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.

[tex]\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 $-\infty$?[/tex]

 

[frenzal_dude]

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

[tex]\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 $-\infty$?[/tex]

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

 

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

[tex] <br /> Yes, it'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's formula then tells you$$e^{zt}=e^{(x+jy)t}=e^{jyt}e^{xt}=\cos(yt)e^{xt}+j\sin(yt)e^{xt}$$<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 $x<0$ 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" cite="https://www.physicsforums.com/goto/post?id=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't look quite right, you'd better show your calculations.[/tex]

 

[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: $$\frac{\omega + j}{(\omega_c-w)^2+1}$$

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:

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

 

[frenzal_dude]

Thanks for the help.

Quick question in regards to $$\lim_{t\to\infty}e^{-zt}=0$$

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.