Convolution problems

Discussion in 'Homework Help' started by JoyAm, Apr 2, 2015.

  1. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Hello, i am trying to solve some convolution problems but since i am not experience with it yet i would like to ask for some help. in this first exercise i am asked for the convolution of two given signals but i am not quiet sure on how to deal with these u(-t) and u(t-3), here is what i tried to do, is it right ? And even if it is do you have a better way to do it ?http://prntscr.com/6onpwi
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    As I interpret those functions, I believe that
    u(-t) = 1 ∀ -∞ < t < 0
    u(-t) = 0 ∀ 0 ≤ t < ∞
    It is the reflection of the unit step about the vertical axis.

    u(t-3) = 0 ∀ -∞ < t ≤ 3
    u(t-3) = 1 ∀ 3 < t < ∞

    It is a unit step function shifted three units to the right of the vertical axis.

    Is that your understanding of those functions?
     
    Last edited: Apr 2, 2015
  3. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Yeah i totally agree with you
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Just use the basic definition of the unit step function:

    u(x) = 1 for x > 0 and 0 otherwise (it is not well defined at x=0, but commonly it is assigned a value of 0.5).

    so if x = -t, then that means that it is 1 for (-t) > o and 0 otherwise. Multiply both sides of the inequality by -1 and you get

    u(-t) = 1 for t < 0 and 0 otherwise.

    Do the same for x = t-3:

    if x = t-3, then that means that it is 1 for (t-3) > 0 and o otherwise. Solve the inequality for t and you get

    u(t-3) = 1 for t > 3 and 0 otherwise.
     
  5. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    thanks for writing WBahn, i have understood what you say but i am not sure how things will work on the convolution formula, have you checked the link i posted ?
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    I usually assign the value 1 to H(0) so that it can be right continuous. Since they are mostly used in integral the value at a single point does not matter.
     
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  7. JoyAm

    Thread Starter Member

    Aug 21, 2014
    112
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    So is my result correct ?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    It looks like you are losing the shift of 3 time units in your convolution integral and then somehow making it magically reappear in the last line. How are you justifying that?

    Just do the math as it is given.

    <br />
y(t) \; = \; x(t) \star h(t) \; = \; \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d \tau<br />

    or, because convolution is commutative,

    <br />
y(t) \; = \; h(t) \star x(t) \; = \; \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d \tau<br />

    Pick the one that looks like it will make life the easiest. Then do what it says.

    In your case, you have

    <br />
x(t) \; = \; e^{2t} u(-t)<br />
\.<br />
h(t) \; = \; u(t-3)<br />

    NOTE: I'm going to treat this as a generic, normalized, entirely dimensionless system.

    Due the exponential in x(t), it will probably be easier if we use the first form above. So at this point it is pure plug and chug.

    What is x(τ)? Just substitute τ for t in x(t):

    <br />
x(\tau) \; = \; e^{2\tau} u(-\tau)<br />

    What is h(t-τ)? Just substitute (t-τ) for t in h(t):

    <br />
h(t-\tau) \; = \; u((t-\tau)-3) \; = \; u((t-3)-\tau)<br />

    Now put those together and then recognize that all the step functions do is turn the integrand on and off in different regions. So identify the regions in which it is on and adjust the integration limits to match and then you can get rid of the step functions. One key point is to realize that in order for the integrand to be turned on, BOTH step functions must be on. When τ is a large negative value, (such as, say, -∞), then the arguments to both step functions is positive and both are on. As τ increases, which step function turns off first? One of them turns off at τ=0, but the other turns off at τ=(t-3). So which one turns off first depends on t, doesn't it? So you have two regions that you need to work with. In one of them, the integral is controlled by the first step function and it independent of t. In the second region, it is controlled by the second step function and IS a function of t.

    Think about the functions involved and what convolution means and see if you can explain what this means physically. and whether it makes sense.
     
  9. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    I was thinking that if i calculate the convolution of x(t) with the unswifted u(t) (lets name the product z(t))then i will be easy to calculate the convolution of x(t) with u(t-3) which will be the z(t-3) since the system is time invariant and linear.
    Is this approach wrong ?
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Do it both ways and see. Or, even better, see if you can prove that your approach is valid.
     
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  11. JoyAm

    Thread Starter Member

    Aug 21, 2014
    112
    4
    I will work on it, thanks
    Btw i have a mistake it should be -2*(t-3)=-2*t+6 and not -2t+3 that i have wrote
     
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