This preview shows page 1. Sign up to view the full content.
Unformatted text preview: EECS 20. Midterm No. 2 Solution
April 5, 2000.
1. (a) The fundamental frequency is !0 = 1. The Fourier series coefﬁcients are A0
A2 = 1 and 1 = ,=2, with everything else having value 0. = A 1 = (b) The output will have Fourier series coefﬁcients Ak scaled by the frequency response
H !, so A0 = 1 and A1 = 1=2, and all others are 0. The phases of the frequency
response add to those of the input, so the output will have 1 = ,=2 + =2 = 0. I.e.,
k = 0 for all k . Thus, the output is yt = 1 + cost=2: (c) The frequency, magnitude and phase responses of the cascade composition are G! = H !;
jG!j = jH !j ;
6 G! = 26 H !:
2 2 Here is a sketch: ∠G(ω) G(ω)
2π 1 0 1 ω 2 0 1 2 ω (d) The Fourier series coefﬁcients of the input will now be scaled by G! instead of H ! ,
getting A0 = 1 A1 = 1=4, and 1 = ,=2 + = =2. Thus, the output is yt = 1 + cost + =2=4 = 1 , sint=4: 2. (a) The sketches are shown below: 1 u(n)
... ... n ... n ... n ... n δ(n)
...
y(n)
...
(b) Note that n = un , un , 1: Since the system is LTI, it must therefore be true that hn = yn , yn , 1 = n , n , 4:
This is sketched below: h(n)
... 3. (a) Note that
2 cos t=6 =
=
= eit= + e,it= =4
i t= + 2 + e,i t= =4
e
1 + cos2t=6=2:
6 2 6 2 6 2 6 Moreover,
sin t=6 = cost=6 , =2: Therefore (b) xt = 0:5 + cost=6 , =2 + 0:5 cos2t=6:
Using the results of part (a), ! = =6; A = 0:5, A = 1, A
k 2; and = ,=2, and k = 0 for k 1.
0 0 1 2 1 2 : = 0 5, and Ak = 0 for (c) Rewriting the result from part (a), :
t=6 , =2 + 0:5 cos2t=6
i t= ,= + 0:5e,i t= ,= + 0:25ei t= + 0:25e,i t= :
= 0:5 + 0:5e
From this, we can read of the Fourier series coefﬁcients, X = 0:5, X = 0:5e,i= =
,j=2, X, = 0:5ei= = j=2, X = X, = 0:25, and Xk = 0 for k 2 or k ,2.
xt = 0 5 + cos 1 4. 2 6 2 6 2 2 0 2 6 1 2 6 2 2 (a) Note from the difference equation that what we need to remember about the past is
yn , 1. Thus, deﬁne the state to be sn = yn , 1: (You could equally well choose to deﬁne the state to be ,0:9y n , 1, among other possible choices.) Thus, this is a onedimensional SISO system. The state update equation
becomes sn + 1 = ,0:9sn + xn
because sn + 1 = y n. Thus, A = ,0:9, a 1 1 matrix, and b = 1. The output equation is (b) yn = ,0:9sn + xn
from which we recognize c = ,0:9 and d = 1.
Let the input x = , the Kronecker delta function, and note that
8 0;
n 0
1;
n=0
yn = ,0:9 n = 1
:81
2
: 0,0:9n n = 2
n
This can be written more compactly as y n = ,0:9n un, where un is the unit
step function. 3 ...
View
Full
Document
This note was uploaded on 05/17/2009 for the course EE 20N taught by Professor Ayazifar during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Ayazifar
 Electrical Engineering, Frequency

Click to edit the document details