- #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.
[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.