# Homework Help: Calculating the antenna power

1. Apr 2, 2014

### Saitama

1. The problem statement, all variables and given/known data
A 600W carrier wave is modulated to a depth of 75%by a 400Hz sine wave. Find the total antenna power.

(1) 769W
(2) 796W
(3) 679W
(4) 637.5W

2. Relevant equations

3. The attempt at a solution
I honestly don't know what to do here. I have tried to find a relevant equation in the chapter related to Communication system present in my textbook but I couldn't find anything useful. The solution uses the following formula:

$$P_T=P_c\left(1+\frac{m^2}{2}\right)$$

Where can I find the above formula?

Any help is appreciated. Thanks!

2. Apr 2, 2014

### rude man

Any textbook discussing "ordinary" AM would do. I'm sure many Websites would also.

By "ordinary" I mean like your AM broadcast band radio. The signal has a carrier and two sidebands. There are other AM signal types, e.g. DSB/SC (double-sideband suppressed carrier) etc.

3. Apr 2, 2014

### Saitama

Hi rude man! :)

Please have a look at this: http://www.ncert.nic.in/NCERTS/textbook/textbook.htm?leph2=7-7

The above link shows the chapter I am currently looking at in my book. I couldn't find that formula.

4. Apr 2, 2014

### rude man

Your textbook is not very detailed. But you could start from equation 15.3. You know that power is the square of amplitude (assume 1 ohm real impedance).

You will start with an expression like (a + b)2. You know how to expand this. Then you will need to tke the time average of the squared expression.

Then, one very important point: assume the modulating signal f(t) varies slowly compared to the carrier. This means that a term like f(t)2cos2(wt) can be time-averaged by considering f(t) constant: < f(t)2 cos2(wt) > = <f(t)2 > * < cos2(wt) >.

By deriving the power expression yourself you will learn a lot more than by just looking the formula up somewhere. But I'm sure you can find an expression for AM power in many places on the Web.

5. Apr 3, 2014

### Saitama

Equation 15.3:
$$c_m(t)=A_c\left(1+\frac{A_m}{A_c}\sin(\omega_m t)\right)\sin(\omega_c t)$$
Squaring both the sides gives:
$$c_m^2(t)=A_c^2\left(1+\frac{A_m}{A_c}\sin(\omega_m t)\right)^2\sin^2(\omega_c t)$$
I first calculate the average of $f^2(t)$ i.e
$$f^2(t)=A_c^2\left(1+\mu^2\sin^2(\omega_m t)+2\mu \sin(\omega_m t)\right)$$
$$\Rightarrow <f^2(t)>=A_c^2\left(1+\frac{\mu^2}{2}\right)$$
Hence,
$$<c_m^2(t)>=\frac{A_c^2}{2}\left(1+\frac{\mu^2}{2}\right)$$
where $\mu=A_m/A_c$. The above looks a lot like the formula I mentioned in #1.

How to proceed now?
If I look it up on the internet, what would be the point of creating this thread? :tongue2:

6. Apr 3, 2014

### utkarshakash

I don't think they will ask these questions.

7. Apr 3, 2014

8. Apr 3, 2014

### rude man

Rewrite as

cm(t) = A{1 + μ sin(ωmt)}sin(ωct)

= Asin(ωct) + f(t) sin(ωct)
where f(t) = μA sin(ωmt)

so cm(t) = Asin(ωct) + μAsin(ωmt) sin(ωct)

from which you can see that < f2(t) > cannot be finite if μ = 0. So work on the last equation above to get < f2(t) > correct, then the total power is two terms, one is the carrier only and the other is the modulation only.

You're absolutely right of course. It's just that most people would choose the textbook way out, which of course teaches them nothing. You are one in ten or twenty, congratulations!

Last edited: Apr 3, 2014
9. Apr 3, 2014

### rude man

what is power, given voltage amplitude? (assume 1 ohm as before).

10. Apr 3, 2014

### Saitama

$$c_m(t)=A\sin(\omega_c t)+\mu A\sin(\omega_mt)\sin(\omega_ct)$$
$$\Rightarrow c_m^2(t)=A^2\sin^2(\omega_c t)+\mu^2A^2\sin^2(\omega_mt)\sin^2(\omega_ct)+2\mu A^2\sin(\omega_mt)\sin^2(\omega_ct)$$
$$\Rightarrow <c_m^2(t)>=\frac{A^2}{2}+\frac{\mu^2A^2}{4}$$

The third term becomes zero, right?

$$\Rightarrow <c_m^2(t)>=\frac{A^2}{2}\left(1+\frac{\mu^2}{2}\right)$$
Isn't this the same expression I wrote before? :uhh:

Thanks!

11. Apr 3, 2014

### rude man

Superb work!

12. Apr 3, 2014

### Saitama

But I don't think that what I have arrived at is correct.

You said that the power is the square of amplitude, from that I get:
$$P_T=\frac{P_c}{2}\left(1+\frac{\mu^2}{2}\right)$$

This is not what I mentioned in #1.

Is the formula mentioned in #1 incorrect?

13. Apr 3, 2014

### rude man

No, the formula in 1 is correct. Yours is not.

Proof: let μ = 0, then what is the power in Asin(ωct)? It sure is Pc, is it not?
So complete the equation "Pc = ........ (a function of A)."

Last edited: Apr 3, 2014
14. Apr 4, 2014

### Saitama

$P_c=A^2/2$? :uhh:

15. Apr 4, 2014

### rude man

Yes! I hope you see why.

16. Apr 4, 2014

### Saitama

Yes, I do, thanks a lot rude man!

17. Apr 4, 2014

### rude man

Happy to have helped someone who really wants to learn as oposed to just passing a course.