Absolute Value Problem (1 Viewer)

frenzal_dude · Oct 22, 2011

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:

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

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

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

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

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

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

AAEldar · Oct 22, 2011

frenzal_dude said:
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:

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

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

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

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

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

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

Fixed it up so far... Not seeing where the $2|cos(\pi tT)|$ is coming from though... Will look at it more.

All I can think of is that it has something to do with there being a square root of the $cos{2\pi}$ , so when doing that you divide through by 2 and it goes.

hup · Oct 22, 2011

i am yr 12 and wat is dis

AAEldar · Oct 22, 2011

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

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

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

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

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

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

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

Because $cos(2\pi tT)=cos^2(\pi tT)-sin^2(\pi tT)$ and $1=cos^2(\pi tT)+sin^2(\pi tT)$

$=|a(t)|\sqrt{4cos^2(\pi tT)}$

$=2|a(t)cos(\pi tT)|$

That's still different though...

frenzal_dude · Oct 22, 2011

do you reckon it's an error in the textbook?

D94 · Oct 22, 2011

Are you sure there's no equal sign or an OR written between $\sqrt{2(1+cos(2\pi tT)}\ and\ 2|cos(\pi tT)|$ ?

After endless working out, I resulted in Wolfram Alpha, and it said the two above are equal to each other, so did you read the answer correctly?

If what I'm saying is right, then look at AAEldar's working because that's correct.

AAEldar · Oct 22, 2011

frenzal_dude said:
do you reckon it's an error in the textbook?

Hard to say. This is from a Uni text book correct? I could be missing something, but from what I've done I can't see anywhere else to go.

D94 raises a good point too.

frenzal_dude · Oct 22, 2011

D94 said:
Are you sure there's no equal sign or an OR written between $\sqrt{2(1+cos(2\pi tT)}\ and\ 2|cos(\pi tT)|$ ?

After endless working out, I resulted in Wolfram Alpha, and it said the two above are equal to each other, so did you read the answer correctly?

If what I'm saying is right, then look at AAEldar's working because that's correct.

It has to be an error in the textbook, another reason to think this is because they even forgot to put the end bracket for the 2(1+cos(.....

Thanks for your help, I went through your working out and it seems fine, they obviously forgot to put "=a(t)" before the "2|cos(PItT)|"

AAEldar · Oct 22, 2011

frenzal_dude said:
It has to be an error in the textbook, another reason to think this is because they even forgot to put the end bracket for the 2(1+cos(.....

Thanks for your help, I went through your working out and it seems fine, they obviously forgot to put "=a(t)" before the "2|cos(PItT)|"

Didn't even notice that! Haha. Sounds like you're right though.

zhiying · Oct 23, 2011

Woah saw this in the MX2 section and nearly had a heart attack, why do this to me night before exam T_T

Absolute Value Problem (1 Viewer)

frenzal_dude

UTS Student

AAEldar

Premium Member

hup

Member

AAEldar

Premium Member

frenzal_dude

UTS Student

D94

New Member

AAEldar

Premium Member

frenzal_dude

UTS Student

AAEldar

Premium Member

zhiying

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)