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

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

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