Profs. Kamalabadi, Katselis, : 5 pm, March 11, 2022
UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering
ECE 310 Digital Signal Processing – Spring 2022
Homework 7 – Solution
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1. The sequence x[n] = cos π3 n, −∞ < n < ∞ was obtained by sampling the continuous-time signal xa(t) = cos(Ω0t), −∞ < t < ∞ at a sampling rate of 1000 samples/sec. What are two possible values of Ω0 that could have resulted in the sequence x[n]? The sampling rate is T = 1 . The sampling points are nT = n . 1000 1000 As cos[ π3 n] = cos[( π3 + 2kπ)n], so some possible choices for signal frequency can be, (π+2kπ)= Ω0 →Ω0=1000(π+2kπ),∀k∈Z+∪{0} (−π+2kπ)= Ω0 →Ω0=1000(−π+2kπ),∀k∈Z+ 3 1000 3 Based on the above expressions, two examples of the candidate solutions are Ω0 = 1000π and Ω0 = 2. The continuous-time signal xa(t) = cos(400πt) is sampled with a sampling period T to obtain a discrete-time signal x[n] = xa(nT ) a) Compute and sketch the magnitude of the continuous-time Fourier transform of xa(t) and the discrete-time Fourier Transform of x[n] for T = 1 ms. b) Repeat part (a) for T = 2 ms. c) What is the maximum sampling period Tmax such that no aliasing occurs in the sampling (a) The continuous-time Fourier Transform of xa(t) is, xa(t) = cos(400πt) ↔ Xa(Ω) = π[δ(Ω − 400π)) + δ(Ω + 400π)] A sketch of the magnitude for Xa(Ω) is given in Figure 1. The discrete-time Fourier Transform of x[n] for ω ∈ [−π, π] is, x[n] = xa(nT) = cos(400πnT) = cos(2πn) ↔ Xd(ω) = πδω − 2π + δω + 2π 555 Figure 1: Continuous-time Fourier transform or more extensively, 1 ∞ ω+2πk π ∞ ω+2πk ω+2πk Xd(ω)=T Xa T =T δ T −400π+δ T +400π k=−∞ k=−∞ = π ∞ δω+2πk−400πT+δω+2πk+400πT T k=−∞ T T =π ∞ δω−2π+2πk+δω+2π+2πk k=−∞ 5 5 where the last step comes from the time-scaling property of δ(ω), i.e. δ(αω) = 1 δ(ω), α ̸= 0, and |α| the sketch of the discrete-time Fourier transform of x[n] is given in Figure 2, Figure 2: Discrete-time Fourier transform with T = 1ms (b) The sampling interval T does not change xa(t), thus the continuous-time Fourier transform is the same as part (a). However, the sampling interval T does affect the DTFT. Using the same procedure as part(a), Xd(ω)=πδω−4π+δω+4π, forω∈[π,π] 55 or more extensively, Xd(ω)=π ∞ δω−4π+2πk+δω+4π+2πk k=−∞ 5 5 The sketch of the discrete-time Fourier transform of x[n] is given in Figure 3, Figure 3: Discrete-time Fourier transform with T = 2ms Note that for both cases, only one period of the DTFT is plotted in the sketches. (c) The maximum sampling period Tmax such that no aliasing occurs in the sampling process is, Tmax= 1 = 1 =1 2fmax 2 × 200 400 Note that 1 corresponds to the Nyquist sampling rate. 3. The continuous-time signal xa(t) has the continuous-time Fourier transform shown in the figure below. The signal xa(t) is sampled with sampling interval T to get the discrete-time signal x[n] = xa(nT). Sketch Xd(ω) (the DTFT of x[n]) for the sampling intervals T = 1/100, 1/200 sec. For T = 1 sec (figure 4) with the dotted black lines corresponding to the shifted copies of Xa(ω) and the solid black line represents Xd(ω). For T = 1 sec, see figure 5. sec does not (in fact, it is critical As seen in Figure 4 and 5, T = 1 sec causes aliasing and T = 1 100 200 sampling interval). Therefore, T = 1 sec avoids aliasing. 200 Figure 4: Discrete-time Fourier transform for T = 1/100s Figure 5: Discrete-time Fourier transform for T = 1/200s 4. Let x[n] = xa(nT). Show that the DTFT of x[n] is related to the FT of xa(t) by where Xd(ω) is the DTFT of x[n] and X(Ω) the FT of xa(t). x[l]e−jωl = xa(lT )e−jωl · · · (a) ∞ 1 ∞ = 1 Xa(Ω) ejlT(Ω−Tω) dΩ···(c) 2π −∞ l=−∞ 1∞ ∞ΩT−ω 2π Xa(Ω) δ l− 2π dΩ···(d) −∞ l=−∞ 1∞ ω+2lπ Xa(Ω)ejΩlTdΩ e−jωl···(b) Xa(Ω)ejΩlT e−jωldΩ = 2π ω Xa(Ω)ejlT (Ω− T )dΩ 1∞∞ ΩT−ω1∞∞ 2πω2π Xa(Ω)δ l − 2π ω + 2 π l dΩ = 2π dΩ Xa(Ω) T δ Ω − T − T l dΩ Xa(Ω)δ Ω− T 1∞ ω+2πl =T Xa(T) l=−∞ where (a) step comes from x[n] = xa(nT), (b) step from the definition of inverse Fourier trans- form, (c) step from swapping the order of summation and integration, (d) step from the fact that ∞k=−∞ e2jπkt = ∞k=−∞ δ(t − k). This equality corresponds to the Fourier series of an infinite-sum of delta functions. 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com