Question

For the sinusoidally modulated DSB-SC waveform shown in Fig: below: a- Find the modulation index. b- Sketch a line spectrum. c- Calculate the ratio of average power in the sidebands to that in the carrier. d- Determine the amplitude of the additional carrier, which must be added to obtain a modulation index of 100%. 

          For the sinusoidally modulated DSB-SC waveform shown in Fig: below:
a- Find the modulation index.
b- Sketch a line spectrum.
c- Calculate the ratio of average power in the sidebands to that in the carrier.
d- Determine the amplitude of the additional carrier, which must be added to obtain a modulation index of 100%.
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Added by Mackenzie R.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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For the sinusoidally modulated DSB-SC waveform shown in Fig: below: a- Find the modulation index. b- Sketch a line spectrum. c- Calculate the ratio of average power in the sidebands to that in the carrier. d- Determine the amplitude of the additional carrier, which must be added to obtain a modulation index of 100%.
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Transcript

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00:01 So here we know that for dsb -l -c, we are having phi of t, that will be equals to ac, multiply by the 1 plus k -a -m, multiply by the t, that is multiplied by the course of 2 pi, f -c of t.
00:16 So from here, if m of t is sinusoidal, so we can say that 5 of t will be equals to ac, multiplied by the 1 plus k a of am multiplied by the m of t that is course of 2 pi j c t of t so from here we can say that k a multiply by the a of m will be equals to mu which is equal to which is a modulation index so from here we can say that phi of t will be equals to ac 1 plus mu of m t multiply by the course of 2 pi f c of t so envelope so envelope of five of t has ac of one plus mu and ac of one minus mu and uh as hundred vault and 50 volt respectively so from here we can say that ac of one plus mu divided by the ac of one minus mu will be close to hundred divided by the 50 so ac to ac to ac cancel that will be two so from here we can say that one plus mu will be equals to 2 minus 2 of mu so 3 of mu is equal to 1 so mu will be 1 divided by the 3 so this is the answer to the part a for part b we can say that for m of t that will be close of 2 pi fm of t so 5 of t will be equals to ac of 1 plus mu course of 2 pi f of x 2 of t multiply by the course of 2 pi fc of t so from here we can say that phi of t so it was like five of t was here so from here we can write it as that two cause of a cause of b will be equals to cause of a plus b plus cause of a minus b so phi of t from here will be equals to ac course of two pi f c plus ac, mu divided by 2 of course of 2 pi t multiplied by the fc minus fm, plus cause of 2 pi t fc plus fm...
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