HKN ECE 310 Exam 2 Review
Corey IUC
April 4, 2020
Copyright By PowCoder代写 加微信 powcoder
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√−jππ ππ 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]=cosπ4n≡cosπ4 +2πkn, 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] = cosπ3n, 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] = cosπ3n, 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