# Communication systems

Communication systems

## An audio is signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as

Q: An audio is signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as

(i) R = 1 k Ω , C = 0.01 μF

(ii) R = 10 kΩ , C = 0.01 μF

(iii) R = 10 kΩ , C = 1  μμF

Sol: Carrier wave frequency , fc = 20 × 106 Hz

Bandwidth required for modulation is , 2 fm = 3 × 103 Hz

Hence , fm = 1.5 × 103 Hz

Demodulated by a diode is Possible if :

$\large \frac{1}{f_c} << R C < \frac{1}{f_m}$

Here , $\frac{1}{f_c} = \frac{1}{20 \times 10^6}$

= 0.5 × 10-7 s

$\frac{1}{f_m} = \frac{1}{1.5 \times 10^3}$

= 0.7 × 10-3 s

(i) R C = 1 × 103 × 0.01 × 10-6

= 10-5

Hence , the condition $\large \frac{1}{f_c} << R C < \frac{1}{f_m}$ is satisfied,.

So, it can be demodulated .

(ii) R C = 10 × 103 × 0.01 × 10-6

= 10-4

Hence , the condition $\large \frac{1}{f_c} << R C < \frac{1}{f_m}$ is satisfied,.

So, it can be demodulated .

(iii) R C = 10 × 103 × 1 × 10-12

= 10-8

Hence , the condition $\large \frac{1}{f_c} << R C$ is not satisfied,.

So, it can not be demodulated .

## Draw the plot of amplitude versus ω for an amplitude modulated were whose carrier wave ( ω_c) is carrying two modulating signals,  ω1 and  ω2 (ω2 > ω1)….

Q: (i) Draw the plot of amplitude versus ω for an amplitude modulated were whose carrier wave ( ωc) is carrying two modulating signals,  ω1 and  ω22 > ω1 ).

(ii) Is the plot symmetrical about ωc ? Comment especially about plot in region ω < ωc .

(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.

(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth ?

## An amplitude modulated wave is as shown in figure. Calculate (i) the percentage modulation, (ii) peak carrier voltage and (iii) peak value of information voltage

Q: An amplitude modulated wave is as shown in figure. Calculate

(i) the percentage modulation,

(ii) peak carrier voltage and

(iii) peak value of information voltage

Sol: (i) From figure ,

Maximum Voltage , $V_{max} = \frac{100}{2}= 50 V$

Minimum Voltage,  $V_{min} = \frac{20}{2}= 10 V$

Percentage modulation , $\mu = \frac{V_{max}-V_{min}}{V_{max} + V_{min}}\times 100$

$\mu = \frac{50-10}{50+10} \times 100$

= 66.67 %

(ii) Peak carrier voltage ,

$V_c = \frac{V_{max} + V_{min}}{2}$

$V_c = \frac{50 + 10}{2}$

= 30 V

(iii) peak value of information voltage,

$V_m = \mu V_c = \frac{66.67}{100} \times 30$

= 20 V

## The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I= I_0 e^(- αx), where I_0 is the intensity at x = 0 and α is the attenuation constant……

Q: The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I= I0 e– αx , where I0 is the intensity at x = 0 and α is the attenuation constant.

(a) Show that intensity reduces by 75 % after a distance of ((In 4)/α).

(b) Attenuation of a signal can be expressed in decibel (dB) according to the relations dB = 10 log10 (I/I0 ). What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50 % over a distance of 50 km ?

Sol: (a) I = I0 e– αx

As Intensity reduced by 75 % , that means $I = \frac{25}{100}I_o = \frac{I_o}{4}$

$\frac{I_o}{4} = I_o e^{-\alpha \; x}$

$\frac{1}{4} = e^{-\alpha \; x} = \frac{1}{e^{\alpha \; x}}$

eαx  = 4

Taking log ,

α x = ln 4

x = ln4/α

Hence at a distance x = ln4/α , the intensity is reduced by 75 %

(b) Here α is the attenuation constant in dB/km . If x is the distance travelled by signal then ,

$10 log_{10}\frac{I}{I_o} = -\alpha \; x$ ; Where Io is initial Intensity .

As 50 % intensity reduced by distance 50 km , so I = Io/2 and x = 50 km

$10 log_{10}\frac{I_o}{2 I_o} = -\alpha \times 50$

$10 log_{10}\frac{1}{2} = -\alpha \times 50$

$– 10 log 2 = -\alpha \times 50$

$\alpha = \frac{10 log 2}{50} = \frac{log 2}{5}$

α = 0.3010/5

## Figure shows a communication system, What is the output power when input signal is of 1.01 mW ?

Q: Figure shows a communication system, What is the output power when input signal is of 1.01 mW ? [gain in dB = 10 log10 (P0/Pi)]