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An AM signal is produced by modulating a carrier signal with peak voltage of 10V and Frequency of 100kHz by an information signal with max modulating frequency of 5KHz and max amplitude 4V. Determine

a.) Frequency limit for lower and upper side band

b.) Bandwidth of AM

c.) Total power of the modulated wave if the load resistance Rl = 10 ohms

d.) Draw the power spectrum.

e.) Calculate the total transmitted current.

Mumbai University > Electronics > Sem 4 > Priniciples of Communication Engineering

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Given :

$f_c$ = 100KHz

$f_{m(max)}$ = 5KHz

a) Frequency limits for upper and lower sideband

$\rightarrow f_{upper} = f_c + f_{m(max)}$ = 105 kHz

$\rightarrow f_{lower }= f_c - f_{m(max)} $ = 95 kHz

b) Bandwidth of AM

$\rightarrow$ Bandwidth = $f_{upper} - f_{lower}$ = 105kHz - 95 kHz = 10kHz.

OR

$\rightarrow$ Bandwidth = $2f_{m(max)}$ = 2 (5 kHz) = 10 kHz

c) Total Power

$ P_{c}=\frac{V_{c}^{2}}{2 R}=\frac{10^{2}}{2(10)}=5 W $

$ P_{L S B}=\frac{m^{2} P_{c}}{4}=\frac{(1)^{2} \times 5}{4}=1 \cdot 25 W $

$ P_{U S B}= \frac{m^{2} P c}{4}=\frac{(1)^{2} \times 5}{4}=1 \cdot 25 w $

$P_t = P_c + P_{L S B} +P_{U S B} $

= 5 + 1.25 + 1.25 = 7.5

d) Power spectrum

enter image description here

e) Current Transmitted

$ P t=P_{c}\left(1+\frac{m^{2}}{2}\right) $

But, $P = I^2 R$

$ I_{t}^{2} R=P_{C}\left(1+\frac{m^{2}}{2}\right) $

= 5 $( 1 + \frac{1}{2} )$

$I_{t}^{2} = \frac{7.5}{10}$

$I_{t} = \sqrt{\frac{7.5}{10}}$

$I_t$ = 0.866 A

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