Name:

SS 202B Signals & Systems Mid-term Exam (Page 1 / 9) Student ID No.:

[Score table]

Chapter 1 Chapter 2 Chapter 3 Matlab Total

Prob. 3 3 3 3 3 10 5 5 10 5 10 10 10 10 10 100
Score

Problems & Solutions

[Chapter 1 | 17 points]
Determine whether or not each of the following statement is true or false. Justify your answers.

 j
n
(1-1) [3 pts] (True/False) is periodic

n
 j     

 j
n j n j ( n8m ) j n j n
(True)  e 2  e 4 e 4  e 4 e j 2 m  e 4 (periodic with fundamental period 8 )
 
 

t
(1-2) [3 pts] (True/False) y (t )  

e  (t  ) x ( ) d is invertible

t t
(True) Assume that y1 (t )   
e (t  ) x1 ( )d and y2 (t )   e(t  ) x2 ( )d . Then from y2 (t )  y1 (t ) 


x2 (t )  x1 (t ) , the system is invertible.

(1-3) [3 pts] (True/False) The energy of a signal x(t ) is equal to the sum of energies of its even and odd functions,
i.e.,
  
 
x 2 (t )dt   xe2 (t )dt   xo2 (t )dt
 

x (t )  2 xe (t ) xo (t )  xo 2 (t ) dt . Because
 
 x 2 (t )dt   2
(True) e xe (t ) xo (t ) is an odd function, its
 
  
integration will be zero. Therefore,  
x 2 (t )dt   xe2 (t )dt   xo2 (t )dt .
 

(1-4) [3 pts] (True/False) The system represented by the following input/output relation is linear.
t
y (t )   x( ) d  1
t 1

t
(False) If we give a  x(t ) as the input signal, y(t )  a   t 1
x( )d  1  a  y (t ) .

(1-5) [5 pts] A complex signal x(t )  Ae j e jt ( A is real and A  0 ). Express the average power of its real part
in terms of A . The average power of a signal f (t ) is given by
 Name:
SS 202B Signals & Systems Mid-term Exam (Page 2 / 9) Student ID No.:

1 T
T  T 0
2
lim f (t ) dt

Answer) A2 / 2

1 T A2
Re  x(t )  A  cos(t   ) , the average power is
T 0
2
A cos( t   ) dt  .
2

[Chapter 2 | 35 points]
Consider a causal linear time-invariant (LTI) system described by the difference equation

y[ n]  e 2  y[ n  2]  x[ n]  e 4  x[ n  4] (  0) --- (1)

(2-1) [10 pts] Find impulse response h[n] of this system.

Answer) h[n]   [ n]  e
2 
 [n  2]
- Impulse response:
y[n] for x[n]   [n]  y[n]  e 2  y[n  2]   [n]  e 4   [n  4]

- Causal LTI system  Condition of initial rest:

 [n]  0 for n  0  y[n]  0 for n  0

- Substitute impulse input:

y[0]   [0]  1
y[1]   [1]  0
y[2]   [2]  e 2  y[0]  e 2  y[0]  e 2 
y[3]  0
y[4]  e 4   [0]  e 2  y[2]  0
and y[n]  0 for n  4

Therefore y[ n]   [ n]  e  [ n  2] and h[n]   [ n]  e  [ n  2]

2  2 

(2-2) [5 pts] Determine frequency response H (e j ) of this system.

Answer) For x[n]  e j n , y[n]  H (e j )e j n  H (e j )  1  e2  e 2 j

(2-3) [5 pts] Describe the characteristic of the frequency response H (e j ) obtained from Prob. (2-2): low-pass,
high-pass, band-pass?
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SS 202B Signals & Systems Mid-term Exam (Page 3 / 9) Student ID No.:

Answer) Band-pass
Solution)
| H (e j ) | (1  e 2  cos(2 ))2  e 4  sin 2 (2 )  1  e 4   2e 2  cos(2 )
 (1  e 2  ) 2  2(1  e 2  cos(2 ))


: has a maximum when    .
2
1
| H (e j ) | increases at   [0,  ]
2
1
| H (e j ) | decreases at   [  ,  ]
2

(2-4) [10 pts] Find impulse response hinv [n] of the causal LTI inverse system of (a) such that h[ n]  hinv [ n ]   [n] .
(Hint: switch x[n] and y[n] of the given differential equation (1) and find its impulse response.)

e   n u[ n] for even n,
Answer) hinv [ n]  
 0 for odd n

Solution) Let switch x[n] & y[n]  x[ n]  e 2  x[ n  2]  y[ n]  e 4  y[ n  4]

- Impulse response:
y[n] for x[n]   [n]  y[n]  e 4  y[n  4]   [n]  e2   [n  2]

- Condition of initial rest:

 [n]  0 for n  0  y[n]  0 for n  0

- Substitute impulse input:

same method as (2-1)

e   n u[ n] for even n,
Therefore hinv [ n]  
 0 for odd n

(2-5) [5 pts] Determine whether the derived inverse system is stable or not.

Answer) Stable
 
1
 | hinv [k ] | e2  k 
k  k 0 1  e 2 
: converges

[Chapter 3 | 40 pts]
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SS 202B Signals & Systems Mid-term Exam (Page 4 / 9) Student ID No.:

(3-1) [10 pts] Suppose that the Fourier series coefficients of a periodic signal xT (t ) with fundamental period
T  4 (Figure 3-1) are given by ak .

Figure 3-1

Express real and imaginary parts of the Fourier series coefficients bk of the following signal in terms of ak .

Figure 3-2

Answer)

Real part: 2sin(k0 )  Im ak 

Imag part: 2cos(k0 )  Im ak  .

Solution) Figure 3-2 is the 1 sec delay of the odd part of Fig. 3-1 ( xT (t  1)  x ( (t  1)) ).

From the properties of Fourier series, F .S . xT (t  1)  e  ak and F .S . xT ((t  1))  e jk0  ak* .
 jk0

Therefore,

F .S . xT (t  1)  xT ( (t  1))  e  jk0  ak  ak* 

 e  jk0  2 j  Im ak  .

 Real part: 2 sin( k0 )  Im ak 

 Imag part: 2 cos(k0 )  Im ak  .
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(3-2) [10 pts] Exponential sequences of period N  10 and their Fourier series coefficients (both real & imaginary
parts) are shown below. Find matching pairs & justify your answers.

Answer) (A)- ( ), (B)-( ), (C)-( ), (D)-( )

 Name:
SS 202B Signals & Systems Mid-term Exam (Page 6 / 9) Student ID No.:

Answer) (A)- (4), (B)-(1), (C)-(2), (D)-(3)

Solution)
By property of Fourier series, if the x[ n] is even function, then ak is a purely real for all k . Therefore, (A) and
(B) must be matched with (1) and (4).

In Fourier series, when k  0 , a0   x[n] . The infinite sum of (A) is larger than (B), so it should be matched
n 

with (4) which has larger ak . Hence, (B) should be matched with (1).

(D) is an odd function, therefore the Fourier series coefficients of (D) are a purely imaginary. As a result, (D)
corresponds to (3), and because (C) is neither even nor odd function, (C) should be matched with (2).
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SS 202B Signals & Systems Mid-term Exam (Page 7 / 9) Student ID No.:

(3-3) [10 pts] Three signals of the same period and area S are given in Figure 3-3. The Fourier series coefficients


a
2
of these signals are denoted by ak . Which one of these signals has the smallest k ? Justify your answer.
k 1

Figure 3-3

Answer) (A)
Solution)

1   2 2
a   ak   a0 
2
For a real signal, k 
k 1 2  k   

1
 
2 2
By Parseval’s relation, x (t ) dt  ak .
T T
k 

1
From the property of Fourier series expansion (temporal mean), a0 
T 
T
x (t ) dt

1 1 
 2
1 
a  T x (t )dt 
2 2
k   x (t ) dt    --- (1)
k 1 2  T T
T 

For a rectangular signal of duration L and area S , the level of signal is S / L . Therefore,
2
1 1 S S2 S

2
x (t ) dt     L  . Likewise, the temporal mean is given by a0  .
T T T L LT T

1  S2 S2  S2 1 1 

2
From (1), ak     .
k 1 2  LT T 2  2T  L T 
 

a a
2 2
The largest L gives the smallest k . Therefore, (A) has the smallest k .
k 1 k 1
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(3-4) [10 pts] Suppose that a periodic function fT (t ) has Fourier series coefficients of ak . Prove that the

maximum real value of the Fourier series coefficients bk of the signal gT (t )  fT (t )
2

2
is equal to ak .
k 

fT (t )*  a* k (conjugation property)

 
gT (t )  fT (t )* fT (t )  a* k  ak  a *
a
 m k m  a *
a
m k m (multiplication  convolution property)
m  m 


Therefore, the problem is asking the maximum real value of bk  a
m 
*
a
m k m .


Consider an inequality  |a
m 
m  ak  m |2  0 .

 a 
 

 |a
2 2
 m  ak  m |2  m  ak  m  am* ak  m  am ak*  m  0
m  m 

   

   (a  am ak* m )   2 Re[am* ak  m ] .
2 2
Let I  am  ak  m , and *
a
m k m
m  m  m  m 

The inequality can now be rewritten as


  
 am  Re   am* ak  m   Re bk 
2

m  m  

   
Re   am* am    am . Therefore, the maximum of Re bk  is
2
The equality (=) holds when k  0 
m   m



2
given by ak
k 

[Matlab problem]
(10 pts) Suppose that a matrix A is given as follows:

1 0 0 0 0
 1 1 0 0 0 
 
 1 1 1 0 0 
 
 1 1 1 1 0 
A
 0 1 1 1 1 
 
 0 0 1 1 1
 0 0 0 1 1 
 
 0 0 0 0 1
Express the output Y of the following operation in Matlab.
 Name:
SS 202B Signals & Systems Mid-term Exam (Page 9 / 9) Student ID No.:

b=1j *(0:4);
c= zeros(1,8);
c(2) = 1;
Y= c*A*b’

Answer) -j

Sol)

c   0 1 0 0 0 0 0 0

 0 
 j 
 
b  0 j 2 j 3 j 4 j   b   2 j  : conjugate transpose
 
 3 j 
 4 j 

c * A  [ 1 1 0 0 0 ]
(2nd row)

 0 
 j 
 
cAb  [ 1 1 0 0 0]  2 j    j
 
 3 j 
 4 j 