How Does Substituting t=∞ Affect the Fourier Transform of g(t)?

[frenzal_dude]

Homework Statement

Hi, I need to find the Fourier Transform of:

g(t) = (e^-t)Sin(Wct)u(t)

where Wc=2πFc
and u(t) is the step function which is equal to 1 if time is +ve and 0 otherwise.

Homework Equations

I know that g(t) = (e^-t)[e^jWct - e^-jWct]/2j = [e^-t(1+jWc) - e^t(-1-jWc)]/(2j)
(0<=t<=∞, because of step function u(t))

The Attempt at a Solution

Therefore the Fourier Transform would be:
[1/(2j)]*∫([e^-t(1+jWc) - e^t(-1-jWc)])(e^-jWt)dt

= [1/(2j)]*∫([e^-t(1-jWc+jW) - e^t(-1-jWc-jW)])dt (limits: t=∞ to t=0)

= [1/(2j)][(e^-t(1-jWc+jW))/(-(1-jWc+jW)) - (e^t(-1-jWc-jW))/(-1-jWc-jW)] (sub in: t=∞ to t=0)

If you sub t=+/-∞, the exponential could be 0 or it could be infinite depending on whether 1-jWc+jW and -1-jWc-jW are -ve or +ve.
How can we know if they are positive or negative?

Hope you guys can help!
David.

 

[gabbagabbahey]

I'd assume that Wc (\omega_c?) is real-valued.

\lim_{t\to\infty}e^{-zt}=0[/itex]<br /> <br /> So long as the real part of z is positive.<br /> <br /> Also, if g(t) is zero for negative t, why is one of your integration limits -\infty?

 

[frenzal_dude]

I'd assume that Wc (\omega_c?) is real-valued.

\lim_{t\to\infty}e^{-zt}=0[/itex]<br /> <br /> So long as the real part of z is positive.<br /> <br /> Also, if g(t) is zero for negative t, why is one of your integration limits -\infty?

<br /> <br /> so is it also true to say: \lim_{t\to\infty}e^{zt}=0[/itex] so long as the real part of z is negative?&lt;br /&gt; &lt;br /&gt; If that&amp;#039;s true that helps me out A LOT! This is something not even my Signal Processing lecturer could explain to me, he told me &amp;#039;he&amp;#039;d get back to me&amp;#039; and he never did.&lt;br /&gt; &lt;br /&gt; Thankyou so much for your help!&lt;br /&gt; &lt;br /&gt; Btw I fixed up the limits, you&amp;#039;re right it&amp;#039;s from t=infinity to t=0

 

[frenzal_dude]

btw this is the answer I got:
\frac{\omega + j}{(\omega_c-w)^2+1}

 

[gabbagabbahey]

so is it also true to say: \lim_{t\to\infty}e^{zt}=0[/itex] so long as the real part of z is negative?

<br /> <br /> Yes, it&#039;s fairly easy to show simply by splitting z into real and imaginary parts. Say z=x+jy, where x and y are real-valued. Euler&#039;s formula then tells youe^{zt}=e^{(x+jy)t}=e^{jyt}e^{xt}=\cos(yt)e^{xt}+j\sin(yt)e^{xt}<br /> <br /> As t\to\infty, both \sin(yt) and \cos(yt) oscillate between 0 and 1 more and more rapidly, and hence are bounded. Meanwhile, e^{xt} goes rapidly to zero if x&amp;lt;0 and so its product with any bounded function will also approach zero.<br /> <br /> <blockquote data-attributes="" data-quote="frenzal_dude" data-source="post: 2694486" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> frenzal_dude said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> btw this is the answer I got:<br /> \frac{\omega + j}{(\omega_c-w)^2+1} </div> </div> </blockquote><br /> That doesn&#039;t look quite right, you&#039;d better show your calculations.

 

[frenzal_dude]

Here's my working out:

g(t)=e^{-t}sin(\omega _ct)u(t)=<br /> e^{-t}(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct})u(t)

 

[frenzal_dude]

\therefore G(f)<br /> =\frac{1}{2j}\int_{0}^{\infty }(e^{j\omega _ct-t}-e^{-j\omega _ct-t})e^{-j\omega t}dt=\frac{1}{2j}\int_{0}^{\infty }e^{j\omega_ct-t-j\omega t}-e^{-j\omega_ct-t-j\omega t}dt

 

[frenzal_dude]

\frac{1}{2j}\int_{0}^{\infty }e^{j\omega_ct-t-j\omega t}-e^{-j\omega_ct-t-j\omega t}dt=\frac{1}{2j}[\frac{e^{t(j(\omega_c-w)-1)}}{j\omega_c-1-j\omega}-\frac{e^{-t(j(\omega_c+w)+1)}}{-j\omega_c-1-j\omega}]t=\infty ,t=0

 

[frenzal_dude]

=-\frac{1}{2j}[\frac{1}{j(\omega_c-\omega)-1}+\frac{1}{j(\omega_c+\omega)+1}]=\frac{\omega_c+\omega+j}{2(\omega_c+\omega)^2+2}+\frac{\omega_c-\omega-j}{2(\omega_c-\omega)^2+2}

 

[gabbagabbahey]

Here's my working out:

g(t)=e^{-t}sin(\omega _ct)u(t)=<br /> e^{-t}(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct})

Ermm... you mean g(t)=e^{-t}\left(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct}\right)u(t), right? You can't get rid of the u(t) until you put it into the integral and use the fact that \int_{-\infty}^{\infty}f(t)u(t)dt=\int_{0}^{\infty}f(t)dt

=-\frac{1}{2j}[\frac{1}{j(\omega_c-\omega)-1}+\frac{1}{j(\omega_c+\omega)+1}]=\frac{\omega_c+\omega+j}{2(\omega_c+\omega)^2+2}+\frac{\omega_c-\omega-j}{2(\omega_c-\omega)^2+2}

This looks fine, but it doesn't simplify to what you wrote originally.

 

[frenzal_dude]

do you mean what I wrote originally by this: <br /> \frac{\omega + j}{(\omega_c-w)^2+1}<br />

I realized where I went wrong and tried it again.

With my final answer, which I hope is now correct, I tried getting a common denominator to make it into one fraction but the fraction ended up being even more complicated, so I thought I'd just leave the answer as how I gave it before

(btw here was the fraction I got when I simplified it: \frac{\omega_c(\omega_c^2+\omega^2+1)-2\omega^2-2\omega j}{(\omega_c^2+\omega^2+1)-4\omega^2})

 

[gabbagabbahey]

Your answer from post#9 is correct.

Personally, I wouldn't worry so much about making the denominator real and instead concentrate on putting it over a common denominator:

\begin{aligned}-\frac{1}{2j}\left[\frac{1}{j(\omega_c-\omega)-1}+\frac{1}{j(\omega_c+\omega)+1}\right] &amp;= -\frac{1}{2j}\left[\frac{j(\omega_c+\omega)+1+j(\omega_c-\omega)-1}{\left(j(\omega_c-\omega)-1\right)\left(j(\omega_c+\omega)+1\right)}\right] \\ &amp;= -\frac{1}{2j}\left[\frac{2j\omega_c}{j^2(\omega_c^2-\omega^2)-2j\omega-1}\right] \\ &amp;= \frac{\omega_c}{\omega_c^2-\omega^2+2j\omega+1} \end{aligned}

 

[frenzal_dude]

Thanks for the help.

Quick question in regards to <br /> \lim_{t\to\infty}e^{-zt}=0<br />

What if z is only imaginary? ie. the real part is 0.

How would you sub t=∞ into this: e^{j(\omega_1 t+\Theta_1)-jwt}

 

[gabbagabbahey]

If the real part is zero, then the limit won't exist. You can use Euler's formula to show this.