Solutions Manual for Introduction to Fiber-Optic Communications 1st Edition by Rongqing Hui Ph.D. (Chapters 2-12)

,Chapter 2, Optical fibers

Problems:
1, Consider a plane wave with a relative phase difference  = /3 between the Ex and Ey
components. If the magnitude of the x component is twice as large as the y component
(E0x = 2 E0y), and assume E0y = 1, please find the orientation angle  of this
polarization ellipse, and find the lengths of the long axis and the short axis of the
ellipse.

Solution:
Based on Equation (2.2.5)
 2 E0 x E0 y cos   1
1
 = tan −1   = tan −1  2  2  0.5  = 1 tan −1  2  = 0.294 rad .

 E0 x − E0 y  2  4 −1  2  3
2 2
2
Since E x = E0 x cos  = 2 cos(0.294 ) = 1.9142
E y = E0 y cos( +  ) = cos(0.294 +  / 3) = 0.2276
The half length of the long axis is
E0 x cos  + E0 y cos( +  ) = 1.9142 2 + 0.22762 = 1.7821
2 2



The half length of the short axis is

E0 x cos( +  / 2 ) + E0 y cos( +  / 2 +  )
2 2




= 4 cos(0.294 +  / 2 ) + cos(0.294 + 5 / 6) = 1.1332
2 2



Therefore the full lengths of the long and the short axes are, 3.564 and 2.2664,
respectively.

2, Consider the air/water interface shown in the following figure.
(a) A collimated light beam launches from the air to the water at an angle  = /6 with
respect to the surface normal as shown in Figure E-2.1(a), where ha = 1.5 m and hb = 1
m. The refractive index of air is n0 = 1, and assume the refractive index of water is n1 =
1.3.
Please find the distance x1 = ?
In order to eliminate reflection from the water surface, what should be the angle , and
in which direction the light should be polarized?
(b) If the light is launched from the bottom of the water tank as shown in E-2.1(b), and
the incidence angle is still  = /6, please find the distance x2 = ?.
What is the angle  to achieve total reflection on the water/air surface?

, x2

ha  ha
n0 n0
n1 n1
hb hb 


x1
(a) (b)

Figure E-2.1

Solution:
x2

ha  ha

n0 x0 n0
x0  n1 n1
hb hb 

x1
(a) x0 = ha tan  = 1.5 tan( / 6) = 0.866m
n   1   
Diffraction angle  2 = sin −1  0 sin   = sin −1  sin    = 0.395rad = 22.6
 n1   1.3  6  
x1 = x0 + hb tan  2 = 0.866 + 1  tan(0.395) = 1.28m
When the incidence angle  is equal to the Brewster angle and the optical field
polarization is parallel to the incidence plane, reflection from water surface can
eliminated. The Brewster angle is,
 =  B = tan −1 (n2 / n1 ) = tan −1 (1.3) = 0.915rad = 52.4 

(b) x0 = hb tan  = 1  tan( / 6) = 0.5774 m
n   1.3    
Diffraction angle  0 = sin −1  1 sin   = sin −1  sin    = 0.708rad = 40.5
 n0   1  6 
x2 = x0 + ha tan  0 = 0.5774 + 1.5  tan(0.708) = 1.86m
Total reflection happens when the incidence angle  is equal to the critical angle,

, = sin −1 (n0 / n1 ) = sin −1 (.3) = 0.878rad = 50.28

3, A light beam is launched to a thin semiconductor film with a thickness of d=2 m. The
refractive index of air is n0 = 1, and assume the semiconductor film has no loss and its
refractive index is n1 = 3.5. The wavelength of the light is  = 633nm which is parallel
polarized on the plane of the paper, and the incidence angle is  = 45°.
Please find the relative amplitude |E1|/|E2| and the phase difference [Phase(E1) -
Phase(E2)] of the two reflected beams at the reference plane.


reference plane
E1 E2

n0
n1 d

Figure E-2.2

Solution:
For the parallel polarization, according equation 2.5, the reflectivity is
− n12 cos1 + n0 ( n12 − n02 sin 2 1 )
 // = = −0.4329
n12 cos1 + n0 ( n12 − n02 sin 2 1 )
which means that the phase shift is .
n 
For the 2nd beam, the beam angle into the glass is  2 = sin −1  0 sin   = 0.2034 rad
 n1 
and the reflectivity on the bottom glass/air interface can be found from
− n02 cos 2 + n1 (n02 − n12 sin 2  2 )
 // = = 0.4329
n02 cos 2 + n1 (n02 − n12 sin 2  2 )
The phase shift is zero. The amplitude ratio is,
E1 0.4329 1
= = = 1.2306
E2 1 − 0.4329  0.4329  1 − 0.4329
2 2 1 − 0.4329 2

2 2d  4d 
The phase difference is,  = − − = − 1 +  = −14
 cos 2   cos  2 

4, A light beam with  = 0.63 m travels inside a slab optical waveguide by bouncing
back and forth between the two interfaces. The refractive index of the waveguide is n1
= 1.5, and the thickness is d = 3 m. Assume the lightwave is vertically polarized (E
field goes into the paper) with respect to the incidence plane. For the incidence angle 

, = 60°, what is the optical phase shift after each roundtrip? What is the evanescent field
penetration depth near the glass/air interface?
n0
n1  d
n0
Figure E-2.3

Solution:
Based on Equation 2.1.8, the optical phase shift of each reflection is
 n0 (n12 / n02 )sin 2  − 1 
 ⊥ = −2 tan  −1  = −1.671 rad .
 n1 cos  
 
and the phase delay due to roundtrip propagation is,
2 4d
 d = 2d = = 116.34 rad .
 cos  cos
Total phase shift is than,  =  ⊥ + 4d = 116.34 − 1.671  2 = 113rad .  36
 cos
The penetration depth can be found as,
1  0.63
ze = = = = 0.121m
 2 n1 sin 1 − n2 5.21
2 2 2




5. A step-index optical fiber has the core refractive index n1 = 1.50 and the cladding
refractive index n2 = 1.497. The core diameter of the fiber is d = 8m.
(a) What is the numerical aperture of this fiber?
(b) What is the cut-off wavelength c of this fiber? (The fiber operates in single mode
when the operation wavelength satisfies  > c)
(c) To carry a blue light at 0.4 m wavelength, approximately how many modes exist in
this fiber?

Solution:
(a) NA = n12 − n22 = 1.52 − 1.497 2 = 0.0948
2a 2
(b) For single mode operation, V = n1 − n22  2.405 , where a = d/2 = 4m.

2a
the cutoff wavelength is, c  n12 − n22 = 0.99m
2.405
2a 2 2  4  10 −6
(c) V = n1 − n22 = 0.0948  6
 0.4  10 − 6
The number of modes is approximately M  V = 18

,6, For a single-mode step-index fiber, the core index n1 = 1.49 and the cladding index n2
= 1.487, and the core radius is a = 4.5 m.
(a) Please find the wavelength at which V = 2.
(b) According to Figure 2.3.5, for V = 2, the normalized propagation constant is
approximately b ≈ 0.4. Please find the longitudinal propagation constant of the
fundamental mode z at this b-value, and compare this z value with free-space  -
value corresponding to the refractive index of the core.
(c) Approximately how far away from the core/cladding interface the field amplitude is
reduced to 1% compared to its value at core/cladding interface?

Solution:
(a) For V = 2, the wavelength is,
2a 2
= n1 − n22 = 4.5 1.49 2 − 1.487 2 = 1.336 m
2

(b) By definition, b =
( z / k )2 − n22
, where k = 2 / 
n12 − n22

2 2 0.4  (1.49 2 − 1.487 2 ) + 1.487 2
z = b(n12 − n22 ) + n22 = = 6.999 m −1
 1.336
2n1 2  1.49
The free-space -value in the core is  = = = 7.0007 m −1 , which is
 1.336
slightly higher than z -value.

(c) Field attenuation in the cladding is proportional to exp(− Wlm r ) . Here

 2n2   2  1.487 
2 2

Wlm =  z2 −   = 6.999 − 
2
 = 0.2815m
−1

    1.336 
Thus, for the field amplitude reduction 99%, the distance r away from the core/cladding
interface, is
ln (0.01) ln (100 )
r=− = = 16.4 m
Wlm 0.2815


7, Again based on Figure 2.3.5, for V = 3, the b-values of LP01 and LP11 modes are 0.7
and 0.16, respectively. Assume the step-index fiber the core index n1 = 1.49 and the
cladding index n2 = 1.487, and the core radius is a = 4.5 m. What is the propagation
constant, z, difference between LP01 and LP11 modes? For a fiber length of L =
10km, what is the arrival time difference between these two modes?

Solution:
For V = 3, n1 = 1.49, n2 = 1.487, and a = 4.5 m, the operating wavelength must be

, 2a 2
= n1 − n22 = 3 1.49 2 − 1.487 2 = 0.8907 m
3

By definition, b =
( z / k )2 − n22 , where k = 2 / 
n12 − n22
For LP01 mode
2 2 0.7  (1.49 2 − 1.487 2 ) + 1.487 2
 z ,01 = b(n12 − n22 ) + n22 = = 10.5044 m −1
 0.8907
For LP11 mode
2 2 0.16  (1.49 2 − 1.487 2 ) + 1.487 2
 z ,11 = b(n12 − n22 ) + n22 = = 10.4930 m −1
 0.8907
Thus, the differential propagation constant is,
 z ,01 −  z ,11 = 10.5044 − 10.4930 = 0.0114 m −1
For a fiber length of 10km, the differential arrival time between these two modes is,
 z L  z L
 0.8907  10 −6
T = =  z L =
= 0.0114  10 6  10 4 = 5.4  10 −8 s = 54ns
  2c 2  3  10 8


8, For a standard single-mode fiber, the core index is n1 = 1.47, the normalized index
difference is  = 0.35%, and the core diameter is d = 8 µm. Please use equation
(2.3.43) to find mode field diameter 2W0 at 1310nm and 1550nm wavelengths. Please
explain why the mode field diameter is smaller at 1310nm wavelength than that at
1550nm wavelength.

Solution: Based on equations 2.3.33, and 2.3.35,
2a 2a 2  4 3.0911
V= NA = n1 2 =  1.47 2  0.0035 =
   
V = 2.36, and 1.99 for  = 1310nm and 1550nm respectively.
Then based on equation 2.3.43, 2W0  2a (0.65 + 1.619V − + 2.879V −6 )
2W0= 8.9µm, and 10.16µm for  = 1310nm and 1550nm respectively. Details can be
plotted in the following figure.

, 13

12.5

12

11.5




2W0 (µm)
11

10.5

10

9.5

9

8.5

8
1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
Wavelength (µm)

9, Optical field propagating along an optical fiber can be expressed as
E ( z, t ) = E0 exp− j (t − z ) exp(−  )z , where  = 1.15  10−4 m −1 .
(1) What is the power attenuation parameter in dB/km?
(2) If the fiber length is 30km and the input optical power is 10dBm, what is the optical
power at the fiber output?

Solution,
Power propagation along the fiber is P ( z ) = P0 exp(− 2 )z , so that
(1) Power attenuation parameter of this fiber is
2dB = 2  4.343 = 2  4.343 1.15  10−4 m−1 = 1  10−3 m−1 = 1dB / km
(2) For a 30km fiber, the total power loss is 30dB. If the input power is 10dBm, the
output power should be -20dBm.

10, A fiber-optic system consists of three fibers with lengths L1 = 10km, L2 = 20km, and
L3 = 30km, and loss parameters 1 = 0.5dB/km, 2 = 0.25dB/km and 3 = 0.2dB/km,
respectively. These three fibers are spliced together to form a 60km optical link.
Neglect the splicing loss between fibers, what is the total loss of this fiber link? If the
input optical power is 4mW, what is the output optical power in mW?

Solution,
The total loss of the fiber link is
Loss = 1L1 + 2 L 2 + 3 L3 = 0.5  10 + 0.25  20 + 0.2  30 = 16dB
The input power is 4mW, which is 6dBm, the output power is 6 - 16 = -10dBm, which is
0.1mW.

11, Chromatic dispersion can be represented by a frequency-dependent propagation
constant (). Suppose the propagation constant of a single-mode fiber is

,  ( ) = A0 + B( − 0 ) + C ( − 0 )2 , where 0 = 1.216  1015 Hz (corresponding
to  = 1550nm wavelength).
(a) If at the optical frequency 0, the fiber has a group velocity vg = 2  108 m / s and
a chromatic dispersion parameter D = 15 ps / (nm  km) , please find the values of B
and C in the () expression.
(b) Assume an optical pulse is simultaneously carried by two wavelengths separated
by 3nm in the 1550nm wavelength window. After 20km transmission through this
fiber, approximately what is the pulse separation in the time domain caused by the
chromatic dispersion?


Solution:
d
( − 0 ) + 1  d 2
2
Since  ( ) =  (0 ) + ( − 0 )2
d  =0 2 d  = 0

1 1
(a) By definition, the group velocity is vg = = = 2  108 m / s
d / d  = 0 B
Therefore, B = 5  10−9 s / m
d 2  ( )
The chromatic dispersion parameter is  2 = = 2C
d 2  =
0

2c 4c
D=  = C , therefore,
2 2 2
2 D (1550  10−9 )
2
10−12
C= =  15  = 9.56  10− 27 s 2 / m
4c 4  3  108 10− 9  103

(b) t = D  L   = 15  20  3 = 900 ps

12, In a single-mode fiber system the light source has two discrete wavelength
components with the same power but separated by 1.2nm. The fiber dispersion
parameter in the laser diode wavelength range is D = 17 ps /{nm  km) .
Single mode fiber
Sinusoid

Light source Optical receiver
Optical
spectrum
 
The light source is amplitude modulated by a sinusoid so that the total optical power
can be expressed as Popt (t ) = P0 (1 + cos(2ft ) ) [note: power carried by each

, wavelength component is P1 (t ) = P2 (t ) = 0.5P0 (1 + cos(2ft ) ) at the fiber input]. After
transmitting for a certain distance along the fiber, the amplitude modulation on the
total optical power may disappear, this is commonly referred to as “carrier fading”.

(a) Explain the physical reason why this carrier fading may occur
(b) If the modulation frequency is f = 10GHz, find the fiber length at which carrier
fading happens
(c) If P0 = 4mW, fiber loss coefficient is  =0.5dB/km, and the fiber length is 50km,
what is the average optical power at the fiber output? Does this power change
with the modulation frequency?

Solution:
(a) the two wavelength components propagate in a slightly different speed due to
chromatic dispersion. This speed difference is
v g = D = 17 ps / nm / km  1.2nm = 20.4 ps / km
When the relative delay is equal to a  phase shift between the two sinusoids, they
cancel each other, and thus, no amplitude modulation remains.
(b) If the modulation frequency is f = 10GHz, for 2ft =  , t = 1 /( 2 f ) = 5  10 −11 [ s ] ,
50 ps
or, v g L = 5  10 −11 , L = 5  10 −11 / v g = = 2.45km ,
20.4 ps / km
Thus, carrier fading happens at L = 2.45km.

(c) The average optical power does not change with modulation frequency, and after
40km, the total fiber loss is 50km  0.5dB / km = 25dB
Input average optical power is 4mW = 6dBm, so that the output average power is
Pout = −19dBm = 10−1.9 = 0.0126mW


13, Sellmeier equation (Eq. (2.5.22)) may be linearized as D ( ) = S0 ( − 0 ) for
simplicity. A single-mode fiber has zero-dispersion wavelength 0 = 1310nm and the
dispersion slop S0 = 0.092 ps/(nm2km) near the zero-dispersion wavelength.
(a) Please find the chromatic dispersion parameters D in [ps/(nm-km)] and 2 in [ps2/km]
at  = 1550nm using Sellmeier equation and the linearized version.
(b) Calculate group delay difference between 1310nm and 1550nm wavelengths using
Sellmeier equation.


Solution,