Answer to Question #247334 in Electric Circuits for Shiril

Question

1. ssuming that an oscilloscope displays a V_max of 5.22V and V_min of 1.33V. Calculate the percentage of modulation. (5 points)
2. Three (3) AM broadcast stations are spaced at 18 kHz, beginning at 73 kHz. Each station is allowed to transmit modulating up to 6 kHz. Compute for the upper and lower sidebands of each station and plot it in the frequency domain. (15 points)
3. Given a V_max of 7.45V and a modulation index of 0.691, calculate for V_min. (5 points)
4. A station is given a carrier frequency of 88 kHz. Having a modulating frequency of 12 kHz, compute for the upper and lower sidebands of the station. (5 points)

Answer 1

Question 1

Percentage modulation M%=

$\frac {V max-V min}{V max+ V min} ×100$%

$\%$ = $\frac{5.22-1 .33}{5.22+1.33} ×100$ %

= 59.4%

Question 2:

For station 1:

f_lsb =73-6 =67kHz

f_usb = 73+6 = 79kHz

For station 2:

f_lsb =[73+18]-6=85kHz

f_usb =[73+18]+6=97kHz

For station 3:

f_lsb =(73+18+18)-6=103kHz

f_usb = (73+18+18)+6=115kHz

Question 3:

Modulation index M= $\frac{V max-V min}{V max+V min}$

0.691=$\frac{7.45-V min}{7.45+V min}$

(0.691×7.45)+0.691(V min)=7.45- V min

1.691V min= 2.3021

V min= 1.3614

Question 4:

Upper side band

=fc+fm=88+12=100kHz

Lower side band= fc-fm=88-12=76kHz

$\%$