NOTES-DTFT

# NOTES-DTFT - Chapters 4 and 5 Discrete Time Fourier...

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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

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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 discrete-frequency 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 0 ] ←→ ? x [ n - n 0 ] ←→ e - j Ω n 0 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

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Frequency Shift x [ n ] ←→ X (Ω) e j Ω 0 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 - θ )) 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 - a - b ) So, Y (Ω) = 1 2 π Z da 2 π Z db 2 π X 1 ( a ) X 2 ( b ) δ - a - b ) dadb = 1 2 π Z 2 π X 1 ( a ) X 2 - a ) da = 1 2 π X 1 (Ω) O X 2 (Ω) 5

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Ex. 1 Find X (Ω) where x [ n ] = a n u [ n ], | a | < 1. What if | a | > 1 ?
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