Discrete-time Fourier transform (DTFT) representation of DT aperiodic signals Section The (DT) Fourier transform (or spectrum) of x[n] is
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1 Discrete-time Fourier transform (DTFT) representation of DT aperiodic signals Section 5. 3 The (DT) Fourier transform (or spectrum) of x[n] is X ( e jω) = n= x[n]e jωn x[n] can be reconstructed from its spectrum using the inverse Fourier transform x[n] = X ( e jω) e jωn dω The above two equations are referred to as the analysis equation and the synthesis equation respectively. 3 These slides are based on lecture notes of Professors L. Lampe, C. Leung, and R. Schober SM 38
2 Notation: X ( e jω) = F{x[n]} x[n] = F {X ( e jω) } x[n] and X ( e jω) form a Fourier transform pair, denoted by x[n] X ( e jω) Main differences from CTFT. X ( e jω) is periodic with period since e jωn is periodic in ω with period. 2. As a result of the above-mentioned periodicity, the integration for the synthesis equation is over a finite interval (of length ). SM 39
3 Example: Determine the DTFT of h[n] = a n u[n], a <. Then, H ( e jω) = = = n= n= a n e jωn ( a e jω ) n a e jω Hence, H ( e jω) = = ( acosω)2 +(asinω) 2 2acosω +a 2 A plot of H ( e jω) for a =.5 is shown in the figure below. SM 4
4 H e jω ( ) = 2 a π Since H ( e jω) = and = = = + a a e jω a ejω a e jω a e jω a e jω 2acosω +a 2 a e jω = acosω jasinω π ω we have H ( e jω) asinω = tan acosω SM 4
5 Example: Determine the DTFT of the rectangular pulse x[n] = {, n N, otherwise. Then, X ( e jω) = = N e jωn n= N 2N e jω(m N) m= (using m = n+n) = ejωn ( e jω(2n+)) e jω = e jωn e jω(n+ 2 ) { e jω(n+ 2 ) e jω(n+ 2 ) } e jω 2 { e j ω 2 e jω 2} {e jω(n+ 2 ) e jω(n+ 2 ) } = { e j ω 2 e jω 2} SM 42
6 = 2jsin[ ω ( )] N + 2 2jsin[ω/2] = sin[ ω ( )] N + 2 sin[ω/2] For N =, X ( e jω) = sin[ ω + ω 2 sin [ ] ω 2 = sinωcos ω 2 +cosωsin ω 2 sin ω 2 ] = 2sin ω 2 cos ω 2 cos ω 2 +cosωsin ω 2 sin ω 2 = 2cos 2 ω 2 +cosω = +2cosω Sketch X ( e jω) for N = in the figure below. SM 43
7 x[ n] n X( e jω ) 3 2 π π ω DTFT convergence issues Recall the DTFT analysis and synthesis equations X ( e jω) = n= x[n]e jωn (4) x[n] = X ( e jω) e jωn dω (5) SM 44
8 The analysis equation (4) contains an infinite sum. Sufficient conditions to guarantee convergence are: () x[n] is absolutely summable, i.e. n= x[n] < or (2) x[n] has finite energy, i.e. n= x[n] 2 < There are no convergence issues associated with the synthesis equation (5) since the integral is over a finite interval. For example, unlike the CTFT case, the Gibbs phenomenon is absent when (5) is used. SM 45
9 DTFT for periodic signals Section 5.2 Similar to the CT case, we can express the FT of DT periodic signals in terms of impulses. It can be shown (see (5.9) and (5.2) in textbook) that if the DT periodic signal x[n] of fundamental period N has as DTFS coefficients a k, i.e. x[n] = k=<n> a k e jk ( N)n then X ( e jω) = k= a k δ ( ω k ). (6) N Furthermore, if we define x[n] as one basic period of x[n], it can be shown that a k = N X (e jk N ). SM 46
10 In words, (6) states that the DTFT of x[n] is a sequence of impulses located at multiples of the fundamental frequency N ; the strength of the impulse located at ω = k N is a k. Example: x[n] = cos(ω n), where ω = 3. cos-----n n { cosω n} π ω ω ω cosω n = 2 ejω n + 2 e jω n SM 47
11 Fact: e jω n l= δ(ω ω l) Proof: The inverse DTFT of the RHS is = l= = e jω n. l= [ δ(ω ω l)e jωn dω ] δ(ω ω l)e jωn dω The last line follows since only one impulse is included in any interval of length. Hence, cos 3 n π + l= l= δ (ω 3 ) l δ (ω + 3 ) l SM 48
12 Exercise: Show that sin 3 n π j δ (ω 3 ) l l= δ (ω + 3 ) l l= Plot sin ( 3 n) and its DTFT in the figures below. sin----- n n { sin ω n} ω ω ω SM 49
13 Example: Determine the DTFT of the impulse train x[n] = k= δ[n kn]. Then, a k = N n=<n> x[n] e jk ( N)n = N and from (6), we have X ( e jω) = N k= δ ( ω k ). N SM 5
14 Some properties of the DTFT Sections Periodicity Section 5.3. The DTFT is always periodic in ω with period, i.e. X (e j(ω+)) = X ( e jω) In contrast, the CTFT is not generally periodic. 2. Linearity Section similar to CTFT Let x[n] X ( e jω), y[n] Y ( e jω). Then, Z z[n] = ax[n] + by[n] ( e jω) ( = ax e jω) + by (e jω) SM 5
15 Proof: follows directly from the defn. of the DTFT. Example: What is the DTFT of the signal, z[n], shown below? z[ n] n SM 52
16 3. Time-shift Section similar to CTFT If x[n] X ( e jω), then Proof: x[n N] e jωn X ( e jω). Note that when a signal is delayed by N, the spectrum amplitude is unchanged whereas the spectrum phase is changed by ωn. SM 53
17 Example: What are the DTFT s of the signals shown below? x [ n] n x 2 [ n] n SM 54
18 4. Conjugation Section similar to CTFT If x[n] X ( e jω), then Proof: x [n] X ( e jω). As a result, if x[n] is real, we have X ( e jω) = X ( e jω) i.e., the spectrum magnitude is an even function of ω and the spectrum phase is an odd function of ω. SM 55
19 5. Differencing and Accumulation Section Differencing: Using the linearity and timeshift properties, if x[n] X ( e jω), then the DTFT of the first difference signal is given by x[n] x[n ] ( e jω) X ( e jω). Accumulation: Let y[n] = n m= x[m]. Then, Y ( e jω) = X( e jω) e jω +πx ( e j) k= δ(ω k). SM 56
20 Example: Determine the DTFT of the unit step sequence u[n] = {, n <, n. u[ n] n SM 57
21 6. Parseval s relation Section similar to CTFT If x[n] X ( e jω), then n= x[n] 2 = X ( e jω) 2 dω. (7) The LHS of (7) represents the total energy in the signal x[n]. The integrand, X ( e jω) 2, in the RHS is the energy density spectrum of x[n]. SM 58
22 7. Convolution Section 5.4 similar to CTFT If y[n] = x[n] h[n] then Y ( e jω) = X ( e jω) H ( e jω). Proof: SM 59
23 Example: Determine the convolution of the DT signals x[n] and h[n] shown below. x[ n] h[ n] n n SM 6
24 8. Multiplication Section 5.5 If y[n] = a[n]b[n] then Y ( e jω) = A ( e jθ) B ( e j(ω θ)) dθ. The RHS of the previous equation is referred to as the periodic convolution of A ( e jω) and B ( e jω). Proof: SM 6
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