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Homework Help: Subbing t=∞ into Integral

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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))

    3. 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!
    Last edited: Apr 28, 2010
  2. jcsd
  3. Apr 28, 2010 #2


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


    So long as the real part of [itex]z[/itex] is positive.

    Also, if [itex]g(t)[/itex] is zero for negative [itex]t[/itex], why is one of your integration limits [itex]-\infty[/itex]?
  4. Apr 28, 2010 #3
    so is it also true to say: [tex]\lim_{t\to\infty}e^{zt}=0[/itex] so long as the real part of [itex]z[/itex] is negative?

    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.

    Thankyou so much for your help!

    Btw I fixed up the limits, you're right it's from t=infinity to t=0
  5. Apr 28, 2010 #4
    btw this is the answer I got:
    [tex]\frac{\omega + j}{(\omega_c-w)^2+1}[/tex]
  6. Apr 29, 2010 #5


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    Yes, it's fairly easy to show simply by splitting [itex]z[/itex] into real and imaginary parts. Say [itex]z=x+jy[/itex], where [itex]x[/itex] and [itex]y[/itex] are real-valued. Euler's formula then tells you


    As [itex]t\to\infty[/itex], both [itex]\sin(yt)[/itex] and [itex]\cos(yt)[/itex] oscillate between [itex]0[/itex] and [itex]1[/itex] more and more rapidly, and hence are bounded. Meanwhile, [itex]e^{xt}[/itex] goes rapidly to zero if [itex]x<0[/itex] and so its product with any bounded function will also approach zero.

    That doesn't look quite right, you'd better show your calculations.
    Last edited: Apr 29, 2010
  7. Apr 29, 2010 #6
    Here's my working out:

    [tex]g(t)=e^{-t}sin(\omega _ct)u(t)=
    e^{-t}(\frac{1}{2j}e^{j\omega _ct}-\frac{1}{2j}e^{-j\omega _ct})u(t)[/tex]
    Last edited: Apr 29, 2010
  8. Apr 29, 2010 #7
    [tex]\therefore G(f)
    =\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[/tex]
  9. Apr 29, 2010 #8
    [tex]\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[/tex]
  10. Apr 29, 2010 #9
  11. Apr 29, 2010 #10


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

    This looks fine, but it doesn't simplify to what you wrote originally.
  12. Apr 29, 2010 #11
    do you mean what I wrote originally by this: [tex]
    \frac{\omega + j}{(\omega_c-w)^2+1}

    I realised 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: [tex]\frac{\omega_c(\omega_c^2+\omega^2+1)-2\omega^2-2\omega j}{(\omega_c^2+\omega^2+1)-4\omega^2}[/tex])
  13. Apr 30, 2010 #12


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    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:

    [tex]\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} [/tex]
  14. May 2, 2010 #13
    Thanks for the help.

    Quick question in regards to [tex]

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

    How would you sub t=∞ into this: [tex]e^{j(\omega_1 t+\Theta_1)-jwt}[/tex]
  15. May 3, 2010 #14


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