Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication

Marks: 10M

Year:May2014

• The photodiode converts the optical bit stream into an electrical time-varying signal. The role of the preamplifier is to amplify the electrical signal for further processing.

• An equalizer is sometimes used to increase the bandwidth. The equalizer acts as a filter that attenuates low-frequency components of the signal more than the high-frequency components, thereby effectively increasing the front-end bandwidth.

  The amplifier gain is controlled automatically to limit the average output voltage to a fixed level irrespective of the incident average optical power at the receiver. The low-pass filter shapes the voltage pulse. Its purpose is to reduce the noise without introducing much Intersymbol interference

• The pulses sent by the transmitter arrive distorted at the receiver. This digital pulse train incident on the photodetector is given by:

  $$P(t)= \sum_{n = -∞}^∞b_n h_p (t-nT_b)$$

  P(t) is the received optical power

  $T_b$ is the bit period

  $b_n $ is the amplitude parameter of the nth message bit

• $h_p t$ is the received pulse shape

$$∫_{-∞}^∞ h_p(t) dt=1$$

• The mean output current from photodiode at time t (neglecting dc components arising from dark current).

$$i(t)= \frac{ηq}{hν } M P(t)= R.M ∑_{n=-∞}^∞b_n h_p (t-nT_b)$$

Where, $R= \frac{ηq}{hν}$ is photodiode responsivity

M = Mean gain of photodetector.

• This current is then amplified and filtered to produce a mean voltage at the output of the equalizer given by convolution of current with amplifier impulse response

  $$v_{out} (t)= A.R.M.P(t)h_B (t)h_{eq} (t) \ \; \ v_{out} (t)= R.G.P(t)* h_B (t)*h_{eq} (t) ....(1) \ \; \$$

  Here, A is the amplifier gain, and G is total gain. ∴ G = AM

  $h_B (t)$ is the impulse response of bias circuit

  $h_eq (t)$ is the equalizer impulse response

• $H_B (f)$ is the Fourier transform of $h_B$ (t)

  $$h_B (t)= F^{-1} [H_B (f)]= ∫_{-∞}^∞H_B (f) e^{j2πft} df$$

  The bias current transfer function $H_B$ (f) is simply impedance of parallel combination of $R_b,R_a,C_d $ and $C_a$

$$H_B (f)= \frac{1}{1⁄R+ j2πfC}$$

Where, $ \frac1{R}= \frac1{R_a} + \frac1{R_b} $

$$C= C_a+ C_b$$

• The mean voltage output from the equalizer can be given as:

• The Fourier transform equation is given as:

  $$H_{out} (f)= ∫_{-∞}^∞h_{out} (t).e^{-j2πft} dt=R.G.H_p f.H_B (f).H_{eq} (f)$$

  Where, $H_p$ (f) is the Fourier transform of received pulse and $H_{eq} (f)$ is the transfer function of the equalizer.