Absolute Value Problem

  • #1

Main Question or Discussion Point

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

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

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

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:

[tex]
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)
[/tex]
[tex]
\therefore |b(t)|=|a(t)|+|a(t)cos(2\pi tT)|+|a(t)sin(2\pi tT)|
[/tex]

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

Answers and Replies

  • #2
22,097
3,282
I need to find the absolute value of b(t):

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

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

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:

[tex]
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)
[/tex]
[tex]
\therefore |b(t)|=|a(t)|+|a(t)cos(2\pi tT)|+|a(t)sin(2\pi tT)|
[/tex]

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:

[tex]|a+bi|=\sqrt{a^2+b^2}[/tex]
 
  • #3
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:

[tex]b(t)=a(t)+a(t)cos(2\pi tT)-a(t)jsin(2\pi tT)[/tex]

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

[tex]=\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)}[/tex]

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

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

[tex]=|a(t)|\sqrt{2[1+\sqrt{a(t)}cos(2\pi tT)]}[/tex]
 
  • #4
22,097
3,282
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:

[tex]b(t)=a(t)+a(t)cos(2\pi tT)-a(t)jsin(2\pi tT)[/tex]

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

[tex]=\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)}[/tex]
That second term should have an [itex]a(t)^2[/itex].
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
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Also [itex]cos^2(2\pi tT)+ sin^2(2\pi tT)= 1[/itex] so what you have reduces to
[tex]a(T)\sqrt{2(1+ cos(2\pi tT))}[/tex]

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.
 
  • #6
Also [itex]cos^2(2\pi tT)+ sin^2(2\pi tT)= 1[/itex] so what you have reduces to
[tex]a(T)\sqrt{2(1+ cos(2\pi tT))}[/tex]

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).
 

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