How to Fourier-transform e^(-2|t|)?

[TheSodesa]

Homework Statement

Determine the Fourier-transfroms of the functions

\begin{equation*}
a) f : f(t) = H(t+3) - H(t-3) \text{ and } g : g(t) = \cos(5t) f(t)
\end{equation*}
and
\begin{equation*}
b) f : f(t) = e^{-2|t|} \text{ and } g : g(t) = \cos(3t) f(t)
\end{equation*}

Homework Equations

The Fourier transformation definition:
\begin{equation}
\digamma\{ f(t) \}(\omega) = F(\omega) = \int_{-\infty}^{\infty} f(u)e^{-j\omega u}\,du
\end{equation}
What to do with cosines:
\begin{equation}

\digamma\{f(t)\cos(\alpha t)\}(\omega) = \frac{1}{2} F(\omega - \alpha) + \frac{1}{2}F(\omega + \alpha)

\end{equation}

The Attempt at a Solution

I managed part a) somewhat:
\begin{align*}
\digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} \left( H(t+3) - H(t-3) \right) e^{-j\omega u} \,du \\
%
&= \int_{-3}^{3} e^{-j\omega u} \,du = \left[ \frac{e^{-j\omega u}}{-j\omega}\right]_{-3}^{3}
= \left( \frac{e^{-j3\omega }}{-j\omega} - \frac{e^{j3\omega}}{-j\omega} \right)\\
%
&= \left( \frac{ e^{j3\omega}}{j\omega} - \frac{e^{-j3\omega }}{j\omega} \right) = \frac{1}{2j\omega}(e^{j3\omega } - e^{-j3\omega })\\
%
&= \frac{\sin(3\omega)}{\omega} = 3 sinc(3\omega)
\end{align*}
Sinc is continuous at $\omega = 0$, so we don't need to handle that case separately.
Then, using equation (2) I wrote
\begin{align*}
\digamma\{ f(t) \cos(5t) \}(\omega) &= \frac{3}{2} sinc(3(\omega-5)) + \frac{3}{2} sinc(3(\omega+5))
\end{align*}

Part b) has me completely stumped. Since
\begin{equation}
|u|=
\begin{cases}
u &, u\geq 0\\
-u &, u < 0,
\end{cases}
\end{equation}
we write, according to equation (1):
\begin{align*}
\digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} e^{-2|u|} e^{-j\omega u} \,du = \int_{-\infty}^{\infty} e^{-2|u|-j\omega u} e^{} \,du\\
%
&= \int_{-\infty}^{0} e^{2u-j\omega u} \,du + \int_{0}^{\infty} e^{-2u-j\omega u} \,du\\
%
&= \lim_{a \to \infty} \int_{-a}^{0} e^{(2-j\omega) u} \,du + \int_{0}^{a} e^{-(2+j\omega) u} \,du\\

%

&= \lim_{a \to \infty} \sij{-a}{0} \frac{e^{(2-j\omega) u}}{(2-j\omega)} + \sij{0}{a} \frac{e^{-(2+j\omega)u}}{-(2+j\omega)}\\

%

&= \lim_{a \to \infty} \left( \frac{e^0}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)

+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} - \frac{e^0}{-(2+j\omega)} \right)\\

%

&= \lim_{a \to \infty} \left( \frac{1}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)

+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} + \frac{1}{(2+j\omega)} \right) \\

%

&= \lim_{a \to \infty} \frac{1}{(2-j\omega)} + \frac{1}{(2+j\omega)} - \frac{e^{-a(2-j\omega) }}{(2+j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)}\\

%

&= \lim_{a \to \infty} \frac{(2+j\omega)}{(2+j\omega)(2-j\omega)} + \frac{(2-j\omega)}{(2+j\omega)(2-j\omega)}\\

&- \frac{(2-j\omega)e^{-a(2-j\omega) }}{(2+j\omega)(2-j\omega)} - \frac{(2+j\omega)e^{-a(2-j\omega)}}{(2+j\omega)(2-j\omega)}\\
\end{align*}

This isn't looking good. Is there a trick that I'm missing? Help would be appreciated.

 

[Ray Vickson]

Homework Statement

Determine the Fourier-transfroms of the functions

\begin{equation*}
a) f : f(t) = H(t+3) - H(t-3) \text{ and } g : g(t) = \cos(5t) f(t)
\end{equation*}
and
\begin{equation*}
b) f : f(t) = e^{-2|t|} \text{ and } g : g(t) = \cos(3t) f(t)
\end{equation*}

Homework Equations

The Fourier transformation definition:
\begin{equation}
\digamma\{ f(t) \}(\omega) = F(\omega) = \int_{-\infty}^{\infty} f(u)e^{-j\omega u}\,du
\end{equation}

The Attempt at a Solution

Part b) has me completely stumped.

If you write $\cos(3t)$ in terms of $e^{\pm 3 j t}$ you have
$$\int_{-\infty}^0 e^{-j w t} e^{\pm 3 j t} e^{2t} \, dt = \frac{1}{2 - j w \pm 3 j} \left. e^{-j w t} e^{\pm 3 j t} e^{2t}\right|_{-\infty}^0,$$
which is convergent because of the exponential damping factor $e^{2t}$ when $t \to -\infty$. Similarly for the other part.

 

[haruspex]

In the $\left[\frac{etc.}{etc.}\right]_0^a$ term you omitted the u from the exponent, but that's just a typo. What I do not get is what happened to that term in the next line. How did (2+jω) turn into (2-jω)? Maybe just another typo, rectified later, but this is making it hard to follow.

 

[TheSodesa]

In the $\left[\frac{etc.}{etc.}\right]_0^a$ term you omitted the unfrom the exponent, but that's just a typo. What I do not get is what happened to that term in the next line. How did (2+jω) turn into (2-jω)? Maybe just another typo, rectified later, but this is making it hard to follow.

Yhea, I just noticed. Fixing.

 

[TheSodesa]

In the $\left[\frac{etc.}{etc.}\right]_0^a$ term you omitted the u from the exponent, but that's just a typo. What I do not get is what happened to that term in the next line. How did (2+jω) turn into (2-jω)? Maybe just another typo, rectified later, but this is making it hard to follow.

Ok, so right now I have
\begin{align*}
\digamma\{f(t)\}(\omega) = \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) }}{(2+j\omega)(2-j\omega)}
+ \frac{(2-j\omega)e^{-a(2-j\omega)}}{(2+j\omega)(2-j\omega)}
\end{align*}
Not compeltely sure what to do here. I'm trying to edit my offline LaTeX document at the same time, so mistakes are bound to be made.

 

[haruspex]

Ok, so right now I have
\begin{align*}
\digamma\{f(t)\}(\omega) = \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) }}{(2+j\omega)(2-j\omega)}
+ \frac{(2-j\omega)e^{-a(2-j\omega)}}{(2+j\omega)(2-j\omega)}
\end{align*}
Not compeltely sure what to do here. I'm trying to edit my offline LaTeX document at the same time, so mistakes are bound to be made.

You still have 2-jω in both exponents.

 

[TheSodesa]

You still have 2-jω in both exponents.

\begin{align*}
\digamma\{f(t)\}(\omega) &= \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) }}{(2+j\omega)(2-j\omega)}
+ \frac{(2-j\omega)e^{-a(2+j\omega)}}{(2+j\omega)(2-j\omega)}\\
&= \lim_{a \to \infty} \frac{(2+j\omega)e^{a(2-j\omega) } + (2-j\omega)e^{-a(2+j\omega)}}{(2+j\omega)(2-j\omega)}
\end{align*}
There. I think it might have been a bad idea to try and learn LaTeX proper by doing all of my coursework with it. Copy-paste can be treacherous.

 

[TheSodesa]

You still have 2-jω in both exponents.

It looks like I have a sum of complex conjugates in the numerator.

 

[TheSodesa]

If you write $\cos(3t)$ in terms of $e^{\pm 3 j t}$ you have
$$\int_{-\infty}^0 e^{-j w t} e^{\pm 3 j t} e^{2t} \, dt = \frac{1}{2 - j w \pm 3 j} \left. e^{-j w t} e^{\pm 3 j t} e^{2t}\right|_{-\infty}^0,$$
which is convergent because of the exponential damping factor $e^{2t}$ when $t \to -\infty$. Similarly for the other part.

It's not the cosine, that scares me. What I'm ahving trouble with is $f(t)$. I also made a few mistakes in the opening post because I'm copyin stuff from my offline LaTeX-document. They should be fixed now.

 

[haruspex]

It looks like I have a sum of complex conjugates in the numerator.

No, there's still a problem.
Look at the last two lines of your original post. In the left hand integral, you had -a in the exponent, but it becomes +a.
As Ray says, it should converge.
I would not have introduced the lim operators, though I know that is technically correct. I would have just integrated still with infinities in the bounds, and used the fact that e^x+iy tends to zero as x tends to minus infinity regardless of what y does.

 

[TheSodesa]

No, there's still a problem.
Look at the last two lines of your original post. In the left hand integral, you had -a in the exponent, but it becomes +a.
As Ray says, it should converge.
I would not have introduced the lim operators, though I know that is technically correct. I would have just integrated still with infinities in the bounds, and used the fact that e^x+iy tends to zero as x tends to minus infinity regardless of what y does.

Alright, let me go over it once more. This might take a moment.

 

[TheSodesa]

No, there's still a problem.
Look at the last two lines of your original post. In the left hand integral, you had -a in the exponent, but it becomes +a.
As Ray says, it should converge.
I would not have introduced the lim operators, though I know that is technically correct. I would have just integrated still with infinities in the bounds, and used the fact that e^x+iy tends to zero as x tends to minus infinity regardless of what y does.

I got it. Just a sec, and I'll clean up my derivation.

 

[Ray Vickson]

It's not the cosine, that scares me. What I'm ahving trouble with is $f(t)$. I also made a few mistakes in the opening post because I'm copyin stuff from my offline LaTeX-document. They should be fixed now.

Your $f(t) = e^{-2|t|}$ is no problem; that is why you used $e^{2t}$ for $t < 0$ and $e^{-2t}$ for $t >0$. You get two convergent integrals. It is even faster and easier to use standard expressions for $\int e^{at} \cos(bt) \, dt$, then use $b = 3$ and
$$ a = \begin{cases} -j\omega + 2, & t < 0\\
-j \omega - 2,& t > 0
\end{cases}
$$

 

[TheSodesa]

Alright, here's the complete derivation this time (hopefully) without errors:

\begin{align*}
\digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} e^{-2|u|} e^{-j\omega u} \,du = \int_{-\infty}^{\infty} e^{-2|u|-j\omega u} \,du\\
&= \int_{-\infty}^{0} e^{2u-j\omega u} \,du + \int_{0}^{\infty} e^{-2u-j\omega u} \,du\\
&= \lim_{a \to \infty} \int_{-a}^{0} e^{(2-j\omega) u} \,du + \int_{0}^{a} e^{-(2+j\omega) u} \,du\\
&= \lim_{a \to \infty} \left[ \frac{e^{(2-j\omega) u}}{(2-j\omega)}\right]_{-a}^{0} + \left[ \frac{e^{-(2+j\omega)u}}{-(2+j\omega)}\right]_{0}^{a}\\
&= \lim_{a \to \infty} \left( \frac{e^0}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)
+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} - \frac{e^0}{-(2+j\omega)} \right)\\
&= \lim_{a \to \infty} \left( \frac{1}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)
+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} + \frac{1}{(2+j\omega)} \right)\\
&= \lim_{a \to \infty} \frac{1}{(2-j\omega)} + \frac{1}{(2+j\omega)} - \frac{e^{-a(2-j\omega) }}{(2+j\omega)} - \frac{e^{-a(2-j\omega)}{(2-j\omega)}\\
&= \lim_{a \to \infty} \frac{(2+j\omega)}{(2+j\omega)(2-j\omega)} + \frac{(2-j\omega)}{(2+j\omega)(2-j\omega)}\\
&- \frac{(2-j\omega) \overbrace{e^{-a(2-j\omega) }}^{\text{Real coefficient goes to zero...}}}{(2+j\omega)(2-j\omega)}
- \frac{(2+j\omega)\overbrace{e^{-a(2-j\omega) }^{\text{... specifically exp(-2a)}}}{(2+j\omega)(2-j\omega)}\\
&= \frac{(2-j\omega) + (2+j\omega)}{(2+j\omega)(2-j\omega)} = \frac{4}{4 + \omega^2}
\end{align*}

Then, if we apply equation $(2)$ to the case where we multiply $f(t)$ by $\cos(3t)$:
\begin{align*}
\digamma\{f(t)\cos(3t)\}(\omega) &= \frac{1}{2} \frac{4}{4 + \left( \omega -3 \right)^2} + \frac{1}{2} \frac{4}{4 + \left( \omega + 3 \right)^2}\\
&= \frac{2}{4 + \left( \omega -3 \right)^2} + \frac{2}{4 + \left( \omega + 3 \right)^2}\\
\end{align*}

 

[Ray Vickson]

Alright, here's the complete derivation this time (hopefully) without errors:

\begin{align*}
\digamma\{ f(t) \}(\omega) &= \int_{-\infty}^{\infty} e^{-2|u|} e^{-j\omega u} \,du = \int_{-\infty}^{\infty} e^{-2|u|-j\omega u} \,du\\

&= \int_{-\infty}^{0} e^{2u-j\omega u} \,du + \int_{0}^{\infty} e^{-2u-j\omega u} \,du\\
&= \lim_{a \to \infty} \int_{-a}^{0} e^{(2-j\omega) u} \,du + \int_{0}^{a} e^{-(2+j\omega) u} \,du\\
&= \lim_{a \to \infty} \left[ \frac{e^{(2-j\omega) u}}{(2-j\omega)}\right]_{-a}^{0} + \left[ \frac{e^{-(2+j\omega)u}}{-(2+j\omega)}\right]_{0}^{a}\\
&= \lim_{a \to \infty} \left( \frac{e^0}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)
+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} - \frac{e^0}{-(2+j\omega)} \right)\\
&= \lim_{a \to \infty} \left( \frac{1}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)
+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} + \frac{1}{(2+j\omega)} \right)\\
&= \lim_{a \to \infty} \frac{1}{(2-j\omega)} + \frac{1}{(2+j\omega)} - \frac{e^{-a(2-j\omega) }}{(2+j\omega)} - \frac{e^{-a(2-j\omega)}{(2-j\omega)}\\
&= \lim_{a \to \infty} \frac{(2+j\omega)}{(2+j\omega)(2-j\omega)} + \frac{(2-j\omega)}{(2+j\omega)(2-j\omega)}\\
&- \frac{(2-j\omega)\overbrace{e^{-a(2-j\omega) }}^{\text{Real coefficient goes to zero...}}}{(2+j\omega)(2-j\omega)}
- \frac{(2+j\omega)\overbrace{e^{-a(2-j\omega) }}^{\text{...specifically exp(-2a)}}}{(2+j\omega)(2-j\omega)}\\
&= \frac{(2-j\omega) + (2+j\omega)}{(2+j\omega)(2-j\omega)} = \frac{4}{4 + \omega^2}
\end{align*}

Then, if we apply equation $(2)$ to the case where we multiply $f(t)$ by $\cos(3t)$:
\begin{align*}
\digamma\{f(t)\cos(3t)\}(\omega) &= \frac{1}{2} \frac{4}{4 + \left( \omega -3 \right)^2} + \frac{1}{2} \frac{4}{4 + \left( \omega + 3 \right)^2}\\
&= \frac{2}{4 + \left( \omega -3 \right)^2} + \frac{2}{4 + \left( \omega + 3 \right)^2}\\
\end{align*}

Answer is correct; derivation did not render properly.

 

[TheSodesa]

There seems to be a math processing error preventing the rendering of the biggest part of this exercise. Looking at the source code, I can't spot the error on my part.

Regradless, if it doesn't decide to resolve itself, the correct answer was
\begin{equation}
\digamma\{ f(t) \} = \frac{4}{4 + \omega^2}
\end{equation}

Thanks for the help, everyone.

 

[TheSodesa]

Answer is correct; derivation did not render properly.

Yeah, I'm wondering why that is. I tried to switch any dollar signs to double pound signs and such, but its still not rendering.

 

[haruspex]

Doesn't seem to like the text imbeds

Alright, here's the complete derivation this time (hopefully) without errors:
$\digamma\{ f(t) \}(\omega) =$ $\int_{-\infty}^{\infty} e^{-2|u|} e^{-j\omega u} \,du =$ $\int_{-\infty}^{\infty} e^{-2|u|-j\omega u} \,du$ $= \int_{-\infty}^{0} e^{2u-j\omega u} \,du + \int_{0}^{\infty} e^{-2u-j\omega u} \,du\\$ $= \lim_{a \to \infty} \int_{-a}^{0} e^{(2-j\omega) u} \,du + \int_{0}^{a} e^{-(2+j\omega) u} \,du\\$ $= \lim_{a \to \infty} \left[ \frac{e^{(2-j\omega) u}}{(2-j\omega)}\right]_{-a}^{0} + \left[ \frac{e^{-(2+j\omega)u}}{-(2+j\omega)}\right]_{0}^{a}\\$ $= \lim_{a \to \infty} \left( \frac{e^0}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right)+ \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} - \frac{e^0}{-(2+j\omega)} \right)\\$ $= \lim_{a \to \infty} \left( \frac{1}{(2-j\omega)} - \frac{e^{-a(2-j\omega)}}{(2-j\omega)} \right) + \left( \frac{e^{-a(2-j\omega) }}{-(2+j\omega)} + \frac{1}{(2+j\omega)} \right)\\$ $= \lim_{a \to \infty} \frac{1}{(2-j\omega)} + \frac{1}{(2+j\omega)} - \frac{e^{-a(2-j\omega) }}{(2+j\omega)} - \frac{e^{-a(2-j\omega)}{(2-j\omega)}\\$ $= \lim_{a \to \infty} \frac{(2+j\omega)}{(2+j\omega)(2-j\omega)} + \frac{(2-j\omega)}{(2+j\omega)(2-j\omega)} - \frac{(2-j\omega) \overbrace{e^{-a(2-j\omega) }}^{\text{Real coefficient goes to zero...}}}{(2+j\omega)(2-j\omega)} - \frac{(2+j\omega)\overbrace{e^{-a(2-j\omega) }^{\text{... specifically exp(-2a)}}}{(2+j\omega)(2-j\omega)}\\$ $= \frac{(2-j\omega) + (2+j\omega)}{(2+j\omega)(2-j\omega)} = \frac{4}{4 + \omega^2}$

Then, if we apply equation $(2)$ to the case where we multiply $f(t)$ by $\cos(3t)$:
\begin{align*}
\digamma\{f(t)\cos(3t)\}(\omega) &= \frac{1}{2} \frac{4}{4 + \left( \omega -3 \right)^2} + \frac{1}{2} \frac{4}{4 + \left( \omega + 3 \right)^2}\\
&= \frac{2}{4 + \left( \omega -3 \right)^2} + \frac{2}{4 + \left( \omega + 3 \right)^2}\\
\end{align*}