Energy harvesting - Based transmission schemes in cognitive radio networks with a power beacon

r of S can be 
formulated as: 
2
2
min , .
pTS
S PT PS
SU
I
P P h
h

 =
 (7) 
SBT Scheme: 
In this scheme, the node S harvests energy from 
the PB as well as PT for its operation. Meanwhile, the 
PT also causes interference to the secondary network. 
Similarly, the transmit power of S after harvesting 
energy from PB and PT as follows: 
2 2
.EHS PB BS PT PSP P h P h = + (8) 
The transmit power of S must satisfy the 
interference constraint required by the primary 
receiver as: 
2
.
pI
S
SU
I
P
h
= (9) 
The transmit power of S can be expressed as: 
2 2
2
min ( ), .
pSBT
S PB BS PT PS
SU
I
P P h P h
h

 = +
 (10) 
3. Performance analysis 
In this section, we analyze the outage 
probability of the system over Rayleigh fading 
channels. The OP of a certain communication system 
can be defined as the probability that the capacity 
falls below a target data rate. The OP of the proposed 
schemes can be expressed as [19]: 
( ) ( )2Pr 1 log 1 ,sch schout S thP R  = − + (11) 
where , ,sch BS TS SBT and ( )th th 0R R is the 
target data rate. 
For ease of presentation and analysis, we use 
some self-defined functions along the developed 
analysis, and they are expressed as follows: 
( )
0
, , exp
ab c
a b c bx dx
x a x
+ 
 = − − 
+ 
 , 
( )
0
, , exp
1
abx c
a b c bx dx
ax x
 = − − 
+ 
 , 
,
p
PT
I
P


= ,SD th
PD
 

 = 


= ; 
, ,
SU BS pSD th BS
PB PB
I
P P
   
 
 
= = 
and ( ) ( )12 2 .x xK x = 
 Journal of Science & Technology 144 (2020) 035-041 
38 
3.1. BS scheme: 
Because only PB transmits power to node S, the 
instantaneous SNR (signalto-noise ratio) can be 
expressed as: 
2 2
2
min , ,
pBS
S PB BS SD
SU
I
P h h
h
 
 =
 (12) 
Now, OP can be calculated as: 
( )2 2 2
1
2
2 2
2 2
2
2 2
2
0
Pr
Pr ,
Pr ,
1
SU SD BS
BS
BS BS
out S th
p th
SU SD
PB BS PB BS
p th SU
BS SD
pPB SU
p th
h h h
PB PB
I
p
h
PB
P
I
h h
P h P h
I h
h h
IP h
I
F F f x dx
P x P x
I
F
P x
 

 



 

+ 
 = 
= 
+ 
= 
+ − 
( )2 2
2
0
SD SU
th
h h
p
I
x
F f x dx
I
+ 
(13) 
where: 
( )1
2 1.
thR
th

−
= − 
The first term of (13) can be expressed as: 
( )
( ) ( ) ( )
1
0
1 exp
1 exp exp
1
SU
SD
BS BS
h p
PB
h th
h h
I
I
P x
x dx
x


 
 

      
+ 
= − − 
 − − − 
 
= − − + +
 (14) 
Next, the second term of (13) can be expressed as: 
( )
( ) ( )
2
0
exp 1 exp
exp
.
BS SD
SU SU
SU
SU
h p h th
PB p
h h
h p
SD th p h
I x
I
P x I
x dx
I
I
  

 

    
  
+ 
= − − − 
 −
= − +
+
 (15) 
Having 
1I and I2 at hands, putting everything 
together (14) and (15), we can obtain the desired OP 
for BS scheme. 
3.2. TS scheme: 
In this case, node S only harvests energy from 
PT, so the instantaneous SNR can be expressed as: 
2
2
2 2
min ,
p SDTS
S PT PS
SU PT PD
I h
P h
h P h
 
 =
 (16) 
Therefore, OP can be calculated as: 
( )2 2
3
2
2
2 2
2
2
2
0
0
Pr
Pr ,
Pr , ,
1
SU PS
PS
TS TS
out S th
pth
SU
PS PT PS
pth PT SU
PS
p PT SU
pth
X h h
PT
I
p t
Xh
PT
P
I
X h
h P h
IP h
X h
I P h
I
F F f x dx
x P x
I
F F
P x
 

 



 


+ 
+ 
 = 
= 
+ 
= 
+ − 
 ( )2
4
,
SU
h PT
h
p
I
P x
f x dx
I
 (17) 
where
2 2
SD PDX h h= . 
The CDF of TS
S can be calculated as: 
( ) ( ) ( )2 2
0
PDSD
X hh
SD
PD SD
F y F yx f x dx
y
y

 
+ 
=
=
+
 (18) 
Plugging ( )XF y into (17) and after some 
manipulates, 
3I can be given by: 
( )
3
0
0
exp
exp
PS PS
p SUPS
PS
PT
x
I dx
x
I
x dx
x P x
 



+ 
+ 
 −
=
+
 
− − − 
+ 
 (19) 
Applying [16, Eq. (3.383.10)] for the first term 
of 
3I , we obtain as: 
( ) ( ) ( )3 exp 0, , ,PS PS PS PS SUI      =     −  
(20) 
Similarly, 
4I can be obtain as: 
( )
4
0
exp
1
, ,
SU PS
SU
SU PS
x
I x dx
x x
  


   
= − − 
+ 
= 
 (21) 
Having 
3I and 4I at hands, putting everything 
together, we can easily obtain the desired OP for the 
TS scheme. 
 Journal of Science & Technology 144 (2020) 035-041 
39 
3.3. SBT scheme 
Node S harvests energy from both the PT and 
PB; thus, the instantaneous SNR can be expressed as: 
( )
2
2 2
2 2
min ,
p SDSBT
S PB BS PT PS
SU PT PD
I h
P h P h
h P h
 
 = +
 (22) 
The OP of SBT scheme can be calculated as: 
( )
( )
( )
5
6
2
2 2
2
2 2
2
2
2 2
2 2
2
Pr
,
Pr
,
Pr
SBT SBT
out S th
SD
PB BS PT PS th
PT PD
p
PB BS PT PS
SU
I
p
th
SU PT PD
p
PB BS PT PS
SU
I
P
h
P h P h
P h
I
P h P h
h
I h
h P h
I
P h P h
h
 
 



 = 
+ 
= 
+ 
+ 
+ 
 (23) 
The first term in the right-hand side of (23) can 
be calculated as: 
( )2
2
2
5 2
0
Pr ,
,
SU
pSD th
SU
PTPD
pth
X Zh
PT
Ih
I h
Z P Zh
I
F F f x dx
x P x

 

 
+ 
= 
= 
 (24) 
where
2 2
SD PDX h h= and 
2 2
BS PSZ h h= + . 
We have the CDF and PDF of Z can be 
calculated respectively as: 
( )
( ) ( )
( ) ( )
( )
( )
( )
( )
2 2
2 2
0 0
0
Pr
exp exp
exp
1 exp
exp
exp
BS PS
BS BS BS PS
BS PS
BS
PS
BS
BS PS
BS PS PS
Z BS PS
z z x
h hx y
z h h h h
x
h h
h
hh
h h h h h
F z h h z
f x f y dxdy
x z
dx
x x
z
z
z z z


   
 


    
−
= =
=
 = + 
=
 − − −
 =
 − +
= − − −
 − −
 −
 − − − +
(25) 
( ) ( )
( ) ( )
( )( )
exp
exp
exp
BS
BS BS
BS PS
PS PS BS PS PS
BS PS PS
h
Z h h
h h
h h h h h
h h h
f z z
z
z

 
 
    
  
= − +
−
 − − + −
 − + −
(26) 
Plugging the CDF of X and PDF of Z into (24) and 
after some manipulations, we obtain: 
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
5 exp 0,
exp 0,
exp 0,
, ,
, , , , ,
BS BS BS
PS PS PSBS
BS PS
BS SU
BS
PS SU SU
BS PS
h h h
h h hh
h h
h h
h
h h h
h h
I   
  
    
  

     
 
=     +
    
 +
− −     
−  −
 −   − 
 −
(27) 
where
BS PS PSh h h
   = + − . 
Similarly, the second term in the right-hand side 
of (23) can be obtained as: 
( )2
1
6
0
0
0
0
1
exp
1
exp
1
exp
1
SU
SU BS
SU
BS PS
SU PS
SU
SU
SU
p th PT
Y X h
PT p
h h f
h
h h
h h
h
h
h
I P x
I F F f x dx
P x I
x
x dx
x x
x
x dx
x x
x
x dx
x x


   

  
  


 


+ 
+ 
+ 
+ 
= − 
= − − + 
+ − 
− − 
+ 
 − − − + 
( )
( ) ( )
,
, ,
, , , , ,
BS
SU BS
BS PS
SU PS SU
h
h h
h h
h h h

   
 
      
= + 
−
 − 
(28) 
Having 
5I and 6I at hands, putting everything 
together (27) and (28), we can obtain the desired OP 
for SBT scheme. 
 3.4. Throughput analysis 
In this section, throughput of three proposed 
schemes are analyzed. At a fixed target data rate 
0R 
(bps/Hz) and the communication time ( )1 T − , the 
throughput in the delay-sensitive transmission mode 
can be defined as: 
0 (1 )(1 ).
sch sch
outR P = − − (29) 
 Journal of Science & Technology 144 (2020) 035-041 
40 
Fig. 2. Effect of 
pI on the system outage probability 
with 1PBP = dB. 
Fig. 4. Effect of on the system throughput 
Fig. 3. Effect of on the system outage probability. Fig. 5. Effect of on the system throughput in SBT 
scheme with different values of 
PI 
4. Results and discussion 
In this section, we present illustrative numerical 
examples to show the achievable performance of the 
proposed schemes. For system settings, we consider a 
two dimension plane, where S, D, PB, PT and PR are 
located at (0,0), (1, 0), (XPB, YPB), and (XPT, YPT), 
(1, 1) respectively. Here, we adopt 0.6 = and 
thR = 1bit/s/Hz. 
We first investigate the effect of 
pI on the 
system outage probability, as shown in Fig. 2. It is 
observed that the OP values of all schemes are first 
reduced with the increase of 
pI , then converged to 
their error floors when 
pI is higher than 5 dB. The 
reason is that the transmit power of all the BS, TS 
and SBT schemes is dominated by the interference 
level in (4), (7), and (10), respectively. Importantly, 
the SBT scheme outperforms the TS one, which by its 
turn outperforms the BS scheme. This observation 
shows the effective design of combiming the energy 
harvested from PB as well as PT for the SBT scheme 
in cognitive radio networks. 
In Fig. 3, we investigate the effect of on the 
system outage performance with 2PBP = dB and 
2pI = − dB. As can be observed, the system OP is a 
convex function with respect to . Thus, there exists 
an optimal value of that minimizes the system OP. 
For the SBT scheme, the optimal value of is about 
0.5 while the TS and BS methods are about 0.6 and 
0.7, respectively. Thus, the SBT scheme is deployed 
will provide the highest system OP, where the system 
consumes about 60% of a coherent block time for 
harvesting energy from the source node and the 
remaining time for data transmisison. Again, the SBT 
scheme provides the highest performance among 
available ones, arising as an efficient strategy for 
CRNs. Moreover, Figs. 2 and 3 also reveal that the 
theoretical results are in excellent agreement with the 
simulation ones, validating the developed analysis. 
 Journal of Science & Technology 144 (2020) 035-041 
41 
In Fig. 4, we investigate the effect of on the 
system throughput of all schemes. As can be 
observed, the SBT scheme achieves the highest 
throughput while the BS scheme is the lowest 
performer. It can be sen that the system throughput is 
shown as a concave function of time switching ratio. 
Thus, there exists an optimal value of that 
maximizes the system OP. 
In Fig.5, we plot the system throughput of SBT 
scheme with different values of
PI . It is observed that 
the system throughput is first increased and reaches 
its highest value, then reduces to its lowest value as 
 is increased. The reason is that the system spends 
too much time for energy harvesting while the data 
transmission time is reduced, leading to the 
throughput degradation. 
5. Conclusion 
In this paper, we proposed the energy 
harvesting-based transmission schemes with power 
beacon to improve the outage and throughput 
performances in cognitive radio networks. In 
particular, we derived the exact closed-form 
expression for the outage probability and the 
throughput of the proposed schemes. The numerical 
results presented that the SBT scheme outperformed 
the TS one, which by its turn outperformed the BS 
scheme. In addition, the optimal time splitting ratio 
can be obtained based on the analytical results. 
Finally, the proposed scheme can be a promising 
design for network planning in future wireless 
cognitive sensor networks. 
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