代写信号分析 – Topics on Data and Signal Analysis

Topics on Data and Signal Analysis

代写信号分析- 本题是一个是比较典型的信号分析等代写方向

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Coursework I

Problem 1.Consider the following three vectors inR^3 :

 1 = [11 2]T,  2 = [2 32]T,  2 = [3 1 1]T.
a) Does the set  ={ 1 , 2 , 3 }form a basis forR^3?
b) If  ={ 1 , 2 , 3 }forms a basis, find its biorthogonal basis  ={ 1 , 2 , 3 }.
c) For an arbitraryx= [x 1 x 2 x 3 ]TinR^3 describe the procedure and the expression for finding coefficients
c 1 ,c 2 andc 3 such that
x=c 1  1 +c 2  2 +c 3  3.
d) Find the largest numberA>0 and the smallest numberB<such that
Ax^2 

^3

i=
|i,x|^2 Bx^2
for allxR^3.

Problem 2.Consider the systems in Figures 1-4.

 2 3   3 2 
 2 4   4 2 
 1  2
 3  4
a) For the signalx[n] which is given in the Fourier domain by the figure below, sketchY(ej) for each of the
four systems in the above.
!"
 3   2   #  3 
1
#  2 
!"


2
  
2
1
b) For any arbitrary signalx[n]  express samples ofy[n] in terms of samples ofx[n] for each of the four
systems in the above.

Problem 3.Consider a filterh[n].

a) Find the Fourier transform of its autocorrelation sequence
a[n] =

k
h[k]h[kn].
b) Show that ifh[kn],h[k]=[n], then
|H(ej)|= 1,.
c) Show that if|H(ej)|= 1,then
h[kn],h[k]=[n].
d) Show that if|H(ej)|= 1,,then{h[kn],nZ}is an orthonormal basis for`^2 (Z).

Problem 4.Consider two waveforms 0 [n] and 1 [n] and two waveform 0 [n] and 1 [n] in`^2 (Z). Leth 0 [n] and h 1 [n] be two filters such thath 0 [n] = 0 [n] andh 1 [n] = 1 [n], andg 0 [n] andg 1 [n] two filters such that g 0 [n] = 0 [n] andg 1 [n] = 1 [n].

a) Show that if 0 [n], 0 [n 2 k]=[k] thenH 0 (z)G 0 (z) +H 0 (z)G 0 (z) = 2 and that if
 1 [n], 1 [n 2 k]=[k] thenH 1 (z)G 1 (z) +H 1 (z)G 1 (z) = 2.
b) Show that if 0 [n], 1 [n 2 k]= 0 for allkZthenH 1 (z)G 0 (z) +H 1 (z)G 0 (z) = 0 and that if
 1 [n], 0 [n 2 k]= 0 for allkZthenH 0 (z)G 1 (z) +H 0 (z)G 1 (z) = 0.
c) Using the results of a) and b) show that if 0 [n], 0 [n 2 k]=[k] , 1 [n], 1 [n 2 k]=[k],
 0 [n], 1 [n 2 k]= 0 for allkZ, and 1 [n], 0 [n 2 k]= 0 for allkZ, then
H 0 (z)G 0 (z) +H 1 (z)G 1 (z) = 2andH 0 (z)G 0 (z) +H 1 (z)G 1 (z) = 0.