Fourier Transform

Discussion in 'Homework Help' started by helloeveryone, May 9, 2015.

  1. helloeveryone

    Thread Starter Active Member

    Apr 8, 2011
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    Hello, given a picture of a sawtooth signal f(t) I am suppose to find the Fourier transform. The picture shows that f(t)=t from 0 to 1 otherwise it is 0.
    I believe another way to represent this signal is f(t)=t[u(t)-u(t-1)].
    The provided solution is below but I don't understand it completely:
    Step 1: f'(t)=u(t)-u(t-1)-delta(t-1)
    Step 2: f''(t)=delta(t)-delta(t-1)-delta'(t-1)
    Step 3: -ω^2F(ω)=1 - e^-jω - jωe^-jω
    Step 4: F(ω)=-(1/ω^2)[1 - e^-jω - jωe^-jω]

    I don't understand how we get the first derivative in step 1. I understand step 2. I don't understand how the Fourier transform of delta'(t-1) is jωe^-jω.

    Please explain step by step.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Well, what do you think the result for the first step should be? Show your work and don't worry is the result is correct. The point is to let us see how you are approaching the problem so that we can spot at what point your thinking is going astray.
     
  3. helloeveryone

    Thread Starter Active Member

    Apr 8, 2011
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    My solution:
    Well, we have f(t)=t[u(t)-u(t-1)] which is the same as f(t)=tu(t)-tu(t-1) so, I think we have two cases where one function multiplies another and the product rule can be applied?
    Product rule: (uv)' = u'v+uv'
    (tu(t))' = (t)'u(t)+(t)(u(t))'=(1)u(t)+tdelta(t)=u(t)+tdelta(t)
    (tu(t-1))'=(t)'u(t-1)+(t)(u(t-1))'=(1)u(t-1)+tdelta(t-1)=u(t-1)+tdelta(t-1)
    f'(t)=u(t)+tdelta(t)-(u(t-1)+tdelta(t-1))
    f'(t)=u(t)+tdelta(t)-u(t-1)-tdelta(t-1)

    f''(t)=u(t)+tdelta(t)-u(t-1)-tdelta(t-1)
    f''(t)=(u(t))'+(tdelta(t))'-(u(t-1))'-(tdelta(t-1))'

    The derivate of u(t) is delta(t). The derivate of u(t-t_0) is delta(t_0).
    Product rule: (uv)' = u'v+uv'
    (tdelta(t))'=(t)'(delta(t))+(t)(delta(t))'=(1)'(delta(t))+(t)(delta(t))'=(delta(t))+(t)(delta(t))'

    (tdelta(t-1))'=(t)'(delta(t-1))+(t)(delta(t-1))'=(1)'(delta(t-1))+(t)(delta(t-1))'=(delta(t-1))+(t)(delta(t-1))'

    f''(t)=delta(t)+(delta(t))+(t)(delta(t))'-delta(t-1)-((delta(t-1))+(t)(delta(t-1))')
    f''(t)=delta(t)+(delta(t))+(t)(delta(t))'-delta(t-1)-(delta(t-1)-(t)(delta(t-1))'
    f''(t)=2delta(t)-2delta(t-1)+(t)(delta(t))'-(t)(delta(t-1))'

    I know how to take the Fourier transform of f''(t)=2delta(t)-2delta(t-1)
    I don't how to take the Fourier transform of (t)(delta(t))'-(t)(delta(t-1))'

    Taking the Fourier transform of both sides yields: -
    ω^2F(ω)=2(1)-2e^-jω(1)+F{(t)(delta(t))'-(t)(delta(t-1))'}
    ω^2F(ω)=2-2e^-jω+F{(t)(delta(t))'-(t)(delta(t-1))'}
    ω^2F(ω)=2(1-e^-jω)+F{(t)(delta(t))'-(t)(delta(t-1))'}
    F(ω)=(2/ω^2)(1-e^-jω)+F{(t)(delta(t))'-(t)(delta(t-1))'}
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Before you go any further, let's look at what you have for f'(t) compared to what they have and see if they really are different.

    You have

    <br />
f^'(t) \; = \; u(t) + t \delta (t) - u(t-1) - t \delta(t-1)<br />

    For each occurrence of the δ() function, note that it is zero except for one particular value of t, which means that you can replace t in that term with the value of t at which the δ() function is non-zero. This is known as the sifting property of the δ() function.

    <br />
f(t)\delta(t-T) = f(T)\delta(t-T)<br />

    If you don't recognize this, then your work should still yield the correct result in the end, but you will be doing a LOT more work since you will have to chase product rules to their conclusion a lot more often.
     
  5. helloeveryone

    Thread Starter Active Member

    Apr 8, 2011
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    0
    The sifting property really does make things a lot easier. I solved the problem myself and got the correct answer. Thank you.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    4,804
    You're welcome. Definitely glad that helped and that you were able to solve it yourself.
     
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