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The equation of AM wave is, $e = 100(1 + 0.6\sin 6280t)\sin 2\pi \times {10^6}t.$ Calculate,
(i) Modulation index
(ii) frequency of carrier wave
(iii) frequency of modulating wave
(iv) frequency of LSB and USB.
Answer
401.1k+ views
Hint: In order to solve this question, we should know about Am waves. AM wave stands for amplitude modulation waves and it is a way to transmit messages as radio waves as their carrier waves by modulating them with respect to amplitude of the modulating and carrier waves in electronic communication. We will use the general expression for an amplitude modulation wave and then compare it with the given wave equation and will find asked parameters in the question.
Complete step by step answer:
The standard form of equation of amplitude wave is written in the form of
$e = A(1 + {m_a}\sin 2\pi {\omega _m}t)\sin 2\pi {\omega _c}t.$
Where, ${m_a}$ is known as the modulation index, ${\omega _m}$ Is the frequency of modulating waves and ${\omega _c}$ Is the frequency of the carrier wave.
So comparing the standard equation $e = A(1 + {m_a}\sin 2\pi {\omega _m}t)\sin 2\pi {\omega _c}t.$ with the given equation in question $e = 100(1 + 0.6\sin 6280t)\sin 2\pi \times {10^6}t$ we can see that,
${m_a} = 0.6$
Hence, (i) modulation index is ${m_a} = 0.6$
Frequency of carrier wave as $2\pi {\omega _c} = 2\pi \times {10^6}Hz$
Hence, (ii) Frequency of carrier wave is ${\omega _c} = {10^6}Hz$
Frequency of modulation wave as $2\pi {\omega _m} = 6280$
${\omega _m} = \dfrac{{6280}}{{2 \times 3.14}}$
$\therefore {\omega _m} = {10^4}Hz$
Hence, (iii) Frequency of modulating wave is ${\omega _m} = {10^4}Hz$
(iv) LSB stands for Lower side Band and calculated as,
$LSB = {\omega _c} - {\omega _m}$
$\Rightarrow LSB = {10^6} - {10^4}$
$\Rightarrow LSB = 99 \times {10^4}Hz = 990KHz$
USB stands for Upper side band and calculated as,
$USB = {\omega _c} + {\omega _m}$
$\Rightarrow USB = {10^4} + {10^6}$
$\therefore USB = 101 \times {10^4}Hz = 1010\,KHz$
Note: It should be remembered that, the SI unit of frequency is called Hertz and denoted by Hz, also the unit of conversion used as $1\,KHz = 1000\,Hz$ also, the symbol $\omega $ represents natural frequency not the angular frequency but the whole term as $2\pi \omega $ represents the angular frequency of the wave.
Complete step by step answer:
The standard form of equation of amplitude wave is written in the form of
$e = A(1 + {m_a}\sin 2\pi {\omega _m}t)\sin 2\pi {\omega _c}t.$
Where, ${m_a}$ is known as the modulation index, ${\omega _m}$ Is the frequency of modulating waves and ${\omega _c}$ Is the frequency of the carrier wave.
So comparing the standard equation $e = A(1 + {m_a}\sin 2\pi {\omega _m}t)\sin 2\pi {\omega _c}t.$ with the given equation in question $e = 100(1 + 0.6\sin 6280t)\sin 2\pi \times {10^6}t$ we can see that,
${m_a} = 0.6$
Hence, (i) modulation index is ${m_a} = 0.6$
Frequency of carrier wave as $2\pi {\omega _c} = 2\pi \times {10^6}Hz$
Hence, (ii) Frequency of carrier wave is ${\omega _c} = {10^6}Hz$
Frequency of modulation wave as $2\pi {\omega _m} = 6280$
${\omega _m} = \dfrac{{6280}}{{2 \times 3.14}}$
$\therefore {\omega _m} = {10^4}Hz$
Hence, (iii) Frequency of modulating wave is ${\omega _m} = {10^4}Hz$
(iv) LSB stands for Lower side Band and calculated as,
$LSB = {\omega _c} - {\omega _m}$
$\Rightarrow LSB = {10^6} - {10^4}$
$\Rightarrow LSB = 99 \times {10^4}Hz = 990KHz$
USB stands for Upper side band and calculated as,
$USB = {\omega _c} + {\omega _m}$
$\Rightarrow USB = {10^4} + {10^6}$
$\therefore USB = 101 \times {10^4}Hz = 1010\,KHz$
Note: It should be remembered that, the SI unit of frequency is called Hertz and denoted by Hz, also the unit of conversion used as $1\,KHz = 1000\,Hz$ also, the symbol $\omega $ represents natural frequency not the angular frequency but the whole term as $2\pi \omega $ represents the angular frequency of the wave.
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