$\pm5.5 vp$ .

Determine

(i) upper and lower side frequency

(ii) modulation coefficient and percent modulation

(iii) Draw output frequency spectrum

(iv) draw modulated wave showing maxima and minima of wave forms

Subject: Priniciples of Communication Engineering

Difficulty : Medium

Marks : 10

Solution:

Given: $f_c$ = 800 KHz

$f_m$ = 10 KHz

$V_c$ = 10 v

$V_m$ = 5.5 v

$\therefore V_m = 5.5 v$

(i) Upper and lower side band frequency

Upper side band frequency

$f_{usb}$ = $f_c$ + $f_m$

= 800 + 10

= 810 KHz

Lower side band frequency

$f_{Lsb}$ = $f_c$ - $f_m$

= 800 - 10

= 790 KHz

(ii) Modulation co efficient and percent modulation

Modulation index/modulation

Coefficient is $m_a$ given by:

$m_a$ = $\frac{v_m}{v_c}$

= $\frac{5.5}{10}$

$m_a$ = 0.55

Percent modulation is

% $m_a$ = 0.55 x 100

% $m_a$ = 55%

(iii) For drawing output frequency spectrum we should know the amplitudes of USB and LSB. They are calculated as follows:

Amplitude of USB = Amp of LSB = $\frac{m_a V_c}{2}$

= $\frac{0.55*10v}{2}$

= 2.75 v

Frequeny spectrum:

Bandwidth B.W = 2 x $f_m$

= 2 x 10 k

= 20 KHz

(iv) Modulated wave showing maxima and minima

$V_{max} = v_m + v_c$

$V_{max} = 5.5 + 10$

$V_{max} = 15.5 v$

$V_{min} = V_{max} – 2v_m$

$V_{min} = 15.5 – 2 * 5.5$

$= 15.5 – 11$

$V_{min} = 4.5 v$