Problem1 - CISL: Communication and Information Systems ...
Transcript of Problem1 - CISL: Communication and Information Systems ...
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Problem1
(a) false
(b) true
(c) true
(d) true
(e) true
(f) false
(g) false
(h) true
(i) true
( j) false
Problem2
(1) epsilon
(2) zeta
(3) eta
(4) lamda
(5) nu
(6) phi
(7) chi
(8) theta
(9) gamma
(10) sigma
(11) Lamda
(12) Pi
(13) phi
(14) Omega
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Problem 3
(a)
(b)
(c)
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(d)
(e)
Problem4
(a)
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(b)
(c)
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(d)
(e)
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Problem 5. (8 points: p. 359, A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd Edition. Prentice Hall, 1996 ) In this problem, we derive the DTFT from the DTFS. Fill in the blanks.
Consider a general sequence x[n] that is of finite duration. That is, for some integers 1N and 2N , x[n]=0 outside
the range 1 2N n N− ≤ ≤ . From this aperiodic signal, we can construct a periodic sequence [ ]x n for which x[n] is
one period. As we choose the period N to be larger, [ ]x n is identical to x[n] over a longer interval, and as N →∞ , [ ] [ ]x n x n= for any finite value of n.
Let us now examine the Fourier series representation of [ ]x n , we have
(2 / )(1) [ ] ,jk N nk
k Nx n a e π
=< >
= ∑
(2 / )1(2) [ ] jk N nk
n Na x n e
Nπ−
=< >
= ∑ .
Since [ ] [ ]x n x n= over a period that includes the interval 1 2N n N− ≤ ≤ , it is convenient to choose the interval of
summation in eq. (2) to include this interval, so that [ ]x n can be replaced by [ ]x n in the summation. Therefore
2
1
(2 / ) (2 / )1 1(3) [ ] [ ] ,N
jk N n jk N nk
n N na x n e x n e
N Nπ π
∞− −
=− =−∞
= =∑ ∑
where in the second equality in eq. (3) we have used the fact that [ ]x n is zero outside the interval 1 2N n N− ≤ ≤ .
Defining the function
(4) ( ) [ ] ,j n
nX j x n e ωω
∞−
=−∞
= ∑
we see the coefficient ka are proportional to samples of ( )X jω , i.e.,
01(5) ( ),jk
ka X eN
ω=
where 0 2 / Nω π= is the spacing of the samples in the frequency domain. Combining eqs. (1) and (5) yields
0 01(6) [ ] ( ) .jk jk n
k Nx n X e e
Nω ω
=< >
= ∑
Since 0 2 / Nω π= , or equivalently, 01/ / 2N ω π= , eq. (6) can be rewritten as
0 00
1(7) [ ] ( ) .2
jk jk n
k Nx n X e eω ω ω
π =< >
= ∑
Each term in the summation in eq. (7) represents the area of a rectangle of height 0 0( )jk jk nX e eω ω and width 0ω . As
0 0ω → , the summation becomes an integral. Furthermore, since the summation is carried out over N consecutive
intervals of width 0 2 / Nω π= , the total interval of integration will always have a width of 2π . Therefore, as
N →∞ , [ ] [ ]x n x n= , and eq. (7) becomes
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2
1(8) [ ] ( )2
j j nx n X e e dω ω
πω
π= ∫
where, since ( )j j nX e eω ω is periodic with period 2π , the interval of integration can be taken as any interval of length 2π . Thus, we have the following pair of equations:
2
1(9) [ ] ( )2
j j nx n X e e dω ω
πω
π= ∫ ,
(10) ( ) [ ] .j j n
nX e x n eω ω
∞−
=−∞
= ∑
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Problem 7. (15 points: p. 355, A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd Edition. Prentice Hall, 1996 ) (a) Design a compensatory system that, when provided with the output of the measuring device, produces an output equal to the instantaneous temperature of the liquid. Sol) We consider a compensatory system that produces an output equal to instantaneous temperature. Since the step response is /2 /2( ) (1 ) ( ) (1 ) 0 ,t ts t e u t e for t− −= − = − ≤ < ∞
the impulse response has to be /2 /21 1( ) ( ) ( 0 ) ( ).2 2
t tdh t s t e for t e u tdt
− −= = ≤ < ∞ =
So, the frequency response of the system is
1/ 2( ) ( ) .1
2
FTh t H jj
ωω
→ =+
We now desire to build an inverse (compensatory) for the above system. Therefore, the frequency response of the inverse system (compensatory system) has to be
1 1( ) 2 .( ) 2
G j jH j
ω ωω
= = +
Taking the inverse Fourier transform we obtain 1( ) ( ) 2 ( )g t t u tδ= + where 1( )u t unit doublet (lecture 4, p8) is. (b) Consider a measuring device whose overall output can be modeled as the sum of the response of the measuring device characterized by eq. (1) and an interfering noise signal ( )n t . Such a model is depicted in Figure (a). Suppose that ( ) sin .n t tω= What is the contribution of ( )n t to the output of the inverse system, and how does this output changes as ω is increased? Sol) When ( ) sinn t tω= passes through the inverse system at problem (a), the output will be
( ) ( ) ( ) sin 2 cos .y t n t g t t tω ω ω= ∗ = +
We see that the output is directly proportional to ω . Therefore, as ω increase the contribution to the output due to the noise also increases. (c) Suppose that we wish to design a compensatory system that will slow down the response to actual temperature variations, but also will attenuate the noise ( )n t . Let the impulse response of this system be ( ) ( ).ath t ae u t−= Choose a so that the overall system of Figure (b) responds as quickly as possible to a step change in temperature, subject to the constraint that the amplitude of the portion of the output due to the noise ( ) sin 6n t t= is no larger than 1/4. Sol) We consider the output due to the noise ( ) sin 6 .n t t= So, the output will be
( ) ( ) ( ) ( ) ( ) ( ).FTy t n t h t Y j N j H jω ω ω= ∗ → = In this case we require the amplitude of the portion of the output,
( ) 1( )( ) 4
Y jH j
N jω
ωω
= ≤ when 6ω = .
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Since 2
22 2( ) ,aH j
aω
ω=
+ we require that
2
2 2
1 .36 16
aa
≤+
Therefore, 6 .15
a ≤