signals and systems

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7/17/2019 Signals and Systems http://slidepdf.com/reader/full/signals-and-systems-568bf72284744 1/3 MEH226 Signals and Systems Final Exam (2014-2015 Summer) 26.08.2015  Name, Surname : 1 (10) 2 (20) 3 (25) 4 (15) 5 (40) Total Student Number Signature : : Q1 1A. For the given 8 () sin 4 ( 0.25) 2 ( 0.25)  x t     compute and draw the spectrum of () ( ).cos400  y t xt   . 15 () =  8 2 4 ∗( − 0.25) =  16 4 4 ∗( − 0.25) = 164 ∗ ( −0.25)  () = [164].  = 2 ∏     = 16,   2  = 4 ===> = 8, =  2   () = 4∏ 8  () = ().400 ↔ () =  1 2 () ∗ [( − 400)+( +400)]  () = 2∏ −400 8    +2∏ +400 8    400π Y(jw) -404π 396π 404π w -400π -396π 400π  Arg(Y(jw)) -404π 396π 404π w -400π -396π π π 2 1B. Find and draw the spectral energy density of  ()  y . 10 () = |()|  = 2∏ −400 8   +∏ +400 8    .2∏ − 400 8    + ∏ + 400 8      () = 4∏ − 400 8 + ∏ + 400 8   /  400π S(w) -404π 396π 404π w -400π -396π 4 1C. Find the energy of  ()  y . 10 = 1 2  ()  = 1 2  4  + 1 2  4 = 1 2(32 +32) = 32  

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Final Exam Questions

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Page 1: Signals and Systems

7/17/2019 Signals and Systems

http://slidepdf.com/reader/full/signals-and-systems-568bf72284744 1/3

MEH226 Signals and SystemsFinal Exam (2014-2015 Summer)

26.08.2015

 Name, Surname : 1 (10) 2 (20) 3 (25) 4 (15) 5 (40) Total

Student NumberSignature

::

Q1 1A. For the given8

( ) sin 4 ( 0.25)2 ( 0.25)

 x t t t 

  

 compute and draw the spectrum of ( ) ( ).cos400 y t x t t   . 15

() =   82 4 ∗( − 0.25) =   164 4 ∗( − 0.25) = 164 ∗ ( − 0.25) 

 ( ) = [164].  = 2 ∏ −     = 16,   2 = 4 ===> = 8, = 2     ( ) = 4∏ 8  

() = ().400

↔ ( ) =  12 ( ) ∗ [( − 400) + ( + 400)]

 

( ) = 2∏ − 4008     + 2∏ + 4008    

400π

Y(jw)

-404π 396π 404πw

-400π -396π

400π

 Arg(Y(jw))

-404π 396π 404πw

-400π -396π

π

π

2

1B. Find and draw the spectral energy density of   ( ) y t  . 10

() = |( )| = 2 ∏ − 4008   + ∏ + 400

8    . 2 ∏ − 400

8    + ∏ + 400

8    ∗

 

() = 4 ∏ − 4008   + ∏ + 4008    / 

400π

S(w)

-404π 396π 404πw

-400π -396π

4

1C. Find the energy of   ( ) y t  . 10

=   12    ()

  =   12     4

  +   12    4 =   12 (32 + 32) = 32

 

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  ( ) z t  6 6( )6 6

w w H jw    

 

 ( ) 2.cos4 . ( 0,5 )n

v t t t n  

( ) 2 ( 0 .5) x t t  

( ) y t 

 2A. Find and draw the spectrum of   ( )v t  . 10

() =    2 4 ( − 0.5) =    2 4 2  ( − 0.5) =    2 (2) ( − 0.5) =     2 ( − 0.5)  

() =     2 ( − 0.5)   ↔ ( ) =     2   − 20.5

  =  2 =   20.5 = 4   ⎭

  ( ) =     8 ( − 4)  

V(jw)

w4   8 8   

 

2B.  ( ) ? z t    10

() = () ∗ ℎ() = [()] ( ) = ( ). ( ) = 8 [( − 4) + ( + 4) + ( − 8) + ( + 8)] () = 8 4 +8 8 

2C. Find and draw of   ( )Y jw . 10

() = ()() = (8 4 +8 8) 2 − 12 = 8 4 12 + 8 8 12  2 − 12 = (8 + 8) 2 − 12 

() = 32 − 12 

( ) = 32  

Y(jw)

w

 Arg(Y(jw))

2πw

-w/2

32

Q2. The system given on the right

side, 

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π

V(jw)

w

3π3π

π

-6π -2π 2π 6π

-2/3

-2π2π

-6π

2/3

-/3

/3

Arg[V(jw)]

w6π

3A. Find the frequency response of the LTI system. Is the LTI system causality? Explain.15

( ) = 3   ↔ ℎ() = 3( − 0.5) 

ℎ() = 2 (2)3 − 14 = 2 2 14   − 14 = 2 12 − 14 = 4 − 14 

ℎ() = ℎ() ∗ ℎ() + ℎ() = 3( − 0.5) ∗ 4 − 14 + 12 + 34 = 12 − 34 + 12 + 34 

 Nedensellik için ℎ() = 0 < 0 olmalı. Burada ise ikinci terim bunu sağlamadığından sistem nedensel değildir.

( ) = 12   + 12  = 24 34  

3B. Find the output signal  y t   for the given system input signal of ( )V jw . 10 

() = 3 2 +

+

6 +

  

() = 3|(2)| 2 + 3 + ((2)) + |(6)| 6 + 23   +((6)) 

|(2)| = 24 34 2 = 0 

|(6)| = 24 34 6 = 24 3 = −24 = 24  

() = 24 6 + 2

3  −  = 24 6 −

Q4. Find the output ( ) y t  when the input signal is 22

( ) 8 sinw j w

V jw e c 

 and the impulse response LTI systems is

( ) (2 3)h t t   . 25

 ∏ −     ↔   2  ( ) = 8   2   ↔ () = 2 ∏ 4 ∗ ( − 0.5) = 2 ∏ − 0.54    

() = () ∗ ℎ() = 2 ∏ − 0.54   ∗ (2 − 3) = 2 ∏ − 0.54   ∗ 12 − 32 = ∏ − 1.5 − 0.54    = ∏ − 24    

Q3. The LTI system given on the right side, 12( ) 3.w

 j H jw e

,

12 4( ) 2 . (2 ). ( )h t Sinc t t     , 3

34

( ) 12 ( )h t t   .

+v(t)

h3(t)

y(t)h1(t) h2(t) +