ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f)...

10
ECE 438 Assignment No. 3 Spring 2018 f

Transcript of ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f)...

Page 1: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as

ECE 438 Assignment No. 3 Spring 2018

f

Page 2: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as

f

f

Note that sinc(20f) has zero crossings at f=k/20 for integer k.

An alternate answer for (c)-i is

𝑋𝑐(𝑓) = sin(2πf10) / sin(2𝜋𝑓2) ∙ {𝑠𝑖𝑛𝑐(2(𝑓 − 1/4)) + 𝑠𝑖𝑛𝑐(2(𝑓 + 1/4))}

Page 3: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as

-8 -6 -4 -2 0 2 4 6 8 n

x[n]

t

x(t)

1/800 2/800 3/800 4/800

Page 4: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as
allebach
Text Box
Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as shown in part (a). So the solution below is not correct.
Page 5: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as
Page 6: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as
Page 7: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as
Page 8: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as
Page 9: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as
Page 10: ECE 438 Assignment No. 3 Spring 2018 - Purdue Engineeringece438/hw_sol/hw3_sol.pdf · Since X_s(f) == 1, X(\omega) == 1. This is consistent with the fact that x[n] = \delta[n], as

Introduction - University of California, Los Angelesbalmer/Pubfile/Koszul.pdf5. The Koszul symmetric space K(n) X over A n X 14 6. The half-Koszul symmetric space E(n) X over U n X

Introduction - University of California, Los Angelesbalmer/Pubfile/Koszul.pdf5. The Koszul symmetric space K(n) X over A n X 14 6. The half-Koszul symmetric space E(n) X over U n X

Matrice densità. 1 elettrone (x)=A(r)ω(s) Funzione densità N elettroni (x 1,x 2,...x N ) Funzione densità Funzione di coppia.

Matrice densità. 1 elettrone (x)=A(r)ω(s) Funzione densità N elettroni (x 1,x 2,...x N ) Funzione densità Funzione di coppia.

4.2B – Finding Binomial Probabilities Binomial Probability Formula nCxpᵡqᶮ ᵡ n MATH PRB nCr x (p)˄x (q)˄(n-x) n=# trials, x=#successes, p=probability of.

4.2B – Finding Binomial Probabilities Binomial Probability Formula nCxpᵡqᶮ ᵡ n MATH PRB nCr x (p)˄x (q)˄(n-x) n=# trials, x=#successes, p=probability of.

n-Butterflies: Modeling Derived Morphisms of Strict n-Groups · Introduction Homotopy Types n-Homotopy Types IA homotopy n-type is an object X of Ho(Top Q) in which p k(X) = 1 for

n-Butterflies: Modeling Derived Morphisms of Strict n-Groups · Introduction Homotopy Types n-Homotopy Types IA homotopy n-type is an object X of Ho(Top Q) in which p k(X) = 1 for

Chapter 6read.pudn.com/downloads137/doc/585947/Chapter6_SM.pdf · 2008-10-29 · Chapter 6 6.1 Therefore, 0 = ∑ = ∑ ∞ = − ∞ =−∞ − n n n X z x n z n x n z = ∑ →∞

Chapter 6read.pudn.com/downloads137/doc/585947/Chapter6_SM.pdf · 2008-10-29 · Chapter 6 6.1 Therefore, 0 = ∑ = ∑ ∞ = − ∞ =−∞ − n n n X z x n z n x n z = ∑ →∞

The z Transform - UTKweb.eecs.utk.edu/~hli31/ECE316_2015_files/Chapter9.pdf · Existence of the z Transform! The z transform of x[n]=αnun−n [0], α∈ is X(z)=αnun−n [0]z−n

The z Transform - UTKweb.eecs.utk.edu/~hli31/ECE316_2015_files/Chapter9.pdf · Existence of the z Transform! The z transform of x[n]=αnun−n [0], α∈ is X(z)=αnun−n [0]z−n

Παρασκευή 17 Νοεμβρίου 2017 Φύλλο 438 | … 438.pdf · 04-2015 (ΦΕΚ 577 Β /09-04-2015). Οι ενδιαφερόμενοι που τηρούν τα κριτήρια

Παρασκευή 17 Νοεμβρίου 2017 Φύλλο 438 | … 438.pdf · 04-2015 (ΦΕΚ 577 Β /09-04-2015). Οι ενδιαφερόμενοι που τηρούν τα κριτήρια

Sparse Fourier Transform (lecture 3)people.csail.mit.edu/kapralov/madalgo15/lec3.pdf · Given x 2Cn, compute the Discrete Fourier Transform of x: bxi ˘ X j2[n] xj! ij, where!˘e2…i/n

Sparse Fourier Transform (lecture 3)people.csail.mit.edu/kapralov/madalgo15/lec3.pdf · Given x 2Cn, compute the Discrete Fourier Transform of x: bxi ˘ X j2[n] xj! ij, where!˘e2…i/n

Solution Manual Chapter 1 - dws/analsoln.pdf · Solution Manual Chapter 1 Exercise 1.1: Since n1 2 (1 + n) 1 2 n= n h f 1 + 1 n 1 2 1 i = 1 n f(0) 1 n; where f(x) = (1 + x)12, the

Solution Manual Chapter 1 - dws/analsoln.pdf · Solution Manual Chapter 1 Exercise 1.1: Since n1 2 (1 + n) 1 2 n= n h f 1 + 1 n 1 2 1 i = 1 n f(0) 1 n; where f(x) = (1 + x)12, the

Formelsammlung M - Federal Council€¦ · Einfaches Trinom: x2 +Sx +P = x2 +(m+n)x +mn =(x +m)(x +n) Betrag jxj= ˆ x, wenn x 0 x, wenn x

Formelsammlung M - Federal Council€¦ · Einfaches Trinom: x2 +Sx +P = x2 +(m+n)x +mn =(x +m)(x +n) Betrag jxj= ˆ x, wenn x 0 x, wenn x

MC 28 Raggi X n 1d

MC 28 Raggi X n 1d

R = { x / x є N ᴧ par, í î ≤ x < í9 }

R = { x / x є N ᴧ par, í î ≤ x < í9 }

andwarck.files.wordpress.com · Web viewDefinisi sebaran peluang binom : p x = n x π x . (1-π) n-x , untuk x=0,1,2,…,n Keterangan : n = banyaknya ulangan x = banyaknya keberhasilan

andwarck.files.wordpress.com · Web viewDefinisi sebaran peluang binom : p x = n x π x . (1-π) n-x , untuk x=0,1,2,…,n Keterangan : n = banyaknya ulangan x = banyaknya keberhasilan

ENSC380 Lecture 28 Objectives: z-TransformUnilateral z-Transform • Analogous to unilateral Laplace transform, the unilateral z-transform is defined as: X(z) = X∞ n=0 x[n]z−n

ENSC380 Lecture 28 Objectives: z-TransformUnilateral z-Transform • Analogous to unilateral Laplace transform, the unilateral z-transform is defined as: X(z) = X∞ n=0 x[n]z−n

34029 Repaso número real. Intervalosyoquieroaprobar.es/_pdf/34029.pdf · Definición de raíz n-ésima n a=x ⇔ Casos particulares de simplificación n x =n (n x =)n Equivalencia

34029 Repaso número real. Intervalosyoquieroaprobar.es/_pdf/34029.pdf · Definición de raíz n-ésima n a=x ⇔ Casos particulares de simplificación n x =n (n x =)n Equivalencia

pdf.binomi.de › fh-probeseiten.pdf · FORMELN + HILFEN HOHERE MATHEMATIK¨BINOMI VERLAG FORMELN + HILFEN HOHERE MATHEMATIK¨ cosh2 x−sinh2 x cos2 x +sin2 x −eiπ X∞ n=1 2−n

pdf.binomi.de › fh-probeseiten.pdf · FORMELN + HILFEN HOHERE MATHEMATIK¨BINOMI VERLAG FORMELN + HILFEN HOHERE MATHEMATIK¨ cosh2 x−sinh2 x cos2 x +sin2 x −eiπ X∞ n=1 2−n

53>I>?;CG Forewn... · 2014. 5. 20. · 13) H [r } X XIñtl3XEX Xôòòíòlíó -09 -îìíïyvv r { 4 x } { N | Ivx r U x }y r h@ swx N | 0 r s r { ~C x rUBx x UE v s r rE r x y

53>I>?;CG Forewn... · 2014. 5. 20. · 13) H [r } X XIñtl3XEX Xôòòíòlíó -09 -îìíïyvv r { 4 x } { N | Ivx r U x }y r h@ swx N | 0 r s r { ~C x rUBx x UE v s r rE r x y

adakdh1/ada/2017/ada04c.pdf · 1 N X i i=1 N ∑ € X≠X IfX i =X,then. Sample Mean : Average of N data points Sample Mean is a statistic. It has a probability distribution, with

adakdh1/ada/2017/ada04c.pdf · 1 N X i i=1 N ∑ € X≠X IfX i =X,then. Sample Mean : Average of N data points Sample Mean is a statistic. It has a probability distribution, with

Dynamics of Nonlinear Di erence Equation x n+1 A Bx Cx · x n+1 = x n+ x n k A+ Bx n+ Cx n k; n= 0;1;2;::: where the parameters , , A, Band Cand the initial conditions x k;x k+1;:::;x

Dynamics of Nonlinear Di erence Equation x n+1 A Bx Cx · x n+1 = x n+ x n k A+ Bx n+ Cx n k; n= 0;1;2;::: where the parameters , , A, Band Cand the initial conditions x k;x k+1;:::;x

David H. Bailey Lawrence Berkeley National Laboratory ... · q x2 n 2x n x n+1 = x n + a n a2 n 2(x n + a n) Then x n converge quartically to p q. For example, for q = 336009;x 0

David H. Bailey Lawrence Berkeley National Laboratory ... · q x2 n 2x n x n+1 = x n + a n a2 n 2(x n + a n) Then x n converge quartically to p q. For example, for q = 336009;x 0