What is the Absolute Value of b(t)?

In summary: Anyway, thanks a lot for your help and pointing out my mistake! In summary, the conversation was about finding the absolute value of b(t) using different methods. The textbook provided a solution using the definition of modulus, while the person asking the question had used a different approach but was unsure if it was correct. After some discussion and corrections, the final answer was found to be |a(t)|\sqrt{2(1+ cos(2\pi tT))}.
  • #1
frenzal_dude
77
0
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.
 
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  • #2
frenzal_dude said:
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
frenzal_dude said:
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
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
HallsofIvy said:
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).
 

1. What is an absolute value problem?

An absolute value problem is a mathematical equation or inequality that involves the absolute value of a number, which is the distance of that number from zero on a number line. It is denoted by two vertical bars surrounding the number, and it always results in a positive value.

2. How do you solve absolute value problems?

To solve an absolute value problem, you need to follow a few steps. First, isolate the absolute value expression by moving all other terms to the other side of the equation or inequality. Then, set up two separate equations, one with a positive value and one with a negative value. Solve both equations to find two possible solutions. Finally, check your solutions by plugging them back into the original equation or inequality to see which one(s) satisfy the given conditions.

3. What are the common mistakes when solving absolute value problems?

The most common mistake when solving absolute value problems is forgetting to consider both the positive and negative solutions. Another mistake is not properly isolating the absolute value expression, which can lead to incorrect solutions. Additionally, mixing up the rules for solving equations and inequalities can also result in mistakes when solving absolute value problems.

4. Can absolute value problems have no solution?

Yes, it is possible for absolute value problems to have no solution. This can occur when the absolute value expression is set equal to a negative number, as there is no real number that can result in a negative absolute value. In other words, there is no number that is equidistant from both zero and a negative number on a number line.

5. How are absolute value problems used in real life?

Absolute value problems have many real-life applications, including in physics, engineering, and economics. For example, the absolute value function is used to calculate distances and magnitudes, as well as to model various phenomena such as temperature changes and stock market fluctuations. Additionally, absolute value equations and inequalities are used to solve real-world problems involving ranges, absolute errors, and absolute deviations.

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