hw6_sol

hw6_sol - Spring 2008 Homework#6 Solutions EE113 Digital...

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Unformatted text preview: Spring 2008 Homework #6 Solutions EE113: Digital Signal Processing Prof. Mihaela van der Schaar Prepared by Martin Andersen and Hyunggon Park Problems from Prof. Sayed’s lecture notes: 12.3 • Let x(n) = sin(nπ/8) nπ ∞ 1 = 8 sinc(n/8). Then, since x(n) is symmetric, ∞ x(n)2 = n=2 = 1 1 x(n)2 − x(0)2 − x(1)2 2 n=−∞ 2 11 · 2 2π π 1 |X (ejω )|2 dω − x(0)2 − x(1)2 . 2 −π The Fourier transform of x(n) is given by ω 2π 1 8 X (ejω ) = rect where rect(t) = 1 |t| ≤ 1 2 0 |t| > 1 . 2 Thus, 1 2π π 1 2π |X (ejω )|2 dω = −π π /8 1 dω = −π/8 1 8 and ∞ x(n)2 = n=2 1 sin(π/8) 11 11 − x(0)2 − x(1)2 = − − 28 2 8 64 π 2 ≈ 0.0398. • Using the identity cos 2θ = cos2 θ − sin2 θ = 1 − 2 sin2 θ , we get ∞ ∞ cos π n 1 4 = −2 n2 n2 n=1 n=1 ∞ = = n=1 π2 6 1 −2 n2 sin π n 8 n ∞ n=1 ∞ − 2π 2 n=1 2 2 sin π n 8 n sin π n 8 nπ 2 1 1 π2 − π2 − ≈ 0.5654. = 6 8 64 12.4 • Since the function sin(4ω ) sin(3ω ) 1−cos2 (ω/2) π 0 is even, we get 1 sin(4ω ) sin(3ω ) dω = 1 − cos2 (ω/2) 2 π −π = π F −1 1 sin(4ω ) sin(3ω ) dω sin2 (ω/2) sin(3ω ) sin(4ω ) sin2 (ω/2) . n=0 Using the transform pair x(n) = 1 0≤n≤L−1 0 otherwise X (ejω ) = ↔ L ω=0 L−1 e−jω 2 sin(ωL/2) sin(ω/2) 0 < |ω | < 2π. we get sin(3ω ) sin(4ω ) sin(ω 6/2) −jω(8−1)/2 sin(ω 8/2) = ej 6ω e−jω(6−1)/2 e 2 sin(ω/2) sin(ω/2) sin (ω/2) and hence x1 (n) = F −1 e−jω(6−1)/2 sin(ω 6/2) sin(ω/2) = 1 0 0≤n≤5 otherwise x2 (n) = F −1 e−jω(8−1)/2 sin(ω 8/2) sin(ω/2) = 1 0≤n≤7 0 otherwise. Now, π 0 sin(4ω ) sin(3ω ) dω = π [δ(n + 6) ⊛ x1 (n) ⊛ x2 (n)]|n=0 1 − cos2 (ω/2) ∞ =π x1 (k)x2 (6 − k) k =−∞ 5 =π x1 (k)x2 (6 − k) = 6π. k =0 • Recall that sinc(x) = ∞ sin πx πx (i.e. the normalized sinc function). sin π n 4 n n=1 3 = = π 4 ∞ 3 sinc3 (n/4) n=1 1 π · 2 4 3 ∞ sinc3 (n/4) . −1 + n=−∞ ∞ F Using the property x1 (n)x2 (n) ↔ 1 jω jω 2π X1 (e ) ∗ X2 (e ), sinc3 (n/4) the summation n=−∞ can be evaluated as 1 2π 2 . F {sinc(n/4)} ∗ F {sinc(n/4)} ∗ F {sinc(n/4)} ω =0 2 The DTFTs of sinc(n/4) and (sinc(n/4)) are shown below. 1 2π F{sinc(n/4)} F{sinc(n/4)} 4 −π 4 ∗ F{sinc(n/4)} 4 π 4 ω 2 −π 2 π 2 ω The top of the triangle has the value π 4 1 2π −π 4 42 dω = 4. Convolving the triangle with F {sinc(n/4)} and evaluating at ω = 0 yields 1 2π 1 F {sinc(n/4)} ∗ F {sinc(n/4)} ∗ F {sinc(n/4)} 2π = ω =0 2 2π π 4 4(4 − 0 8 ω ) dω π = 3. ∞ sinc3 (n/4) = 3 and Thus, n=−∞ ∞ 3 sin π n 4 n n=1 π 4 = 3 . ∞ x(n) = X (ejω ) 13.6 (a) n=−∞ ∞ (b) =0 ∞ x(n)e−jπn = X (ejω ) n (−1) x(n) = n=−∞ (c) ω =0 =0 ω =π n=−∞ • The energy of x(n) can be found using Parseval’s theorem as follows: ∞ |x(n)|2 = Ex = n=−∞ = = π /4 1 π 0 1 π 1 2π 16 ω π 4π + 4π 3 = |X (ejω )|2 dω 2π 2 π /2 dω + 42 dω π/4 16 . 3 • The energy of nx(n) can be found by using the properties of the DTFT as follows: – Let y (n) = nx(n), then j 16 0<ω< π π 4 d Y (ejω ) = j X (ejω ) = −j 16 − π < ω < 0 π 4 dω π 0 4 < |ω | < π. – Using Parseval’s theorem, ∞ |y (n)|2 = Ey = n=−∞ = 1 π 16 π 3 2 1 2π π 64 = 2. 4 π |Y (ejω )|2 dω 2π (d) x(0) = 1 2π X (ejω )ejωn dω 2π = n=0 (e) Let y (n) = x(n) cos(nπ/4), then Y 1 2π (ejω ) X (ejω ) dω = 2π 3 2 is given by π π 1 1 Y (ejω ) = X ej (ω− 4 ) + X ej (ω+ 4 ) . 2 2 The sum ∞ n=−∞ y (n) evaluates to ∞ x(n) cos(nπ/4) = Y (ejω ) ω =0 = 4. n=−∞ The DTFT of Y (ejω ) is shown below: |Y (ejω )| 4 2 π − 34 (f) −π 4 π 4 ω 3π 4 • The DTFT of x(−n) is X (e−jω ). Since X (ejω ) is an even function, then X (e−jω ) = X (ejω ). • Let y (n) = nx(−n), then j 16 0<ω< π π 4 d Y (ejω ) = j X (ej ) = −j 16 − π < ω < 0 π π4 dω 0 4 < |ω | < π. • Let z (n) = nx(−n) sin(nπ/2), then Z (ejω ) = π π 1 Y ej (ω− 2 ) − Y ej (ω+ 2 ) 2j . • Finally, let v (n) = (−1)n nx(−n) sin(nπ/2), and note that (−1)n = e−jπn . Using the frequency shift property, we get V (ejω ) = Z (ej (ω−π) ) = π π 1 Y ej (ω+ 2 ) − Y ej (ω− 2 ) 2j . The DTFT is shown in the figure below. |V (ejω )| 8 π π − 34 π 4 −π 4 4 3π 4 ω ∠V (ejω ) π π − 34 π 4 −π 4 ω 3π 4 −π 13.7 Let x1 (n) be the output of the lowpass filter, x2 (n) the output of the first modulator, and x3 (n) the output of the second modulator. The frequency responses H1 (ejω ), H2 (ejω ), and H3 (ejω ) are shown below: X (ejω ) 4 −π 2 π 4 −π 4 ω π 2 X1 (ejω ) 4 2 −π 2 −π 8 π 8 ω π 2 X2 (ejω ) 2 1 −π 2 −π 4 π 4 ω π 2 X3 (ejω ) 4 2 −π π − 34 −π 2 π 2 5 3π 4 π ω The energy of X3 (ejω ) is 1 2π π |X3 (ejω )|2 dω = 4 −π 1 2π π /8 0 2 ω π/8 2 16 π 2 π dω = 2 1π 38 3 1 =. 3 13.8 We see from the figure that X (ejω ) can be written as π π ej (π−2ω) 4 ≤ |ω | < 2 π X (ejω ) = 2ej (π−2ω) π ≤ |ω | ≤ 34 2 0 otherwise. Notice that the magnitude of X (ejω ) is an even function of ω and the phase of X (ejω ) is an odd function of ω . Hence, x(n) = 1 2π 1 = 2π + π X (ejω )ejωn dω −π −π 2 π − 34 j (π −2ω ) jωn 2e 1 2π π 2 π 4 e 1 dω + 2π ej (π−2ω) ejωn dω + −π 4 −π 2 ej (π−2ω) ejωn dω 3π 4 1 2π π 2 2ej (π−2ω) ejωn dω. These integral are of the same form, b b ej (π−2ω) ejωn dω = =− = ej (n−2)ω dω a a a b ej (π+(n−2)ω) dω = − 1 ej (n−2)ω j (n − 2) ej (n−2)a ej (n−2)b − j (n − 2) b ω =a . Using this result we get π π π ej (n−2) 4 − ej (n−2) 2 2(ej (n−2) 2 − ej (n−2) 1 + x(n) = π (n − 2) 2j 2j + 2(e−j (n−2) 3π 4 π π 3π 4 ) π − e−j (n−2) 2 ) e−j (n−2) 2 − e−j (n−2) 4 + 2j 2j 3π 1 π π − 2 sin (n − 2) sin (n − 2) + sin (n − 2) π (n − 2) 4 4 2 1 n−2 3(n − 2) n−2 3 1 = sinc − sinc + sinc 4 4 2 4 2 2 = where sinc(x) = sin(xπ) . Since X (ejω ) is periodic, the DTFT of x(n) over the interval [0, 2π ] xπ is simply as shown below. 6 |X (ejω )| 2 1 π 2 π 2 π 3π 2 2π π 3π 2 2π ω ∠X (ejω ) π 2 ω −π 2 13.9 Multiplying by e−j (2ω+π/2) yields e−j (4ω−2π/3) X ′ (ejω ) = 2e−j (4ω−2π/3) 0 π 4 π 2 ≤ |ω | < π 2 π ≤ |ω | ≤ 34 otherwise. Thus, the magnitude response is the same (see above), but the phase response changes (see below). ∠X ′ (ejω ) π 2π 3 π 3 −π 3 π − 23 π 2 π 3π 2 2π ω −π Problems from Mitra: 3.32 Recall that an even real-valued sequence has a real-valued DTFT, and an odd real-valued sequence has a imaginary-valued DTFT. (a) y1 (n) must be an even sequence since Y1 (z ) is real-valued. (b) y2 (n) must be an odd sequence since Y2 (z ) is imaginary-valued. (c) y3 (n) must be an odd sequence since Y3 (z ) is imaginary-valued. 7 3.36 (a) Let |X (ejω )|2 = 4 5+4 cos ω . Then we have 4 4 2 4 2 = = = · jω + e−jω ) jω )(1 + 2e−jω ) jω 1 + 2e−jω 5 + 4 cos ω 5 + 2(e (1 + 2e 1 + 2e X (ejω ) 1 and x(n) = − 2 π 0 n u(n). From Parseval’s relation we get 2 ∞ ∞ π 1 |X (e )| dω = π 2π jω jω 2 2 |x(n)| = π |X (e )| dω = π −π n=−∞ 3.39 Given x(n) = {4, 2, −1, 5, −3, 1, -2 , 4, 2} evaluate the following: ∞ (a) X (ej 0 ) = x(n) = 12 n=−∞ ∞ ∞ x(n)(−1)n = −12 x(n)e−jπn = (b) X (ejπ ) = n=−∞ n=−∞ π (c) X (ejω ) dω = 2πx(0) = −4π −π π ∞ jω |x(n)|2 = 160π 2 |X (e )| dω = 2π (d) −π π (e) −π 2 dX (ejω ) dω X (e−jω ) n=−∞ ∞ |nx(n)|2 = 1972π dω = 2π n=−∞ 8 n=0 1 4 n = 4π . 3 ...
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This note was uploaded on 10/03/2011 for the course EE 113 taught by Professor Walker during the Spring '08 term at UCLA.

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