Calculating Total Antenna Power for Modulated Carrier Wave

[Saitama]

Homework Statement

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

Homework Equations

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!

 

[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.

 

[Saitama]

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.

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.

Can you please share a link?

 

[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.

 

[Saitama]

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 take 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) >.

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?

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.

If I look it up on the internet, what would be the point of creating this thread?

 

[utkarshakash]

I don't think they will ask these questions.

 

[Saitama]

I don't think they will ask these questions.

I don't know if the test paper will include this kind of question but previous year too there was a weird kind of question (which, I feel is based on this chapter) . Problem 12: http://www.resonance.ac.in/jee-main-2013-answer-key-solutions/jee-main-paper-1-solutions-2013-english-1.1.pdf

 

[rude man]

Equation 15.3:
$$c_m(t)=A_c\left(1+\frac{A_m}{A_c}\sin(\omega_m t)\right)\sin(\omega_c t)$$

Rewrite as

c_m(t) = A{1 + μ sin(ω_mt)}sin(ω_ct)

= Asin(ω_ct) + f(t) sin(ω_ct)
where f(t) = μA sin(ω_mt)

so c_m(t) = Asin(ω_ct) + μAsin(ω_mt) sin(ω_ct)

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.

If I look it up on the internet, what would be the point of creating this thread?

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!

 

[rude man]

When you're ready for numbers:
what is power, given voltage amplitude? (assume 1 ohm as before).

 

[Saitama]

Rewrite as

c_m(t) = A{1 + μ sin(ω_mt)}sin(ω_ct)

= Asin(ω_ct) + f(t) sin(ω_ct)
where f(t) = μA sin(ω_mt)

so c_m(t) = Asin(ω_ct) + μAsin(ω_mt) sin(ω_ct)

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.

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

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?

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!

Thanks!

 

[rude man]

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

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?


Thanks!

Well, your answer is correct. Looks like we had different definitions of f(t).
Superb work!

 

[Saitama]

Well, your answer is correct. Looks like we had different definitions of f(t).
Superb work!

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?

 

[rude man]

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?

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

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

 

[Saitama]

So complete the equation "P_c = ... (a function of A)."

##P_c=A^2/2##?

 

[rude man]

##P_c=A^2/2##?

Yes! I hope you see why.

 

[Saitama]

Yes! I hope you see why.

Yes, I do, thanks a lot rude man!

 

[rude man]

Yes, I do, thanks a lot rude man!

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