Mumbai University > Electronics and telecommunication > Sem 7 > optical communication and networks

Marks: 10

Years: MAY 2014

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.

$\Rightarrow$ 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.

$\Rightarrow$ 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 Inter symbol interference.

$\Rightarrow$ The pulses sent by the transmitter arrive distorted at the receiver. This digital pulse train incident on the photo detector is given by:

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

P(t) is the received optical power

$T_b$ is the bit period

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

$\Rightarrow h_p (t)$ is the received pulse shape

$∫\limits_{-∞}^∞ h_p (t) dt=1$

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

$i(t)= \dfrac {ηq}{hν} M \space P(t)= R∙M ∑\limits_{n=-∞}^∞ b_n h_p (t-nT_b) $

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

M = Mean gain of photo detector.

$\Rightarrow$ 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

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

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

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

$H_B (f)= \dfrac 1{1⁄R+ j2πfC}$

Where, $\dfrac 1R= \dfrac 1{R_a} +\dfrac 1{R_b} \\ C= C_a+ C_b$

$\Rightarrow$ The mean voltage output from the equalizer can be given as:

$h_{out} (t)=R∙G∙h_p (t)*h_B (t)*h_{eq} (t) ….(2)$

Where, $h_{out} (t)$ is the shape of an isolated amplified and filtered pulse.

$\Rightarrow$ The Fourier transform equation is given as:

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

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