代写代考 ECE 310 Digital Signal Processing – Spring 2022

Profs. Kamalabadi, Katselis, : 5 pm, April 1, 2022
UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering
ECE 310 Digital Signal Processing – Spring 2022
Homework 9 – Solution

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1. Let {X[k]}20 and Xd(ω) respectively be the 21-point DFT and DTFT of a real-valued sequence k=0
{x[n]}7n=0 that is zero-padded to length 21. Determine all the correct relationships in the following and justify your answer.
(a) X[19] = Xd(−4π ). 21
(b) X[2]=X∗(−4π) d 21
(c) X[12] = Xd(−4π ) 21
(d) X[4]=X∗(−4π) d 21
(a) Correct. By definition of DFT:
21 􏰅 X[19] = 􏰃 x[n]e−j 2πkn 􏰅
= 􏰃 x[n]e−j 38πn 21
􏰅k=19 −j 38πn +j2πn
where (a) comes from the 2π periodicity and (b) holds by the definition of DTFT. (b) Correct. Again, by definition of DFT:
X[2] = 􏰃 x[n]e−j 2πkn 􏰅 = 􏰃 x[n]e−j 4πn
n=0 􏰅k=2 n=0
􏰈 20 􏰉∗ 􏰈 20 􏰉∗
(a)􏰃∗ j4πn (b)􏰃 j4πn
x[n]e 21 = x[n]e 21 n=0
where (a) comes from the property of complex conjugate, (b) comes from the fact that x[n] is real- valued therefore x[n] = x∗[n], ∀n.
(c) Incorrect. Again, by using the same derivation as in part (a):
21􏰅 21d21 X[12] = 􏰃x[n]e−j2πkn 􏰅 = 􏰃x[n]e−j24πn = X 24π
n=0 􏰅k=12 n=0
j 4πn x[n]e 21

Then there is no clue that Xd 􏰀 24π 􏰁 = Xd 􏰀− 4π 􏰁. Indeed, suppose the sequence is x[n] = δ[n − 1],
then one can see that Xd 21 = e 21 two values are not equal.
21 􏰀24π􏰁 −j24π
and Xd −21 = e 21 , which is a counter-example that the
(d) Incorrect. Again, by using the same derivation as in part (b):
20 􏰅20 􏰈20 􏰉∗􏰆􏰇
21 􏰅 21 21
x[4] = 􏰃x[n]e−j2πkn 􏰅 = 􏰃x[n]e−j8πn = 􏰃x[n]ej8πn = X∗ −8π
There is no clue that X∗ 􏰀−8π 􏰁 = X∗ 􏰀−4π 􏰁. Indeed, if we still plug-in x[n] = δ[n − 1] then we have
􏰅 d21 n=0 k=4 n=0 n=0
d 21 d 21 ∗􏰀 8π􏰁 −j8π ∗􏰀 4π􏰁 −j4π
Xd −21 =e 21 andXd −21 =e 21,whicharenotequal.
2. Given that the DFT of x = [2,0,6,4] is X = [X0,X1,X2,X3], determine the DFT of y = [2,1,0,3] and express the result in terms of X0, X1, X2, X3.
Hint: relate the two sequences using transformations we discussed in class and use the corresponding properties of the DFT.
Solution: By observation, one can convert x to y by first scaling x by 21 , which yields an intermediate sequence z such that
z = 12x = [1,0,3,2].
Next, we (circularly) shift z to the right by 1 to obtain y. That is,
y[n] = z[⟨n − 1⟩4].
In terms of the relation on DFT, first by linearity we have the relation between Z[k] and X[k] that
Z [ k ] = 12 X [ k ]
where Z[k] is the DFT of z at the k-th frequency. Then, by the (circular) time shift property of
In this way, we can express the DFT of y as Y = [Y0, Y1, Y2, Y3] such that
3. You are given two sequences x = [1,2,3,4,5,6] and y = [4,5,6,1,2,3]. It is known that Y[k] =
words, y[n] = x[⟨n − 3⟩6]. Then by the shifting property of DFT we have 6
422 Y [k] = Z[k] · e−j 2πk = Z[k]e−j πk = 1X[k]e−j πk
−j 2π kn0 X[k]e 6
, for k = 0, 1, . . . , 5. Find two values of n0 consistent with this information.
Solution: One can observe that y is obtained by (circularly) shifting x to the right by 3. In other
Y [k] = X[k]e−j 2π k·3+2πk = X[k]e−j 2π k·9 66
Y0 = 21X0, Y1 = −j12X1, Y2 = −12X2, Y3 = j12X3.
Y [k] = X[k]e−j 2π k·3.
Therefore, a clear choice of n0 is n0 = 3. To obtain another valid n0, by using the periodicity we

which leads to n0 = 9. In general, one can see that any n0 that satisfies n0=3+6m, m∈Z
is valid (try to understand this from the perspective of time shift).
4. Let X be the 6-point DFT of x = [1,2,3,4,5,6]. Determine the sequence y whose DFT Y[k] = X[⟨−k⟩6], for k = 0,1,…,5.
Solution: By definition of DFT:
Y [k] = X[⟨−k⟩6] 55
􏰃 −j 2πn ·⟨−k⟩ = x[n]e 6
(a) 􏰃 −j 2πn ·(−k) 6= x[n]e 6
􏰃 j 2πkn 􏰃 −j 2πk(−n) (b) 􏰃 −j 2πk(6−n)
= x[n]e 6 = x[n]e 6 = x[n]e 6 n=0 n=0 n=0
= x[0] + x[5]e−j 2πk + x[4]e−j 2π2k + x[3]e−j 2π3k + x[2]e−j 2π4k + x[1]e−j 2π5k (1)
−j 2πn ·(−k+6m) where (a) holds by using the fact that ⟨−k⟩6 = −k + 6m for some integer m, and e 6 =
−j 2πn ·(−k)
e 6 by periodicity; (b) uses the periodicity again; the last equality comes by expanding the
summation. On the other hand, we have
6 Y [k] = 􏰃 y[n]e−j 2πkn
= y[0] + y[1]e−j 2πk + y[2]e−j 2π·2k + y[3]e−j 2π3k + y[4]e−j 2π4k + y[5]e−j 2π5k , (2)
Then, by matching each term between Eq. (1) and Eq. (2) we get
y = [x[0], x[5], x[4], x[3], x[2], x[1]] = [1, 6, 5, 4, 3, 2]
Remark: In general, one can show that for any sequence x and y of length N that satisfy Y [k] = X[⟨−k⟩N ], we have
y = [x[0], x[N − 1], x[N − 2], . . . , x[1]] y[n] = x[⟨−n⟩N ].
or in other words,
The proof will be a simple generalization/extension of the one given above.
5. A continuous-time signal xc(t) = cos (12πt) is sampled at a rate of 60 Hz for five seconds to produce a discrete-time signal x[n] with length L = 300.
(a) Let X[k] be the length-L DFT of x[n]. At what value(s) of k will X[k] have the greatest magnitude?
(b) Suppose that x[n] is zero-padded to a total length of N = 512. At what value(s) of k does the length-N DFT have the greatest magnitude?

(a) The sequence after sampling is x[n] = Xc(nT) = cos(12πnT) = cos􏰀πn􏰁. Then, there will be
two peaks in the DTFT Xd(ω) between [0, 2π) which locate at π and 9π respectively. Regarding the 55
DFT, since it only gives samples of DTFT at frequencies 2πk for k = 0,1,,L−1, we see that k = 30 L
and k = 270 correspond to the two peaks respectively. Therefore, both k = 30 and k = 270 will have the greatest magnitude of X[k]
(b) After zero-padding, now the length-N DFT have samples at frequencies 2πk for k = 0, 1, , N − 1. N
As a result, we don’t have the samples right at the peaks π and 9π anymore. Instead, the largest 55
magnitude of X[k] is located at the frequency which is closest to the peaks. This leads to 2π · 51 512
(that is closest to the peak at π ) and 2π · 461 (that is closest to the peak at π ), and the corresponding 5512 5
kchoicesare k=51 and k=461.
Remark: One might be confused that how to show both k = 51 and k = 461 share the same
amount of magnitude. So there are two ways: (1) By symmetry, since x[n] is real, we have |X[k]| =
|X[N − k]| and therefore |X[51]| = |X[461]|; (2) The difference between each peak and its closest
sample frequency is 2π· 0.2 , which is the same for both k = 51 and k = 461, therefore their magnitude 512
should also be identical.
6. A scientist is performing spectroscopy to identify the chemical composition of a material. They calculated the DFT magnitude plot shown below using data collected at a rate of 32 Terahertz for one picosecond. Assume that the measured electromagnetic signal took the form xa(t) = A0 cos(Ω0t) + A1 cos(Ω1t).
(a) Assume that Ω0 and Ω1 are both less than the Nyquist frequency. Find A0, A1, Ω0, and Ω1. (b) Suppose that xa(t) were instead sampled at 64 Terahertz for one picosecond to generate 64
samples. Sketch the new DFT magnitude plot and clearly label all nonzero values.

(a) By observation, the two peaks should exactly correspond to the two frequencies Ω0 and Ω1. Then, by assuming that Ω0 < Ω1 and denote fs = 32 THz, we have Ω0 · 1 = 2π · 4 ⇒ Ω0 = 2π · 4 THz fs 32 Ω1 · 1 = 2π · 12 ⇒ Ω0 = 2π · 12 THz . fs 32 To get the magnitude, given the fact that A0 and A1 in the signal xa(t) will lead to magnitudes A0N and A1N respectively with N = 32 as the length of sequences. Then we have 2 A0N =32, A1N =16 ⇒ A0 =2, A1 =1. 22 (b) The sketch of the (magnitude of) DFT is given as follows: Figure 1: Problem 6(b). The sketch of the new DFT magnitude. Here,thepeaksforΩ0 locateatk=4andk=60,andthepeaksforΩ1 locateatk=12andk=52. The magnitude for Ω0 and Ω1 are A0N = 64 and A1N = 32 respectively. 22 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com