# Single-tone AM

### single tone AM:-

The expression for conventional AM is $S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$

now if the message signal is a single-tone    $i.e,&space;m(t)&space;=&space;A_{m}cos&space;2\pi&space;f_{m}t$

$S_{Single-tone&space;AM}(t)=A_{c}(1+k_{a}A_{m}cos2\pi&space;f_{m}t)cos&space;2\pi&space;f_{c}t$

where $\mu&space;=k_{a}A_{m}$ is called as modulation index

$S_{Single-tone&space;AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\mu&space;A_{c}cos2\pi&space;f_{m}tcos&space;2\pi&space;f_{c}t$

this equation can be further simplified as follows     $S_{Single-tone&space;AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}+f_{m})t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}-f_{m})t$

that is by taking the fourier transform

$\dpi{150}&space;S_{Single-tone&space;AM}(f)=\frac{A_{c}}{2}\left&space;\{&space;\delta&space;(f-f_{c})+\delta&space;(f+f_{c})&space;\right&space;\}+\frac{\mu&space;A_{c}}{4}\left&space;\{&space;\delta&space;(f-(f_{c}+f_{m}))+\delta&space;(f+(f_{c}+f_{m}))&space;\right&space;\}+\frac{\mu&space;A_{c}}{4}\left&space;\{&space;\delta&space;(f-(f_{c}-f_{m}))+\delta&space;(f+(f_{c}-f_{m}))&space;\right&space;\}$

from the above expression the amplitude spectrum can be drawn as follows

from the spectrum single tone AM consists of 6 impulse functions located at frequencies $\pm&space;f_{c}$ , $\pm&space;(f_{c}&space;+&space;f_{m})$ and $\pm&space;(f_{c}&space;-&space;f_{m})$ respectively.

### Power content in AM/ Conventional AM:-

$S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$ represents the AM signal , here m(t) is  some arbitrary signal , then the power of this signal can be calculated from its Mean Square value $\overline{}{m^{2}(t)}$

i.e, message signal power = $\overline{}{m^{2}(t)}$ Watts.

Carrier signal is  $C(t)=A_{c}cos&space;2\pi&space;f_{c}t$ and it’s power is $\frac{A_{c}^{2}}{2}$ Watts.

Now the total power available in the signal $S_{AM}(t)=A_{c}(1+k_{a}m(t))cos&space;2\pi&space;f_{c}t$   will be  $P_{TOTAL}$ .

$S_{AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t&space;+A_{c}k_{a}cos&space;2\pi&space;f_{c}t&space;.&space;m(t)$

$P_{TOTAL}&space;=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}k_{a}^{2}}{2}&space;X&space;message&space;signal&space;power$

$P_{TOTAL}&space;=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}k_{a}^{2}}{2}&space;X\overline{m(t)^{2}}$ Watts.

Total Side Band power can be calculated from the term   $A_{c}k_{a}cos&space;2\pi&space;f_{c}t&space;.&space;m(t)$ can be denoted as $P_{SB}$ that would be $\frac{A_{c}^{2}k_{a}^{2}}{2}&space;X\overline{m(t)^{2}}$ Watts.

from these power calculations transmission efficiency of AM can be obtained as $\eta&space;=&space;\frac{P_{SB}}{P_{Total}}&space;X100$ %

$\eta&space;=&space;\frac{\frac{A_{c}^{2}k_{a}^{2}}{2}&space;.\overline{m(t)^{2}}}{\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}k_{a}^{2}}{2}&space;.\overline{m(t)^{2}}}$ X 100%.

$\eta&space;=&space;\frac{k_{a}^{2}.\overline{m(t)^{2}}}{1+k_{a}^{2}.\overline{m(t)^{2}}}$ X 100%.

### Power content in Single-tone AM:-

In single tone AM message signal is $i.e,&space;m(t)&space;=&space;A_{m}cos&space;2\pi&space;f_{m}t$, then power of the message signal is $\frac{A_{m}^{2}}{2}$ watts

carrier signal is $C(t)&space;=&space;A_{c}cos&space;2\pi&space;f_{c}t$ implies the carrier power is $\frac{A_{c}^{2}}{2}$ watts.

$S_{Single-tone&space;AM}(t)=A_{c}cos&space;2\pi&space;f_{c}t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}+f_{m})t+\frac{\mu&space;A_{c}}{2}cos2\pi&space;(f_{c}-f_{m})t$

then the  total power of the  single-tone AM signal is from the  above equation given as

$P_{Total}=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}\mu&space;^{2}}{8}+\frac{A_{c}^{2}\mu&space;^{2}}{8}$

PTotal = Pc +PUSB+PLSB

$P_{Total}=\frac{A_{c}^{2}}{2}+\frac{A_{c}^{2}\mu&space;^{2}}{4}$

$P_{Total}=\frac{A_{c}^{2}}{2}(1+\frac{\mu&space;^{2}}{2})$ Watts.

USB and LSB has same power $P_{USB}=P_{LSB}=\frac{A_{c}^{2}\mu&space;^{2}}{8}$ watts.

Now total side band power is $P_{SB}=P_{USB}+P_{LSB}=\frac{A_{c}^{2}\mu&space;^{2}}{4}$

from these power calculations transmission efficiency of AM can be obtained as $\eta&space;=&space;\frac{P_{SB}}{P_{Total}}&space;X100$ %

$\eta&space;=&space;\frac{\frac{A_{c}^{2}\mu&space;^{2}}{4}}{\frac{A_{c}^{2}}{2}(1+\frac{\mu&space;^{2}}{2})}&space;X&space;100$%

$\eta&space;=&space;\frac{\mu&space;^{2}}{(\mu&space;^{2}+2)}&space;X100$%.

Note:- Effeciency (or) Transmission efficiency of AM is only 33.3% only i.e, $\eta$ value  calculated when $\mu$ =1.

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