300
7 Optical Receiver Operation
normalization condition
∞
−∞
h p (t)dt = 1
(a) Rectangular pulse (α = constant)
h p (t) =
1
αT b
for −
αT b
2
< t <
αT b
2
0 otherwise
(b) Gaussian pulse
h p (t) =
1
√
2π
1
αT b
e
−t
2 /2(αT b )
2
(c) Exponential pulse
h p (t) =
1
αT b
e
−t/αT b
for 0 < t < ∞
0 otherwise
7.3 Use Eq. (7.8) to derive the error probability expression given by Eq. (7.16).
7.4 Consider an optical receiver that has a high-impedance amplifier with an
input resistance of R a = 3 M. Suppose it is matched to a photodetector
bias resistor that has a value R b = 3 M. (a) If the total capacitance is C =
6 pF, show that the maximum bandwidth achievable without equalization is
17.3 kHz. (b) Now consider the case when the high-impedance amplifier is
replaced with a transimpedance amplifier that has a 100 k feedback resistor
and a gain G = 350. Show that the maximum achievable bandwidth without
equalization in this case is 92.8 MHz.
7.5 Consider the probability distributions shown in Fig. 7.7, where the signal
voltage for a binary 1 is V 1 and υ th = V 1 /2.
(a) If σ = 0.20V 1 for p(y|0) and σ = 0.24V 1 for p(y |1), use Eqs. (7.10) and
(7.11) to show that the error probabilities
P 0 (υ th ) = 0.5 [1−erf (1.768)] = 0.0065
and P 1 (υ th ) = 0.5 [1 − erf (1.473)] = 0.0185
(b) If a = 0.65 and b = 0.35, show that P e = 0.0143.
(c) If a = b = 0.5, show that P e = 0.0125.
7.6 An LED operating at 1300 nm injects 25 μW of optical power into a fiber.
Assume that the attenuation between the LED and the photodetector is 40 dB