1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution and Transfer Function

Tags:
  1. Mar 12, 2016 #1
    1. The problem statement, all variables and given/known data

    The impulse response ##h(t)## of a linear time invariant system is real valued. Where * denotes the complex conjugate, the transfer function satisfies:

    $$H(- \nu) = H^* (- \nu)$$

    Use this result to show that for such a system, given the input ##f(t)= \sin(2 \pi \nu t)##, the output will be:

    $$g(t)=(f*h)(t) = |H(\nu)| \sin (2\pi \nu t +arg(H(\nu)))$$

    2. Relevant equations

    Convolution integral is given by: ##g(t)= \int^\infty_{- \infty} f (\tau) h(t-\tau) d \tau##

    For input ##f(t)=\exp (j2 \pi \nu t)##:

    ##g(t)=(f*h)(t)= (\int h(\tau) \exp (- j 2 \pi \nu \tau) d \tau) \ exp (j 2 \pi \nu t) = H (\nu) \exp (j 2 \pi \nu t)##

    3. The attempt at a solution

    So, starting from the definition of a convolution:

    $$\int^{\infty}_{-\infty} \sin (2\pi \nu \tau) . h(t- \tau) d \tau \ \ (1)$$

    We must show that this is equal to:

    $$g(t)= |H(\nu)| \sin (2\pi \nu t +arg(H(\nu)))$$

    I know that ##H(\nu)= \int^{\infty}_{-\infty} h(t) e^{-2 \pi j \nu t} dt##, so the expression above is equal to:

    $$g(t)= (\int^{\infty}_{-\infty} h(t-\tau) e^{-2 \pi j \nu \tau} d \tau) \sin (2\pi \nu t +arg(H(\nu)))$$

    So, how can I manipulate (1) further to get to this result? :confused:

    Any explanation is greatly appreciated.
     
  2. jcsd
  3. Mar 12, 2016 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Are you sure the minus signs should appear in the argument in both sides of the equation? If yes, that tells you that ##H(\nu)## is real, but then the placement of the minus signs is superfluous.
    Further consequence is that ##\textrm{arg }H(\nu)=n\pi## with ##n## integer numbers.
     
  4. Mar 12, 2016 #3
    Oops, that was a typo, sorry. I meant to say:

    $$H(- \nu) = H^* (\nu)$$

    How do we use this to get to the given expression for ##g(t)##?
     
  5. Mar 12, 2016 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Start by expressing ##\sin(2\pi \nu t)## in terms of complex exponentials and ##h(t-\tau)## in terms of its Fourier transform, then plug in to equation (1).
     
  6. Mar 12, 2016 #5
    What do you mean by expressing ##h(t-\tau)## in terms of its Fourier transform? In terms of complex exponentials ##\sin(2\pi \nu t) = \frac{e^{j 2\pi \nu t} - e^{-j 2\pi \nu t}}{2 j}##. So plugging this into Eq (1):

    $$g(t) = \int^{\infty}_{-\infty} \frac{e^{j 2\pi \nu \tau} - e^{-j 2\pi \nu \tau}}{2 j} . h(t-\tau) \ d \tau$$

    $$= \frac{-j}{2} \left( \underbrace{\int^{\infty}_{-\infty} h (t-\tau) e^{+ j 2 \pi \nu \tau} \ d \tau}_{\text{inverse transform}} - \underbrace{\int^{\infty}_{-\infty} h (t-\tau) e^{-j 2 \pi \nu \tau} \ d \tau}_{\text{Fourier transform of h}}\right)$$

    We can rewrite this in terms of the transfer function as:

    $$= \frac{-j}{2} \left( \int^{\infty}_{-\infty} h (t-\tau) e^{+ j 2 \pi \nu \tau} \ d \tau - H(\nu) \right)$$

    Is that right? I am stuck here, I don't know how to progress from here. :confused:
     
  7. Mar 12, 2016 #6

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    No, that's not exactly the Fourier transform of ##h(t)##, instead they are the FT of the shifted ##h(t)##, and obviously this is different from ##H(\nu)##.
    I mean the inverse of ##H(\nu)= \int^{\infty}_{-\infty} h(t) e^{-2 \pi j \nu t} dt##.
     
  8. Mar 12, 2016 #7
    According to the shifting property of the Fourier transform: ##f(t-t_0) \iff exp(-j2 \pi \nu t_0) F(\nu)##, the FT of h(t-τ) becomes:

    $$H(t- \tau) = \exp(-j 2 \pi \nu \tau) \left( \int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t \right)$$

    Substituting into (1) we find the equation:

    $$g(t) = \int^{\infty}_{- \infty} \frac{e^{j 2 \pi \nu t} - e^{-j 2 \pi \nu t}}{2 j} . \left( \exp(-j 2 \pi \nu \tau) \left( \int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t \right) \right) dt$$

    The last term, ##\int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t##, is equal to ##H(t)##. But what can I do with the other terms?

    How can we get to ##\sin(2 \pi \nu t + arg(H(\nu)))##? :confused:

    The inverse of ##H(\nu)## is the complex conjugate given by:

    $$H^* (\nu)= H(- \nu) = \int^{\infty}_{- \infty} h(-t) e^{-2 \pi j \nu t} dt$$

    I'm not sure how we can introduce this into the equation for convolution.
     
  9. Mar 12, 2016 #8

    Charles Link

    User Avatar
    Homework Helper

    Perhaps I can give a helpful input here. In F.T. theory, when ## g(t)=\int h(t-t')f(t')dt' ## in F.T. space the result is ## g(\omega)=h(\omega)f(\omega) ## Your notation is slightly different=you're using ## H(\nu) ## instead of ## h(\omega) ## but I think you should be able to follow this part. When ## f(t) ## is a sinusoid, ## f(\omega) ## becomes a delta function expression. (Comes from ## \delta(\omega)=(1/2\pi)\int exp(i \omega t) d t ## ) Once you get ## g(\omega) ## you need to take the inverse transform to get the expression for g(t)
     
  10. Mar 13, 2016 #9

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    No, you have got it wrong there, You were on the right track when introducing the use of shifting theorem, but you used it in a wrong way.
    Start from the inverse transform
    $$
    h(t)= \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu t} d\nu
    $$
    Since you what you want is ##h(t-\tau)##, you have got to replace ##t## in the above equation with ##t-\tau##. Having done this, plug in to equation (1).
    No, it's not equal to ##H(t)##, it's equal to ##H(\nu)##.

    EDIT: There is a mistake in the inverse integral transform above - the integral element ##dt## has been replaced by ##d\nu##, which is the correct one.
     
    Last edited: Mar 13, 2016
  11. Mar 13, 2016 #10
    Thank you very much for the hint. But I'm not sure what is wrong here because when I plug this into (1):

    $$g(t)= \frac{-j}{2} \int^{\infty}_{-\infty} \left( e^{j2 \pi \nu t} - e^{-j2 \pi \nu t} \right) \left( \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu (t - \tau)} d(t-\tau) \right) d \tau,$$

    The expression for the inverse transform goes to infinity when I integrate over ##(-\infty, \infty)##, because we will have something like ##\frac{H(\nu)}{e^{2 \pi j \nu}} [e^{\infty} - e^{-\infty}]##.

    And also where can we use the H(-ν)=H*(ν) property?
     
  12. Mar 13, 2016 #11

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    I edited my post #9, there is a small mistake there.
    There is still a mistake originating from your side in the above equation, remember that you should be working with ##\sin(2\pi \nu \tau)##, not ##\sin(2\pi \nu t)##. (I have purposely corrected another mistake which is due to me, namely the integral element from ##d(t-\tau)## to ##d\nu## for you).

    The next step is to reverse the integration order, where you will integrate over ##d\tau## first. In order to do this, collect all terms which contains ##\tau## in one.
    That comes into play later.
     
  13. Mar 13, 2016 #12

    Charles Link

    User Avatar
    Homework Helper

    The final result of the calculation is a very useful result in linear response theory, that the response g(t) of any linear system to a sinusoidal input f(t) is a sinusoidal g(t) at the same frequency that has an amplitude factor and phase change that can be found from the F.T. of the response function h(t). This result has many applications including AC circuit theory where the output voltage of a sinusoidal input voltage is a sinusoidal output voltage with the various impedance factors making up the ## H(\nu) ##. A couple of algebraic errors made your calculations difficult to follow (e.g. in post #7 ## H(t-\tau) ## should read ## H(\nu,\tau) ##) , but suggest you use the convolution theorem result for Fourier transforms mentioned in post #8. This one should only need a couple of lines of algebra and the final result should emerge.
     
    Last edited: Mar 13, 2016
  14. Mar 13, 2016 #13
    What do you mean by collecting τ terms in one?

    So we had:

    $$g(t) = \frac{1}{2j} \int^{\infty}_{-\infty} (e^{j 2 \pi \nu \tau} + e^{-j 2 \pi \nu \tau}) \left( \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu (t-\tau)} d \nu \right) d \tau$$

    Do you mean that I should bring H(ν) out of the integral and then integrate ##e^{2 \pi j \nu (t-\tau)}## over ##d\tau##?

    We were told that we can use a property like this: ##A e^{j 2 \pi \nu t} + B e^{-j 2 \pi \nu t} \to A H(\nu) e^{j 2 \pi \nu t} + B \underbrace{H(- \nu) e^{-j 2 \pi \nu t}}_{\text{H*(ν)}}##. Is that right?

    I'm not sure how to justify this result. But if we can use that, then we can introduce ##H^*(\nu)## into the first part of the equation like this:

    $$\frac{1}{2j} \left( H(\nu) e^{j 2 \pi \nu t} - H^* (\nu) e^{-j 2 \pi \nu t} \right)$$

    If this the correct idea?
     
  15. Mar 14, 2016 #14

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Before proceeding, I forgot to warn you to use different notation between the frequency of ##f(t)## and the frequency components of ##h(t-\tau)##. With this taken into account, the equation becomes
    $$
    g(t) = \frac{1}{2j} \int^{\infty}_{-\infty} (e^{j 2 \pi \nu \tau} - e^{-j 2 \pi \nu \tau}) \left( \int^{\infty}_{-\infty} H(\nu') e^{2 \pi j \nu' (t-\tau)} d \nu' \right) d \tau
    $$
    Can you please stop making typos?
    For example, from the above integral you will get two terms, the first one being ##\int\int e^{j2\pi\nu \tau}H(\nu')e^{2\pi j \nu'(t-\tau)}d\nu' d\tau##. Rearrange the order to obtain
    $$
    \int H(\nu') e^{2\pi j \nu' t}\left( \int e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu'
    $$
    Then solve the integral inside the bracket using the definition of a delta function.
    Yes, that's correct. But you don't instantly arrive at that equation, do you?
     
  16. Mar 14, 2016 #15
    Thank you, I understand now. The entire expression for g(t) would then be:

    $$\frac{1}{2j} \left( \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t}\left( \int^{\infty}_{-\infty} e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) - \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} \left( \int^{\infty}_{-\infty} e^{- j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) \right)$$

    Could you please explain how the delta function factors into the equation?

    Do you mean the Dirac delta function ##\delta(t)= \left\{\begin{matrix}\infty, \ \ t=0 \\ 0, \ \ else \end{matrix}\right.##?
     
  17. Mar 14, 2016 #16

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Take a look at Charles's post #8.
     
  18. Mar 14, 2016 #17
    I can't follow Charles's post. He says that if f(t) is sinusoidal f(ω) becomes a delta function expression, so that means you would rewrite g(ω)=h(ω)f(ω) as h(ω)δ(ω)?

    A little bit more explanation would be appreciated because I am not sure how exactly it's applicable to this problem (solving the integral ##\int^{\infty}_{-\infty} e^{j2 \pi \tau (\nu-\nu')}##).
     
  19. Mar 14, 2016 #18

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    It's this
    Probably you can neglect the constant prefactor for the present purpose.
    Seems like your sloppy nature in manipulating mathematical expression cannot disappear anytime soon. Check again the second term.
     
  20. Mar 14, 2016 #19

    Charles Link

    User Avatar
    Homework Helper

    By delta function, I meant an expression containing delta functions. In this case I think you're going to get a ## \delta(\nu-\nu_o) ## plus a ## \delta(\nu+\nu_o) ## . In the original sinusoid in time you need to call the frequency ## \nu_o ## to indicate it is a constant. You sort of got around this by introducing ## \nu' ## but the prime is a little clumsy. You are better off from the beginning to call the constant sinusoidal input frequency ##\omega_o ## or ## \nu_o ##. When you have a ## \delta(\nu) ## instead of ## \delta(\omega) ## where ## \omega=2 \pi \nu ## I believe blue_leaf77 is correct, the ## 1/(2 \pi) ## factor will be absent from the delta function formula. Suggestion is to work the Fourier transform of ## \sin(2 \pi \nu_o t) ## and then I think you will find it all comes together. (Express the sine function in terms of complex exponentials and then take the F.T.)
     
    Last edited: Mar 14, 2016
  21. Mar 16, 2016 #20
    There was a sign error, this should be correct:

    $$g(t)=\frac{1}{2j} \left( \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t}\left( \int^{\infty}_{-\infty} e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) - \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} \left( \int^{\infty}_{-\infty} e^{- j2\pi \tau(\nu+\nu')} d\tau \right) d\nu' \right) \right)$$

    According to my post #13 the equation has to be reduced to the form ##\frac{1}{2j} \big[H(\nu)e^{+...}-H(\nu) e^{-...} \big]##.

    I wasn't sure how to use the delta property but I found these properties in a textbook:

    $$\int^{\infty}_{-\infty} e^{-j 2 \pi (\nu-\nu_0)t} dt = \delta (\nu - \nu_0) \ \ (i)$$

    $$FT(\delta(t-t_0)) = e^{-j2 \pi \nu_0 t_0} \ \ (ii)$$

    For instance using (i), I think the second term in the equation becomes:

    $$- \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} . \delta(\nu + \nu') d \nu'$$

    This now looks similar to the inverse FT of ##\delta(\nu + \nu')##. If we had a ##e^{-2\pi j \nu' t}## instead we could use property (ii) and write:

    $$=- H(\nu) e^{-j 2 \pi \nu' t}$$

    So what do I need to do in this case? This is very confusing and unfortunately the textbook does not explain this.

    Likewise, when we deal with the first term in the equation, we can't use (i), because the exponent in ##e^{j2\pi \tau(\nu-\nu')}## is positive. How should I modify these properties?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted