Can you please help me with these problems?

Question

David · Accepted Answer

SOLUTION 4.1: 
For PM:     cos 2c c Ps t A f t K m t   
Now,  
 
2
i c P
c c
t K m t
f
 
 
 

 
Therefore, Instant Frequency, 
  
 
 0
0 0
1
2
1
2
2
2
2 2
i
i
c P
P
c
P P
c
d
f
dt
d
f t K m t
dt
dm tK
f
dt
AK AK
f t nT
T T






 
 
  
 
 
 
   
 
Now, phase shift 
0 02 2
P P
c i c c
AK AK
f f f f
T T 
 
       
 
        0
0
0
2
PAK t nT
T


 
  
 

For FM, 
 
   
0
cos 2 2
i c f
t
c c f
f f K m t
s t A f t K m t 
 
 
  
 


SOLUTION 4.4: 
(a) The envelope of the FM Wave  s t  is: 
   2 21 sin 2c ma t A f t    
The maximum value of envelope: 
2
max 1ca A    
The minimum value of envelope: 
min ca A  
Therefore, 
2max
min
1
a
a
   
This ratio is shown plotted below for 0 0.3  .
 
 
(b) Expressing  s t  in terms of its frequency components: 
         
1 1
cos 2 cos 2 cos 2
2 2
c c c c m c c ms t A f t A f f t A f f t          
Therefore, mean power of  s t  is therefore, 
2 2 2 2 2
1
2 2
2 8 8
1
2 2
c c c
c
A A A
P
A
 

  
 
  
 
 
The mean power of unmodulated carrier is: 
2
2
c
c
A
P   
Therefore, 
2
1 1
2c
P
P

   
which is shown plotted for 0 0.3 
 
 
(c)  The angle  i t , expressed in terms of the in – phase component,  Is t , and the 
quadrature component  Qs t , is:

Can you please help me with these problems?

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