x 1 [ n ] = 1 2 π Z 2 π X 1 ( a ) e jan da x 2 [ n ] = 1 2 π Z 2 π X 2 ( b ) e jbn db Y (Ω) = X n 1 2 π Z da X 1 ( a ) e jan da 1 2 π Z db X 2 ( b ) e jbn db e j Ω n = 1 2 π 2 Z da Z db X 1 ( a ) X 2 ( b ) X n e j ( a + b ) n e j Ω n dadb Now, X n e j ( a + b ) n e j Ω n is just the DTFT of e j ( a + b ) n that is, e j ( a + b ) n ↔ 2 πδ (Ω a b ) So, Y (Ω) = 1 2 π Z da 2 π Z db 2 π X 1 ( a ) X 2 ( b ) δ (Ω a b ) dadb = 1 2 π Z 2 π X 1 ( a ) X 2 (Ω a ) da = 1 2 π X 1 (Ω) O X 2 (Ω) 5 Ex. 1 FindEx....
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This note was uploaded on 09/10/2010 for the course ECE 107 taught by Professor Fullterton during the Spring '07 term at UCSD.
 Spring '07
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