- #1

- 77

- 0

## 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.

[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.