Problem 1. Define Fourier transform of function and Fourier inverse transform as:8F ( ? ) = ( ) j t...

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Problem 1. Define Fourier transform of function and Fourier inverse transform as:8F ( ? ) = ( ) j t f t e dt - ? ?-81 8 j t ; ( ) f t ( ) 2 F e d ? = ? ? ? p -8Prove the following theorems:Name of Theorem signal Fourier Transform Superposition (a1f1(t)+a2f2(t)) (a1F1(?)+a2F2(?)) Time Delay (f(t-t0)) F(?)exp(-j?t0) Time Reversal (Here, f(t) is real.) (f(-t)) F(-?)=F*(? ) Scale (f(at)) F(?/a)/|a| Frequency Translation (f(t)exp(j?0t)) F(?-?0) Convolution 8 ? - f 1 ( t ) f 2 ( t t ) d t -8 F1(?)F2(?) Multiplication (f1(t) f2(t)) 8 F ? w F w dw XXXXXXXXXX ? - 2 p -8 Parseval 8 * ? f 1 ( t ) f 2 ( t ) dt -8-8 8 * 1 ( ) ( ) ? F ? F ? d ? 1 2 2 pProblem 2. Prove the following theorems:System FunctionInputOutputTime Domain Descriptionh(t): impulse responsex(t)8y ( t ) = ? x (t ) h ( t -t )dt-8Problem 3. A pulse train with N half sine pulse as1 ? p f t - g t nT g t N sin( ) t 0 Find the Fourier transform of f(t)Frequency Domain DescriptionH(?): transfer functionX(?)Y(?)=X(?)Y(?)
	
Document Preview: Problem 1.  
Define Fourier transform of function and Fourier inverse transform as: 
8
8
1
-jt? jt?
 F? = f()te dt;  f()tF=?ed? 
() ()
??
-8
2p
-8
 
Prove the following theorems:  
 
Name of Theorem signal Fourier Transform 
Superposition (a f (t)+a f (t)) (a F (?)+a F (?)) 
 XXXXXXXXXX
Time Delay (f(t-t )) 
F(?)exp(-j?t ) 
0 0
*
F(-? )=F (? ) 
Time Reversal (Here, f(t) is (f(-t)) 
real.) 
Scale (f(at)) F(?/a)/|a| 
Frequency Translation 
(f(t)exp(j?t)) F(?-? ) 
0 0
8
Convolution F (?)F (?) 
1 2
 
ff ()tt (t- )dt
12
?
-8
8
Multiplication (f (t) f (t)) 
1 2 1
 
F()?-wF(w)dw
12
?
2p
-8
8
8
Parseval 
1
* *
 
 F()?Fd (?? )
ft ()f ()tdt
12 12
??
2p
-8-8
 
Problem 2.  
Prove the following theorems:  
 
 Time Domain Description Frequency Domain 
Description 
System Function h(t):  impulse response 
H(?):  transfer function 
Input x(t) X(?) 
8
Output Y(?)=X(?)Y(?) 
yt ()=- x(tt )h(t )dt 
?
-8
 
Problem 3. A pulse train with N half sine pulse as 
N-1
sin(ptt ) 0

Robert · Accepted Answer

Sol: (1) Superposition property: 
           We know that Fourier transform is given by, 
                                       ,j tF f t e dt



   
Replace       1 1 2 2.  by ,f t a f t a f t we get
      
      
      
       
       
     
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 1 2
1 1 1 2
.  by ,
 ,
  ,
,
,
,
j t
j t j t
j t j t
j t j t
f t a f t a f t
F a f t a f t e dt
F a f t e a f t e dt
F a f t e dt a f t e dt
F a f t e dt a f t e dt
F a F a F

 
 
 




  




 

 
 
 
 
 
 

 
 
 
 
 


 
 
 
Hence it shows that, 
                          1 1 1 2F a F a F     
Time delay property: 
         We know that inverse Fourier transform is given by, 
                                     
1
 
2
j tf t F e d 



   
         Substituting 0t t t  in above equation, we get
     
   
   
   
0
0
0
0
0
0
0
0
1
 
2
1
 
2
1
 
2
j t t
j tj t
j t j t
j t
f t t F e d
f t t F e e d
f t t F e e d
f t t F e


 

 

 

 


 








 
 
    
 



 
Hence it shows that,             
                           00
j t
f t t F e
      
Time Reversal property: 
              We know that Fourier transform is given by, 
                                       
       
 
 ,
 ,
 ,
j t
j t
j t
F f t f t e dt
F f t f t e dt
f t e dt








  



   
    
 



 
Replacing t  by t , we get
     
 
     
 ,
  ,
,
j t
j t
f t f t e dt
f t e dt
f t f



 







     
 
    


F
F
            
 Hence it shows that, 
                                     ,f t f     F
Scale property: 
                 Here consider only 0,a  because if use 0,a  then proof will be change. 
So we use the modulus. 
We know that Fourier transform is given by, 
                          
   
     
 ,
  ,
j t
j t
F f t e dt
F f at f at e dt











   


 
Let          ,at x    
Now differentiate on both sides,
,
,
 ,
at x
adt dx
dx
or dt
a



 
  Now,
     
 
   
 ,
1
 ,
1
,
j t
x
i
a
F f at f at e dt
f x e dx
a
F f at F
a a










   

 
     
 

  
 Hence it shows that,

Problem 1. Define Fourier transform of function and Fourier inverse transform as: 8 F ( ? ) = ( ) j t f t e dt - ? ? -8 1 8 j t ; ( ) f t ( ) 2 F e d ? = ? ? ? p -8 Prove the following theorems: Name...

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