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**Unformatted text preview: **UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Homework 2 Solutions Due Wednesday, September 10, 2008 Prof. Bresler / Prof. Jones 1. Find a closed-form expression (no sums) for the DTFT of the following sequences: (a) , where | | 1 1 1 (b) 5 , where | | 1 , | | || 1 , | | || 1 1 (c) 10 1
/ 1 1
/ 1 1 sin 5 sin 2 , 0 / / 0 ∑ 10, 0 or expressing the result in terms of sinc functions, we have:
/ sin 5 sin 2 1 / 5ω /2 5 2 10 / 5 2 , (d) 4 , where | | 1 1 , 1 || 1 || 1 (e) 8 0.6 cos 1 1 2 0.6 0.6 cos 4 8 0.6 cos 4 8 0.6 8 1 1 0.4243 1.2 0.36 Alternate solution: x[n] = 8(-0.6)n cos(πn/4)u[n] (e- jπn/4 + e jπn/4 ) x[n] = 8(-0.6) u[n] 2 X d ( w) = DTFT {(4(-0.6)n (e- jπn/4 )u[n]} + DTFT {4(-0.6)n (e jπn/4 )u[n]}
n 4 4 + − j ( w −π / 4 ) 1 − (−.6)e 1 − (−.6)e− j ( w +π / 4 ) 2 2. The DTFT of x[n] is as given in the following figure. Determine x[n] for each case. a) For this problem, we can use the modulation property of the cosine and then integrate: cos 1 2 This plot can be further broken down into two parts, and we can use the principle of superposition: +
Xd(ω) 2 1 = ‐π − 3π − 4 π
2 ω 0
π 2 3π 4 π Using both principles, we can find x[n] with this formula: 2 where and are shown below: cos 3 4 2 cos 5 8 3 Now, all we need to do is compute First of all, we note that: 1 2 Note that 1 sin 2 and : cos 12 sin 2 1 sin , 0 / , so we can simplify the above as: 1 2 where we used the fact that: sin 1 Applying this result, we see that: 1 2 Similarly, we have: 1 2 Putting it all together, we have: 2 3 sinc cos 4 4 4 or by simplifying, 1 3 sinc cos 2 4 4 1 5 sinc cos 8 4 8 , 2 5 sinc cos 8 8 8 , / / 0 0 / 1 4 4 / 1 8 8 Just to check directly, for n=0, we have: 1 2 1 2 2
/ 0 1 1 2 2 / 2 / 1 4 1 2 3 4 4 Alternatively, we can directly compute the inverse DTFT via brute force: 1 2 1 2
/ / / 1
/ 2
/ 2
/ 1 3 4 2 sin 12 sin 2 4 Using the fact that 4 sin 2 2 sin sin 1 sin 2 0 0 1 we have: 3 4 3 4 3 4 2 3 4 Although the two solutions obtained using the two different methods mathematically look different, they are indeed the same result. This can be checked by plotting the resulting discrete time sequences for x[n]. As a sanity check, we notice that x[0] = 3/4, immediately from the above equation for x[n]. Notice that often, it is easier and more intuitive to work with sinc() functions. To mathematically prove that the two results are indeed equivalent, we will show that indeed the first result is exactly equal to the second result: 2 1 sin sin 4 cos 4 1 1 Finally, we have: 3 4 3 4 3 4 2 3 4 3 4 sin sin 3 4 sin sin 2 2 sin 4 2 sin 2 8 cos 3 4 1 sin 5 8 1 3 4 , sin 0 8 sin 2 5 8 sin 8 5 8 3 4 5 b) We can use the modulation property of the cosine in this part as well, we can express x[n] as follows: 2 Notice that cos 2 can be computed using the following DTFT: Proceeding: 1 2
/ 2 1 2 2 sin /2 cos 2 2 sin 2 / sin 2 sin 2 2 sin 2 Finally, putting it all together: 2 sin 2 2 2 0 cos 2 cos 2 where we used the fact that: sin and sin 0 , and 2 0 2 . For n = 0, we have: 0 1 2 2 1 2 2 2 1, 0 6 Alternatively, we can directly use the formula for the inverse DTFT: 1 2 sin 1 2 cos 2 2 sin 2 2 cos 1, 0 where we used the fact that sin Using the result that 0 0 for all integer n. 1, we can express x[n] as follows: 1 1 1 0 0 which can further be simplified to: 2 0 1 0 To show that the two answers using the two different methods are indeed equivalent, we only need to consider the case when n is even. So, we want to show: 2 This can be easily shown by noticing that: cos 2 1, 2 cos 2 It turns out that this is exactly the case for n = 2, 4, 6, etc. 7 3. Find a closed-form expression for the inverse DTFT of the following sequence: 4 2 sin 2 10cos 6 4 2 sin 2 4 10 cos 6 5 4 2 5 2 5 10 2 4 5 By looking at the form of this DTFT is the following: , we can directly infer that the discrete-time sequence corresponding to 5 1 4 1 5 0 6 2 0 2 6 8 4. Let x[n] be an arbitrary signal, not necessarily real-valued, with DTFT for the following y[n]: terms of a) . Determine in Yd (ω ) =
b) n = −∞ ∑ x [n]e
* ∞ − jωn =( ∑ x[n]e jωn )* =( ∑ x[n]e − j ( −ω ) n )* = X d * (−ω ) n = −∞ n = −∞ ∞ ∞ Yd (ω ) =
c) 3 n = −∞ ∑ x[−n]e ∞ − jωn = ∑ x[m]e − j ( −ω ) m = X d (−ω )
m = −∞ ∞ Using the shifting property of the DTFT, we have: Yd (ω ) = (1 + e − jω 3 ) X d (ω ) Long method: Yd (ω ) = Yd (ω ) = n = −∞ ∞ ∑ x[n]e − jωn + ∞ n = −∞ ∑ x[n − 3]e
∞ m = −∞ ∞ − jωn n = −∞ ∑ x[n]e − jωn + e − jω 3 ∑ x[m]e − jωm Yd (ω ) = (1 + e − jω 3 ) X d (ω ) 9 5. Let be a signal with DTFT as shown in the following figure. Determine and sketch the DTFT of cos . Using the modulation property of the cosine in the frequency domain: cos 4 1 2 4 4 This is simply two shifts and one scaling operation in the frequency domain, which results in the following DTFT : 1 0 6. Consider the complex sequence (a) Calculate
N −1 n =0 . | for the case of large N. How does the plot change for small N? and sketch |
− jω n X d (ω ) = ∑ e − jω 0 n e X d (ω ) = 1 − e − j ( ω +ω 0 ) N 1 − e − j (ω + ω 0 )
− j (ω + ω 0 ) N −1 2 X d (ω ) = e and sin((ω + ω0 ) N ) 2 , forω ≠ −ω 0 sin((ω + ω0 ) / 2) X d (ω0 ) = ∑1 = N , forω = −ω0
n=0 N −1 Using a similar computation as in Problem 2(a) to convert the sin()to s inc() function, we have : N N −1 sin c ((ω + ω0 ) ) − j (ω + ω 0 ) 2 , forallω 2 X d (ω ) = Ne sin c((ω + ω0 ) / 2) As N gets large, the plot will look more and more like a delta function. For small N, the lobes and zero crossings are spaced farther apart. The amplitude of the main lobe also decreases. 1 1 (b) Calculate the N-point DFT for the case of large N. of the finite length sequence
2π kn N , 0,1, … , 1. Sketch | | X [k ] = ∑ e
n=0 N −1 − jω 0 n e −j X [k ] = and 1 − e− j ( 2πk / N +ω 0 ) N , forω0 ≠ −2πk / N 1 − e− j ( 2πk / N +ω0 )
N −1 n=0 X [k ] = ∑ 1 = N , forω0 = −2πk / N X [k ] = e
− j ( 2πk / N + ω 0 )( N −1 ) 2 N ) 2 , forω ≠ −2πk / N 0 sin((2πk / N + ω0 ) / 2) sin((2πk / N + ω0 ) N ) 2 , forallω sin c((2πk / N + ω0 ) / 2) sin c((2πk / N + ω0 ) Using a similar computation as in Problem 2(a) to convert the sin()to s inc() function, we have : X [k ] = Ne
− j ( 2πk / N + ω 0 )( N −1 ) 2 From our solution, we see that the DFT is a sampled version of the DTFT. The DFT samples the DTFT at points ω = 2πk/N. We can easily determine the DFT using our solution from part (a). (c) Find the DFT of for the case of , where is an integer. Sketch | |. 1 − e − j ( k + k 0 ) 2π X [k ] = 1 − e − j ( k + k 0 ) 2π / N X [k ] = e and X [k0 ] = ∑1 = N , fork = −k0
n=0 N −1 − j 2π / N ( k + k 0 )( N −1 ) 2 sin(π (k + k0 )) , fork ≠ −k0 sin(π (k + k0 ) / N ) Using a similar computation as in Problem 2(a) to convert the sin()to s inc() function, we have : X [k ] = Ne
− j 2π / N ( k + k 0 )( N −1 2 ) sin c(π (k + k0 )) , forallk sin c(π (k + k0 ) / N ) ⎧N X [k ] = ⎨ ⎩0 for k = (− k 0 ) mod N = N − k 0 for k = 0,... N - 1 k ≠ −k 0 1 2 1 3 7. Calculate the DFT of the following finite-length sequence of length 0, 1, You can assume that is even. ,0 ,0 1 1 : 0 1 1 1 1 1 2 , 0
/ / 1
/ / sin sin To consider the case of 0, we can re-write the result above in terms of sinc functions using a similar computation as in Problem 2(a): 2 Alternatively, we can directly compute 2 at 0: 2 2 , 0 1 1 We can also directly compute at /2: 2 , 0 0 , 0 1 1 1 2 1 4 To summarize in concise form, we have: /2 /2 0 0 /2 1 5 ...

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