One input to a conventional AM modulator is a 500 KHz carrier with an amplitude of $20V_P$. The second input is a 10 KHz modulating signal that is of sufficient amplitude to cause a change in the output wave of $±7.5V_P$. Determine - 1. Upper and lower side frequencies - 2. Modulation coefficient and percent modulation - 3. Peak amplitude of the modulated carrier and upper and lower side frequency voltages - 4. Expression for the modulated wave. - 5. Draw the output spectrum -

Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals

Marks: 10 Marks

Year: Dec 2013

1.The upper and lower side frequencies are simply the sum and difference frequencies, respectively:

$$f_{usf}=500kHz + 10 kHz = 510 kHz \\ f_{lsf}= 500 kHz - 10 kHz = 490 kHz$$

2.The modulation coefficient is determined from equation below

$$m=\dfrac{7.5}{20}=0.375$$

Percent modulation is determined from equation below

$$M=100 \times 0.375=37.5 \%$$

3.The peak amplitude of the modulated carrier and the upper and lower side frequencies is

$$E_v\text{(modulated)}=E_v\text{(unmodulated)}=20V_P \\ E_{ud}=E_{ld}=\dfrac{mE_v}{2}=\dfrac{(0.375)(20)}{2}=3.75V_P$$

4.The maximum amd minimum amplitudes of the envelope are determined as follows:

$$V_{\max}=E_v+E_m=20+7.5=27.5V_P \\ V_{\min}=E_v-E_m=20-7.5=12.5V_P$$

5.The expression for the modulated wave follows the format of equation below:

$$V_{am}(I)=20 \sin(2\pi 500 kt)-3.75 \cos(2 \pi 510kt)+3.75 \cos(2\pi 490kt)$$

6.The output spectrum is shown in figure 4-9.

7.The modulated envelope is shown in figure 4-10