What is the Absolute Value of b(t)?

[frenzal_dude]

I need to find the absolute value of b(t):

$$b(t)=a(t)[1+e^{-j2\pi tT}]$$

Here is the answer in the textbook:
$$\therefore \left | b(t) \right |=\left | a(t) \right |\sqrt{2(1+cos(2\pi tT)}2|cos(\pi tT)|$$

However I got a different answer when working it out, and can't understand how they got to that result from the textbook.
Here is my working out:

$$b(t)=a(t) + a(t)e^{-j2\pi tT}=a(t)+a(t)[cos(2\pi tT) - jsin(2\pi tT)]=a(t)+a(t)cos(2\pi tT)-a(t)jsin(2\pi tT)$$
$$\therefore |b(t)|=|a(t)|+|a(t)cos(2\pi tT)|+|a(t)sin(2\pi tT)|$$

Please let me know where I'm going wrong.
Thanks.

 

[micromass]

I need to find the absolute value of b(t):

$$b(t)=a(t)[1+e^{-j2\pi tT}]$$

Here is the answer in the textbook:
$$\therefore \left | b(t) \right |=\left | a(t) \right |\sqrt{2(1+cos(2\pi tT)}2|cos(\pi tT)|$$

However I got a different answer when working it out, and can't understand how they got to that result from the textbook.
Here is my working out:

$$b(t)=a(t) + a(t)e^{-j2\pi tT}=a(t)+a(t)[cos(2\pi tT) - jsin(2\pi tT)]=a(t)+a(t)cos(2\pi tT)-a(t)jsin(2\pi tT)$$
$$\therefore |b(t)|=|a(t)|+|a(t)cos(2\pi tT)|+|a(t)sin(2\pi tT)|$$

Please let me know where I'm going wrong.
Thanks.

You can't do that. It's not true that |a|=|b|+|c|... It's not true at all. You need to use the very definition of modulus:

$$|a+bi|=\sqrt{a^2+b^2}$$

 

[frenzal_dude]

Thanks for the reply, I tried it with that formula, however I still don't get the same answer as the textbook. Here's my working out:

$$b(t)=a(t)+a(t)cos(2\pi tT)-a(t)jsin(2\pi tT)$$

$$\therefore |b(t)|=\sqrt{[a(t)+a(t)cos(2\pi tT)]^{2}+[a(t)sin(2\pi tT)]^{2}}$$

$$=\sqrt{a^{2}(t)+2a(t)cos(2\pi tT)+a^{2}(t)cos^{2}(2\pi tT)+a^{2}(t)sin^{2}(2\pi tT)}$$

$$=\sqrt{a^{2}(t)[1+2\sqrt{a(t)}cos(2\pi tT) + cos^{2}(2\pi tT)+sin^{2}(2\pi tT)]}$$

$$=|a(t)|\sqrt{2+2\sqrt{a(t)}cos(2\pi tT)}$$

$$=|a(t)|\sqrt{2[1+\sqrt{a(t)}cos(2\pi tT)]}$$

 

[micromass]

Thanks for the reply, I tried it with that formula, however I still don't get the same answer as the textbook. Here's my working out:

$$b(t)=a(t)+a(t)cos(2\pi tT)-a(t)jsin(2\pi tT)$$

$$\therefore |b(t)|=\sqrt{[a(t)+a(t)cos(2\pi tT)]^{2}+[a(t)sin(2\pi tT)]^{2}}$$

$$=\sqrt{a^{2}(t)+2a(t)cos(2\pi tT)+a^{2}(t)cos^{2}(2\pi tT)+a^{2}(t)sin^{2}(2\pi tT)}$$

That second term should have an $a(t)^2$.

 

[HallsofIvy]

Also $cos^2(2\pi tT)+ sin^2(2\pi tT)= 1$ so what you have reduces to
$$a(T)\sqrt{2(1+ cos(2\pi tT))}$$

By the way, this has nothing to do with your question but that "tT" looks very strange to me. If it were t/T, then your sin and cosine functions would have a period of T, which I would interpret as a period of time. The way you have it their period is 1/T. It strikes me as strange to all something with units of "1/time", T.

 

[frenzal_dude]

Also $cos^2(2\pi tT)+ sin^2(2\pi tT)= 1$ so what you have reduces to
$$a(T)\sqrt{2(1+ cos(2\pi tT))}$$

By the way, this has nothing to do with your question but that "tT" looks very strange to me. If it were t/T, then your sin and cosine functions would have a period of T, which I would interpret as a period of time. The way you have it their period is 1/T. It strikes me as strange to all something with units of "1/time", T.

You're completely right, the T there is actually frequency which is 1/T, I should have used a different letter other than T (which ofcourse usually means period).