Solutions manual for communications systems 4th edition latest update 100%% answers//

Solutions Manual for: n n




Communications Systems, n n




5th edition
n




by

Karl Wiklund, McMaster University,
n n n n




Hamilton, Canada n




Michael Moher, Space- n n




Time DSPOttawa, Canada
n n n




and

Simon Haykin, McMaster University,
n n n n




Hamilton, Canada n




Published by Wiley, 2009. n n n




Copyrightn©n2009nJohnnWileyn&nSons,nInc.n AllnRightsnReserved.

, Chaptern2

2.1n(a)

−Tn Tn
g(t)n=n Ancos(2n fn t) tnn ,nn 

2 2n
c

1
f =n
c
T

Wencannrewritenthennhalf-cosine
tn  nas:

Ancos(2n fnct)nnrectn  
T 
Usingnthenpropertynofnmultiplicationninnthentime-domain:
G(nfn )n=nG1(nfn )nnG2n(nfn)
1n
=n n(nfn −n fn )n+nn(nfn +n f sin(n fTn)
c 
) nATn
n fT
c
2
Writingnout nthenconvolution:
n ATn nsin(Tn)n

G(nfn)n=n n  T n(n−n(nfn +n fcn)n+nn(n−n(nfn −n fcn)  d 
2 n
 
−
1
An sin(n(nfn +n fcn)Tn)n sin(n(nfn −n fcn)Tn)n
 fn =n
=n n +nn
n


2n fn +n f fn −n f  c
 c c  2T
 
An ncos(n fTn)n cos(n fTn)n
=n  −n 1 
2n fn −n 1 f n +n
 
 2T 





2T

(b) Bynusingnthentime-shiftingnproperty:
T
g(tn−ntn )n\nexp(−nj2n ftn ) tn =n
0 0 0
2
 
G(nfn)n= An ncos(n fTn)n −n cos(n fTn)nnnexp(−njn fTn)
n
2n fn −n 1
n
1 
f n +n
 

 2T 2T 




Copyrightn©n2009nJohnnWileyn&nSons,nInc.n AllnRightsnReserved.

,(c) Then half-sinen pulsen isnidenticaln ton then half-
cosinen pulsen exceptn forn then centren frequencynandntime-shift.

1
fcn =
2Ta
An ncos(n fTa)n cos(n fTa)nn
G(nfn)n= −n n(cos(n fTa)n−n jnsin(n fTa))
n
2n fn −n f
n
fn +n f 
 c c 
An cos(2n fTa)n cos(2n fTa)n sin(2n fTa)n sin(2n fTa)n
=n  −n +n jn −n jn
4n fn −n f f n +n f fn −n f fn +n f 
 c c c c 
An nexp(−n j2n fTa)n exp(−n j2n fTa)n
=n  −n
4n fn −n f fn +n f 
 c c 

(d) Then spectrumn isn then samen asn forn (b)n exceptn shiftedn backwardsn inn timen andn multipliednb
yn-1.

 
G(nfn)n= An ncos(n fTn)n −ncos(n fTn)nnnexp(njn fTn)
n
2n fn −n 1
n
1 
f n +n
 

 2T 2T 
 
An nexp(nj2n fTn)n exp(nj2n fTn)n
=n  −n 
4n fn −n
1
fn +n
1
 
 2T 





2T

(e) Becausen then Fouriern transformn isn an linearn operation,n thisn isn simplyn then summationn ofnt
henresultsnfromn(b)nandn(d)
 
G(nfn)n= An exp(n j2n fTn)n+nexp(−n j2n fTn)n −n exp(n j2n fTn)n+n(−n j2n fTn)n
n
 1 n
1 
4n  fn −n f n +n 
 





2T 2T
 
An ncos(2n fTn)n cos(2n fTn)n
=n  1 −n 1 
2n fn −n fn +n
 
 2T 





2T




Copyrightn©n2009nJohnnWileyn&nSons,nInc.n AllnRightsnReserved.

, 2.2

g(t)n =n exp(−t)nsin(2n fct)u(t)
=n (exp(−t)u(t))(sin(2n f c t) )
nn 1 
n f  2nj (n(nfn −n f cn)n−nn(nfn +n fc ))
1
G(nfn)n= 1+ n j2
 
1n  1 1 
=n  −
2njn 1+n j2n(nfn −n fn ) 1+ n j2n(nfn +n f 
 c cn) 




2.3n(a)

g(t)n=n gen(t)n+n gon(t)n
1n
gn (t)n =n  g(t) n+n g(−t)
e
2
n tn 
g e (t)n=n Arectn  
 2T 

1n
gn (t)n =n
o
 g(t) − g(−t)
n n

2
 ntnn−n12nTn  ntn+n12nTn  
=   −nrect  
gon(t) Anrectn   
  T   T 
    




Copyrightn©n2009nJohnnWileyn&nSons,nInc.n AllnRightsnReserved.