Expert: Alon Mandes - 3/1/2009

Question
QUESTION: hello,
for the fourier transform attatched I need to determine expressions for the amplitude and phase of g(w). Could Thanks keith

ANSWER: G(w)=[20/√(2π)]*[1/(401-w²+2jw)] is a complex variable. To find
absolute value & phase, we perform the following steps :

1. |G(w)|=[20/√(2π)]*[1/|401-w²+2jw|]. Let's find |401-w²+2jw| :
|401-w²+2jw|=√ [ (401-w²)²+(2w)² ] =
√ [ 160801-802w²+w^4+4w² ] = √ [160801-798w²+w^4].
So, |G(w)|=[20/√(2π)]/√[160801-798w²+w^4].

2. Ө{G(w)}=Ө {[20/√(2π)]*[1/(401-w²+2jw)]}=Ө{ 1/(401-w²+2jw) }.
In order to calculate a phaze of this form , we need to convert it
to normal shape like a+jb. To do so we multiply the fraction by the
lower part complex conjugate. Meaning :
1/(401-w²+2jw)=[401-w²-2jw]/[(401-w²+2jw)*(401-w²-2jw)]=
[401-w²-2jw]/[160801-798w²+w^4]. Now we know that
Ө{a+jb}=Arctang[b/a], & Ө{(a+jb)/c}=Arctang[b/a].
Thus  Ө { [401-w²-2jw]/[160801-798w²+w^4] } = Ө { [401-w²-2jw] } =
Arctang[-2w/401-w²].
Note that both |G(w)| & Ө{G(w)} are both real function of w.

Alon.

---------- FOLLOW-UP ----------

QUESTION: Hello Alon,

thanks for your fast response. The fourier transform in the previous question was derived from the signal

g(t) = {(e^-t) sin 20t   for t>0
      {0          for t<0

could you show me how you get from the signal to the fourier transform equation.
Thanks keith

Answer
Sure, first of all let's review some of relevant facts :
1. G(w)=(1/√2π)*∫f(t)e^(-iwt) dt (t goes from 0 to ∞).
  This is the definition of Fourier Transform.

2. ∫e^(ax)sin(bx) dx = [e^(ax)/(a²+b²)]*[asin(bx)-bcos(bx)]
  This can be easy looked up in integral tables, or proved by
  Integration by parts.

Now, let's get to work :

(1/√2π)*∫f(t)e^(-iwt) dt = (1/√2π)*∫[(e^-t)sin(20t)]e^(-iwt) dt =
(1/√2π)*∫[e^(-t-iwt)]sin(20t) dt=(1/√2π)*∫e^[-t(1+iwt)]sin(20t) dt =
According to (2) :
=(1/√2π)*{e^[-t(1+iwt)]/[(1+iw)²+400]}*{(-1-iw)sin(20t)-20cos(20t)}.
& the limits go from ∞ to 0. So that leaves us :
(1/√2π)*{1/[(1+iw)²+400]}*{(-1)*[-20]}=
(1/√2π)*{1/[2wi-w²+401]}*20.

Alon.