Answers Sample Exam Questions Set 2
OPTICAL COMMUNICATION ENGINEERING
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
1
(a) The power incident on a detector of light in 100 nW.
(i) Determine the number of photons per second incident on the detector if the
wavelength is 800 nm.
N / t == 4 × 1011 photons/sec.
(1.5 mark)
(ii) If we carried out the above calculation for a longer wavelength, briefly
state if the number of photons per second goes up or down, and why?
The longer wavelength requires higher photons/sec, because the energy per
photon decreases the higher the wavelength via inspection from E = hf and c =
fλ.
(1 mark)
(b) A T3 system running at 45 Mbps has a BER of 10-9, compute the number of
errors per minute.
Number of errors per minute = 2.7 errors/minute.
(0.5 mark)
(c) To operate properly, a fibre optic receiver requires -34 dBm of power. The
system losses are 31 dB in total, from the light source to the receiver. Compute
the power in mW that the light source needs to emit to meet the requirement.
PT = -34 + 31 = -3 dBm
In mW, PT = 10-3/10 = 0.5 mW.
(2 marks)
(d) A cable contains 144 single-mode fibres, each operating at 2.3 Gb/s. How
many digitised 64 kbps voice messages can be transmitted along this cable?
⇒ Total messages = 5.175 million messages. (1 mark)
(e) What is the different (in Watts) between -65 dBm and 65 dBm?
As -65 dBm is negligible compared to 65 dBm,
⇒ the difference ≈ 1065/10 - 0 = 4 ×106 mW = 4000 W
(1 mark)
(f) A fibre system operates with a carrier wavelength of 1.55 µm. Suppose that
the system can handle digital information at a rate equal to one-hundredth of one
percent of the optical frequency. How many 20 Mbps HDTV compressed video
channels can be multiplexed onto this fibre system?
R/RHDTV = 950 Video Channels
1290
(2 marks)
Page 2 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
(g) Prove that the power change γ in dB/km and the attenuation coefficient α are
related by γ = -8.685α, where α is given in the units of km-1.
The travelling wave equation, with loss is: E = Eoexp(-αz)sin(ωt-kz)
⇒ The intensity is: I = Eo2exp(-2αz)sin2(ωt-kz)
Dividing the average intensity at z = L by the average intensity at z = 0 yields the
fractional loss:
Loss = exp(-2αL)/exp(0) = exp(-2αL), and by converting to dB we get,
Loss (in dB) = 10log exp(-2αL) = -20αL log e = -8.685αL
If α is given in the units of km-1, then L must be in km, and so,
γ = Loss (dB)/L (km) = -8.685α
(3 marks)
(h) Derive the dynamic range of a conventional compact disc (CD) in dB.
Dynamic range is 65536:1
Dynamic range in dB = 20log(65536) = 96 dB
(2 marks)
A CD has several levels of error detection and correction. Provide a very brief
description of the following, and explain their purpose:
(i)
Eight to Fourteen Modulation (EFM)
Codes each octet (8 bits) using 14 bits to provide a base level of error
detection and correction. Corrects and detects on an octet-by-octet basis.
(2 marks)
(ii)
Reed-Solomon Code
A very powerful parity code for error correction. It is needed because a
CD is a storage medium, and thus a resend operation is not possible. It is
intended to correct long bursts of errors and a second layer of R-S coding
is used on CD-ROM to further improve error correction.
(2 marks)
(iii) Interleaving
Errors on a CD tend to be caused by debris or scratches; hence errors
occur in long bursts. Reed-Solomon codes work better on randomly
distributed errors, so by interleaving the bit stream a burst error
effectively becomes distributed, improving performance and error
correction.
(2 marks)
1290
Page 3 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
2. Consider a heterodyne receiver for a digital optical fibre communications system.
(a) Briefly explain how a heterodyne receiver detects phase, given that
photodetectors can only detect amplitude or optical power.
A heterodyne system detects phase with an amplitude detector via mixing the
signal with that of a local oscillator (LO) to produce beats. Thus phase
changes in the optical carrier are converted to changes in optical intensity,
which can then be measured by a photodetector.
(2 marks)
(b) State the key advantage of phase detection over amplitude detection.
Phase detection is less prone to errors due to non-linearities.
(1 mark)
(c) What type of optical modulation scheme can a heterodyne system permit that
is not otherwise achievable with direct detection using a standard receiver?
Heterodyning makes optical FM detection possible.
(1 mark)
(d) Heterodyne receivers offer increased sensitivity. Briefly state why.
The IF signal is proportional to local oscillator power. Thus, in effect, the LO
acts as a signal amplifier, increasing sensitivity of the receiver.
(1 mark)
(e) Using a heterodyne receiver, compute the local-oscillator (LO) power
required to make the SNR 1 dB less than the quantum limit. You may assume
the IF bandwidth is 500 MHz and the received optic power is constant at 5
nW when a binary “1” is received. The dark current of the photodetector is ID
= 5 nA, and its responsivity is ρ = 0.4 A/W. Assume the temperature is 27ºC
and a load resistance of 100 Ω.
The quantum limited SNR of the heterodyne receiver can be written as,
2 ρ 2 PS PL
SNR =
(2 marks)
2e∆f [ I D + ρ ( PL + PS )]
For large local oscillator power ( PL >> PS ) and negligible dark current yields
a quantum limited SNR of
SNR = 14dB
(2 marks)
The question wants an SNR 1 dB less than this limiting SNR, thus, in linear
units, this gives an
SNR = 1013 / 10 = 19.9
1290
(1 mark)
Page 4 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
To calculate the LO power needed to give us an SNR of 19.9, we can write
down a simplified expression for total SNR (containing thermal plus shot
noise) with a simplification assuming PL >> PS and that the dark current, ID, is
negligible. This gives:
SNR =
2 ρ 2 PS PL RL
.
2eRL ∆f ρ PL + 4 kT ∆f
(2 marks)
Rearranging to obtain PL, gives:
∴ PL = 5mW
(2 marks)
Our assumption that PL >> PS is confirmed, as 5 mW >> 5 nW.
(1 mark)
Also our assumption that dark current can be neglected is justified because
5 nA << ρPL = 2 mW.
(1 mark)
(f) If this were not a heterodyne system, then the receiver’s bandwidth could be
as small as 250 MHz. For this case determine the signal power required to
achieve a SNR equal to that in part (e).
For a direct detection, and neglecting dark current and rearranging to get
signal power gives a quadratic in PS:
ρ 2 RL PS 2 − SNR × 2eRL ∆f ρ PS − SNR × 4kT ∆f = 0
(1 mark)
The three quadratic coefficients come to:
a = 16 (no units)
b = −6.4 × 10 −8 (no units)
c = −8.2 × 10 −11 W
The second coefficient is negligible and taking just the positive solution of the
quadratic (as negative power has no meaning in this context), gives a source
power of,
Ps = 2.3µW
(2 Marks)
Our assumption that dark current can be neglected is justified because
5 nA << ρPS = 1 µW.
(1 marks)
Note that − b × PS = 6.4 × 10 −8 × 2.3 × 10 −6 = 1.5 × 10 −13 W is the shot noise
power and this is considerably lower than the thermal noise which is of the
order of |c| =8.2×10-11 W. Thus the system is thermal noise limited, whereas
the heterodyne system in part (e) was shot (quantum) noise limited. This
makes sense because the LO power in part (e) is over 200 times larger than PS,
here, thereby having many more photons to create shot noise (3 bonus marks)
1290
Page 5 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
3. (a) A long fibre, of arbitrary length, has 10 optical amplifiers equally spaced along
its length. The amplifiers are used to compensate loss due to fibre attenuation. You
may assume that the amplifier gain exactly equals the loss due to fibre attenuation.
Each amplifier has a 3 dB noise figure, the SNR at the transmitter is 108, and there is a
30 dB transmission loss between amplifiers along the fibre. Compute the SNR at the
output of the fibre.
The formula for noise figure of an amplifier chain is:
F=
F3
F1
F2
+
+
+ ... +
α 1 α 1G1α 2 α 1G1α 2 G2α 3
FN
N −1
α N ∏ (α i Gi )
(1 bonus mark)
i =1
If amplifier gain exactly compensates loss ⇒ α i Gi = 1
⇒ The noise factor formula for a chain reduces to: F = NFi Gi
(It is ok if this was memorized, but they get the bonus mark if they do it from the full
equation).
A 30 dB loss ⇒ α i = 10 −3 ⇒ Gi = 10 3
Given that N = 10 and Fi = 10 3 / 10 = 2 ⇒ F = 10 × 2 × 10 3 = 20,000
Finally, (SNR)out = (SNR)in/F = 108/20,000 = 5000 or 37 dB.
(3 marks)
(b) A 1-Mb/s NRZ pulse train is transmitted along a shot-noise limited system at λ =
0.82 µm. The receiver has negligible dark current and an ideal quantum efficiency of
unity. How many photons per bit must be incident on the photodetector if the desired
BER is 10-4? Compute the optic power incident upon the detector.
For a shot-noise limited system with negligible dark current the probability of
error, Pe = e − ns ,
⇒ The number of photons, ns = -ln(10-4) = 9.2 ≈ 10 photons
We round off to an integer because we must have an integral number of photons.
(1 mark)
We round up (not down) otherwise 9 photons will result in a BER that is too high.
(1 bonus mark).
As the quantum efficiency is unity, 10 photons/bit are required.
(1 mark)
An NRZ pulse at 1-Mbps is going to have a bit interval of τ = 10 −6 s and so the optic
power is:
hfn s hcns 6.63 × 10 −34 × 3 × 10 8 × 10
=
=
= 2.4pW
P=
ητ
ηλτ
1 × 0.82 × 10 −6 × 10 −6
1290
(1 mark)
Page 6 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
(c) Prove Johnson’s thermal noise formula en2 = 4 kTR∆f , where the symbols have
all their usual meanings, by taking the following steps:
(i)
Consider a resistor in parallel with a capacitor. You may assume the noisy
resistor can be modelled by a single noise source in series with a lumped
ideal resistor. Let the mean square voltage noise of the source be en2 . Let
the mean square noise across the capacitor be vn2 . Let the mean square
noise current through the loop be in2 . The angle brackets represent
temporal averages. By KVL, over a time window τ, we can
write vn (t ,τ ) + in (t ,τ ) R = en (t ,τ ) . You may assume that the windowed
Fourier transform of this equation is: Vn (ω,τ ) + I n (ω ,τ ) R = En (ω,τ ) .By
substituting in the relation I n (ω,τ ) = jωCVn (ω ,τ ) and then applying
Plancherel’s theorem (i.e. the energy theorem), prove that,
2
n
e
1
= lim
τ →∞ 2π
+∞
∫2
En (ω,τ )
τ
0
2
dω
Substituting in the relation I n (ω,τ ) = jωCVn (ω ,τ ) into the circuit equation
Vn (ω,τ ) + I n (ω ,τ ) R = En (ω,τ ) gives,
Vn =
En
.
1 + jω RC
(1 mark)
As the energy theorem requires squared variables, we must now multiply
through by complex conjugates giving,
2
Vn =
En
2
(1 mark)
1 + (ω RC )2
Plancherel’s theorem states that,
+∞
+∞
1
2
(1 mark)
∫−∞ e (t,τ )dω = 2π −∞∫ En (ω,τ ) dω
This is simply a statement of the conservation of energy: the total energy
in the time domain equals the total energy in the frequency domain.
(2 bonus marks)
2
n
2
As En is an even function, we can write the one-sided form,
+∞
2
∫ en (t,τ )dω =
0
1290
1
2π
+∞
∫ 2 E (ω,τ )
n
2
dω .
0
Page 7 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
(1 mark)
By definition of temporal average,
en2 = lim en2
τ →∞
τ
1 +∞ 2
∫ en (t,τ )dt
τ →∞ τ 0
= lim
(1 mark)
and therefore,
2
n
e
1
= lim
τ →∞ 2π
+∞
∫2
En (ω,τ )
τ
0
2
dω
(1 mark)
(ii)
Explain why we cannot take the limit inside the integral without first
taking ensemble averages. Then by assuming the system is ergodic, i.e.
2
lim en
τ →∞
τ
= en2 , where the overbar indicates ensemble averaging, show
that,
+∞
1
S (ω )d ω
2π ∫0
where S (ω ) is the one-sided power spectral density (PSD).
en2 =
As we have a random process, the limit would be indeterminate had the
ensemble average not been performed first.
(1 mark)
We now take the limit inside the integral and invoke ergodicity giving,
en2 =
1
2π
+∞
2 En2
dω
∫0 τlim
→∞ τ
(1 mark)
By definition, the one-sided power spectral density of en is,
2 En2
and therefore
τ →∞ τ
S (ω ) = lim
(iii)
en2 =
1
2π
+∞
∫ S (ω )dω
(1 mark)
0
Explain why for thermal noise, we can simply rewrite this expression
1
as en2 =
S0 ∆ω , where S0 is a constant PSD.
2π
We assume that we have a finite measurement bandwidth ∆ω , and that it
is sufficiently narrow that we can assume a flat spectrum (i.e. white noise)
(1 mark)
thus we can replace S (ω ) with a constant PSD S0 .
1290
Page 8 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
We also know from practice that for frequencies such that, kT >> hf
thermal noise is white.
(1 bonus mark)
(iv) By identical arguments, we can write down a similar expression for the mean
square noise voltage across the capacitor as
v
2
n
1
=
2π
+∞
2Vn2
dω .
∫0 τlim
→∞ τ
En
2
Now by substituting this into Vn =
2
1 + (ω RC )
2
, show that vn2 =
1
π
S0
.
2π 2 RC
The substitution gives,
v
2
n
1
=
2π
+∞
∫
0
2
lim 2 En d ω
2
1+ (ω RC ) τ →∞ τ
∴ vn2 =
∴ vn2 =
1
S0
2π
+∞
dω
∫ 1+(ω RC )2
0
S0 1
S π
[arctan(ω RC )]∞0 = 0
.
2π RC
2π 2 RC
(2 marks)
1
1
(v) From the equipartition theorem you may assume that C vn2 = kT . Use
2
2
this relation to now finally arrive at the thermal noise formula.
1
S0 ∆ω , obtained from part (iii), into the expression
2π
obtained in part (iv) gives,
Substituting en2 =
2
C vn2 R∆ω
π
Substitution in the expression for capacitor energy finally gives,
en2 =
en2 = 4 kTR∆f
(2 marks)
1290
Page 9 of 11 Pages
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
4.
(a) Prove that the maximum value of a/λ is 1.6 times larger for a single-mode
parabolic index GRIN fibre than for a single mode SI fibre, where the
symbols have their usual meaning.
The step-index single mode condition is:
2.405
a
  <
 λ  SI 2π n1 2 − n2 2
The GRIN single mode condition is:
1.4
a
<
 
 λ  GRIN π n1 (n1 − n2 )
Then,
(a / λ ) GRIN 2 × 1.4 × (n1 + n 2 )(n1 −n 2 )
=
(a / λ ) SI
2.405 n1 (n1 − n 2 )
(3 marks)
Now, n1 ≈ n2 in fibres, because this is the condition for minimising modal distortion.
(1 bonus mark)
Invoking n1 ≈ n2 , we can put n1 + n2 = 2n1 , thus
(a / λ ) GRIN
= 1.164 2 = 1.646 .
(a / λ ) SI
(2 marks)
(b) The equilibrium length of a multimode fibre is 2 km. The modal pulse spread is
25 ns for a 1 km length. The light source emits at 800 nm and has a spectral
width of 50 nm. Compute the optical 3 dB bandwidth of a 5 km length of
this fibre. You may assume that at λ0 = 800 nm, M = 115 ps/nm/km.
Equilibrium length, Le = 2 km.
Modal pulse spread per unit length ∆(τ /L)mod = 25 ns/km.
∴ Modal pulse spread over 5 km is ∆τ mod =
LLe ∆ (τ / L) mod = 79.1 ns
(1 mark)
Now, λo = 800 nm, M = 115 ps/mn/km and ∆λ = 50 nm hence material
dispersion pulse spread per unit length is ∆(τ /L)dis = -M∆λ = 5.75 ns/km,
(1 mark)
so for 5 km, ∆τdis = 28.75 ns
Total pulse spread:
∆τ = ∆τ mod + ∆τ dis = 84.4 ns
2
2
(1 mark)
These pulse spreads were added in quadrature as they are orthogonal
(uncorrelated) quantities.
(1 mark)
1290
Pages
Page 10 of 11
Optical Comm. Engineering and Photonics for Comms. SOLUTIONS, 2005 (contd.)
∴ f 3dB ( opt ) =
1
= 5.94 MHz
2∆τ
(1 mark)
(c) Briefly explain why mode mixing reduces modal distortion, and what happens
when the equilibrium length is reached.
Different modes have different path lengths in the fibre. Fibre defects cause
energy to couple between modes. Thus if a fibre is very long, enough defects
are encountered that each ray has experienced all modes. Thus on average
each ray travels the same path length and hence modal dispersion is reduced.
(3 marks)
When the equilibrium length is reached, mode mixing is complete, and
further fibre length does not improve modal distortion.
(1 mark)
This is analogous to stirring milk in a cup of tea. Initially, the tea becomes
lighter brown, but there comes a point when further stirring no longer
changes the colour. This is the equilibrium mixing condition.
(1 bonus mark)
(d) A fibre has a numerical aperture, NA = 0.2588. A light source is coupled to it
which emits 75% of its light into a 60 degree full-cone angle, 50% into a 30
degree cone and 25% into a 15 degree cone.
(i)
What is the coupling efficiency when this source and
fibre are connected?
NA = sin α = 0.2588
∴ Acceptance angle, α = 15°. This is the half cone angle.
(1 mark)
The fibre accepts light over a cone with a full angle of 30° ⇒ 50% of source
emission is captured.
(1 mark)
∴ Coupling efficiency, η = 0.5 and coupling loss L = 3 dB.
(ii)
If the refractive index of the core is 1.45, what is the loss due to
reflections?
Let us assume α is small enough that we can use the normal incidence
reflection formula,
(1 mark)
2
2
 n − n2 
 1 − 1.45 
 = 
R =  1
(1 mark)
 = 0.03
 1 + 1.45 
 n1 + n 2 
assuming an air-glass interface with n1 = 1 and n2 = 1.45 as given.
∴ Transmission loss due to reflection, L = -10 log10(1-R) = 0.15 dB.
(2 marks)
1290
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Page 11 of 11