程序代写 ECE 310 Exam 2 Review

HKN ECE 310 Exam 2 Review
Corey IUC
April 4, 2020

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DTFT Exercise 1: Part (a)
Let our signal be h[n] = {1, 2, 1}. ↑
Compute the DTFT of h[n]:
F(h[n]) = 􏰃 h[n]e−jωn n=−∞
= 1 + 2e−jω + e−j2ω
= e−jω(2 + ejω + e−jω) = e−jω(2 + 2 cos(ω)).

DTFT Exercise 1: Part (b)
Let our signal be h[n] = {1, 2, 1}. ↑
Plot the magnitude response:
Hd(ω) = e−jω(2 + 2 cos(ω))
|Hd(ω)| = 􏰒Hd(ω)Hd∗(ω) = |2 + 2 cos(ω)|.
Figure 1: |Hd(ω)|

DTFT Exercise 1: Part (c)
Let our signal be h[n] = {1, 2, 1}. ↑
Plot the phase response:
Recall that we may decompose frequency responses as follows:
Hd(ω) = Re{Hd(ω)} + jIm{Hd(ω)} = |Hd(ω)|ej∠Hd(ω) Thus, we have
∠Hd(ω) = −ω (±π jumps where Hd(ω) changes signs.)
Figure 2: ∠Hd(ω)

DTFT Exercise 2: Part (a)
Given h[n] real-valued and
π ≤ ω < 3π 2 π ≤ ω < 3π 2 Plot |Hd(ω)| ∈ [−π, π]: , ∠Hd(ω) = 242 We are given h[n] is real-valued, thus Hermitian symmetry is our best friend here. Hd(ω) = Hd∗(−ω), |Hd(ω)| = |Hd(−ω)|, ∠Hd(ω) = −∠Hd(−ω) Also, utilizing the 2π periodicity of the DTFT, we have: Figure 3: |Hd(ω)| DTFT Exercise 2: Part (b) Given h[n] real-valued and π ≤ ω < 3π 2 π ≤ ω < 3π 2 , ∠Hd(ω) = 242 Plot ∠Hd(ω) ∈ [−π, π]: Similarly as in Part (a), Figure 4: ∠Hd(ω) Sinusoidal Response Exercise 1: Part (a) Given LSI system y[n] = x[n] − 2x[n − 1] + x[n − 2]: Find transfer function H(z): y[n] = x[n] − 2x[n − 1] + x[n − 2] Z{·} −1 −2 ↔ Y(z)=X(z)(1−2z +z ) H(z)= Y(z) = 1 − 2z−1 + z−2. Sinusoidal Response Exercise 1: Part (c) Given LSI system y[n] = x[n] − 2x[n − 1] + x[n − 2]: Find frequency response Hd(ω): The solution in part (a) is clearly BIBO stable since we have no poles. Thus, 􏰅 Hd(ω) = H(z)􏰅􏰅z=ejω = 1 − 2e−jω + e−j2ω = e−jω(ejω + e−jω − 2) = e−jω(2 cos(ω) − 2) = ej(−ω+π)(2 − 2 cos(ω)) Sinusoidal Response Exercise 1: Part (c) Given LSI system y[n] = x[n] − 2x[n − 1] + x[n − 2]: Find the output y[n] for signals: From our solution in (b), we have: |Hd(ω)| = |2 − 2 cos(ω)|, ∠Hd(ω) = ej(−ω+π). x1[n] = 2 + cos(πn) : |Hd(0)| = 0, ∠Hd(0) = π, |Hd(π)| = 4, ∠Hd(π) = 0 y1[n] = 2(0) cos(0n + π) + (4) cos(πn + 0) = 4 cos(πn) Sinusoidal Response Exercise 1: Part (c) Given LSI system y[n] = x[n] − 2x[n − 1] + x[n − 2]: Find the output y[n] for signals: From our solution in (b), we have: |Hd(ω)| = |2 − 2 cos(ω)|, ∠Hd(ω) = ej(−ω+π). x2[n]=ejπ4n +sin􏰀−π2n􏰁: 􏰋π􏰌√−jπ􏰅􏰋π􏰌􏰅 􏰋π􏰌π Hd 4=(2−2)e4,􏰅􏰅Hd−2􏰅􏰅=2,∠Hd−2=−2. y [n] = 􏰋(√2 − 2)e−j π 􏰌 ej π n + (2) sin 􏰋− π n − π 􏰌 √ππ π2π2 =( 2−2)ej(4n−4)+2sin􏰋− n− 􏰌 Sampling Exercise 1 For analog signal xa(t) = cos (Ω0t) sampled at T = 1 s to obtain x[n] = cos 􏰀 π4 n􏰁 1000 What are possible values for Ω0? Always remember ωd = ΩaT! Moreover, by 2π periodicity of the x[n]=cos􏰋π4n􏰌≡cos􏰋􏰋π4 +2πk􏰌n􏰌, k∈Z. Thus, Ω 0 T = 􏰋 π4 + 2 π k 􏰌 Ω0 =250π+2000πk, k∈Z. Possible values for Ω0 are then 250π ( choice a), -1750π (choice c) and 4250π (choice d). Sampling Exercise 2: Part (a) The maximum frequency of Xa(Ω) is 4000π. Sketch DTFT spectrum when T = 1 In each part, we should focus on where the maximum frequency maps. This will help us identify aliasing. If there is no aliasing, the shape of the spectrum will not change. By ωd = ΩaT, ωd = 4000π 8000 Figure5:T1 = 1 s 8000 Sampling Exercise 2: Part (b) The maximum frequency of Xa(Ω) is 4000π. Sketch DTFT spectrum when T = 1 By ωd = ΩaT, ωd = 4000π 4000 Figure6:T1 = 1 s 4000 Sampling Exercise 2: Part (c) The maximum frequency of Xa(Ω) is 4000π. Sketch DTFT spectrum when T = 1 By ωd = ΩaT, ωd = 4000π 2000 Figure7:T1 = 1 s 2000 DFT Exercise 1: Part (a) Given x[n] = cos􏰀π3n􏰁, 0 ≤ n < 18, For which value(s) of k is X[k] largest? ωk = 2πk N π3 = π9 k k1 = 3. We also have energy at ω = − π3 : − π3 = π9 k Instead by N periodicity of the DFT, we should say k2 ≡ k2 + N = 15 DFT Exercise 1: Part (a) Given x[n] = cos􏰀π3n􏰁, 0 ≤ n < 18, For which k is X[k] largest if x[n] is padded with 72 zeros? New length is N = 90: ωk = 2πk N k1 = 15. −π = π k 3 45 k2 = −15 =⇒ k2 = 75. Thanks everyone! Good luck studying! 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com