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ENGR 451 Problem Set #1 1.1 Find and plot()H?and()H?(, the magnitude and phase of the DTFT of the pulse,[]hn, shown below. 01201[]hnnMake sure your plot of()H?(is appropriately phase wrapped. 1.2 Find...

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ENGR 451 Problem Set #1 1.1 Find and plot()H?and()H?(, the magnitude and phase of the DTFT of the pulse,[]hn, shown below. 01201[]hnnMake sure your plot of()H?(is appropriately phase wrapped. 1.2 Find and plot()H?and()H?(, the magnitude and phase of the DTFT of the sequence,[][ 1] [ 3]hnnndd= XXXXXXXXXXMake sure your plot of()H?(is appropriately phase wrapped. 1.3 Given a system with impulse response, [] [] [ 1] [ 2]hn n nnddd=--+-, find the response of the system when 23[] cosxnnp=. 1.4 A system is defined by the linear constant coefficient differential equation12[][ 1] [ 1]yn yn xn--= -. Find the output,[]yn, when the input14[][]nxnun=. 1.5 a) Given the system below, find()H?. b) Find[]ynwhen2[] cosxnnp=. c) Find[]ynwhen4[] cosxnnp=. Delay by 2+Delay by[][]xnyn2-1.6 Consider a system with input12[] [][ 1]xn nndd=- -, and output is14[][][]nynn und=- +, find[]hn[]xn[]hn[]yn
1.7 For the following system, assume that[]xnhas the spectrum shown below a) Make a fully labeled sketch of()X?,()Q?,()R?and()Y?. b) Find[]ynin terms of[]xn. 1()H??cos(2)n?[]xn[]qn[]rn[]yn×cos(2)n?×2p-2p01?pp/20p/2p()X?
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ENGR 451 Problem Set #1 1.1 Find and plot H()? and( H()? , the magnitude and phase of the DTFT of the pulse,hn [] , shown below. hn [] 1 0 n 0 1 2 Make sure your plot of( H()? is appropriately phase wrapped. 1.2 Find and plot H()? and( H()?, the magnitude and phase of the DTFT of the sequence,hn [] =-d[n +1]-d[n -3] . Make sure your plot of( H()? is appropriately phase wrapped. 1.3 Given a system with impulse response, hn []=d[n]--dd [n 1]+ [n-2], find the response of the system 2p when x[]nn=cos . 3 1.4 1 A system is defined by the linear constant coefficient differential equationyn []- yn [ -= 1] x[n-1]. Find 2 n 1 the output, y[] n , when the inputx[]nu= []n . 4 1.5 a) Given the system below, findH()? . p b) Find y[] n whenx[]nn=cos . 2 p c) Find y[] n whenx[]nn=cos . 4 Delay by 2 x[] n yn [] + Delay by-2 1.6 n 1 1 Consider a system with inputxn []=-dd []n [n-1] , and output isyn [] =-d[]n + u[]n , findhn [] 2 4 x[] n hn [] yn []1.7 For the following system, assume thatx[] n has the spectrum shown below a) Make a fully labeled sketch ofX()? ,Q()? ,R()? andY()? . b) Find y[] n in terms ofx[] n . H()? qn [] rn [] 1 x[] n yn [] ×× ? -p 2p 2 cos(n? 2) cos(n? 2) ? X() 1 ? 0 pp/2 0p/2p

Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
122 Votes
Sol: (1.1)
DTFT of pulse  h n is,
    j n
n
H h n e 



 
From the figure, DTFT of pulse  h n is,
   
       
 
     
     
2
0
2
2
0 1 2
1
1 cos sin cos 2 sin 2
1 cos cos 2 sin sin 2
j n
n
j j
j j
H h n e
H h h e h e
H e e
H j j
H j

 
 



    
    


 
 

  
  
    
    


Now taking a magnitude of  H  ,
     
     
     
   
   
2 2
2 22
2 22 2
2 2
1 cos cos 2 sin sin 2
1 cos 2cos 1 sin 2sin cos
cos 1 2cos sin 1 2cos
1 2cos cos sin
1 2cos
H
H
H
H
H
    
     
    
   
 
    
     
   
  
 

And phase,
 
 
1
1
2
1
sin sin 2
tan
1 cos cos 2
sin 2sin cos
tan
1 cos 2cos 1
sin 1 2cos
tan
cos 1 2cos
 

 
  

 
 

 



 
    
 
     
 
  
  

1
sin
tan
cos



     


 1tan tan 
 
 
 

Now plot the magnitude and phase,
Magnitude plot
Phase plot
Sol: (1.2)
DTFT of pulse  h n is,
    j n
n
H h n e 



 
Given that
     1 3h n n n     
Now taking a DTFT of  h n ,
 
     
     
3
cos sin cos3 sin 3
cos cos3 sin sin 3
j jH e e
H j j
H j
 
    
    
  
    
     

Now taking a magnitude of  H  ,
     
     
       
   
    
    
 
2 2
2
2 2 2 2
2 2 2 2
2
cos cos3 sin sin 3
cos cos 3 2cos cos3 sin sin 3 2sin sin 3
cos sin cos 3 sin 3 2 cos cos3 sin sin 3
1 1 2 cos 4
2 1 cos 4
2 1 2cos 2 1
2cos 2
H
H
H
H
H
H
H
    
        
        
 
 
 
 
     
     
     
  
 
  


And phase,
 
 
 
 
1
1
sin sin 3
tan
cos cos 2
sin sin 3
tan
cos cos 2
 

 
 

 


  
  
  
 
  
 


 1tan tan 
 
 
 

Now plot the magnitude and phase,
Magnitude plot
Phase plot
Sol: (1.3) Given that,
Impulse response        1 2h n n n n      
On taking the DTFT of  h n ,
  21 j jH e e     
Input  
2
cos
3
x n n


On taking the DTFT of  x n ,
   
 
2
cos
3
2 2
3 3
X DTFT x n DTFT n
X


 
     
 
     
 
    
       
    

We know that,
 
 
 
Y
H
X




Or
     Y H X  
Now substitute the value of     and H X  in above equation,
   
 
2 2 21
3 3
2 2 2 2 2 2
3 3 3 3 3 3
j j
j j
Y e e
Y e e
 
 
 
     
     
             
 
 
    
         
    
                 
                   ...
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