What is the value of the impulse function?

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SUMMARY

The discussion centers on the properties of the impulse function, specifically the Dirac delta function, and its representation in Fourier transforms. The delta function is characterized by an infinite amplitude but an infinitesimal width, resulting in a finite area under the curve. The correct expression for the Fourier transform of a shifted impulse function is confirmed as 3δ(f-5) + 3δ(f+5), leading to an inverse Fourier transform that yields a cosine wave with a frequency of 5 Hz and an amplitude of 6.

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  • Familiarity with Fourier transforms and their applications
  • Knowledge of complex exponential notation in signal processing
  • Basic concepts of amplitude and frequency in waveforms
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clw
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Homework Statement


I am confused with the idea that the value of a delta function is infinity but when given a graph why is there an amplitude/magnitude value for the delta function. I have attached a graph. Assuming the phase is zero I'm trying to write a Fourier transform for it.

Homework Equations



G(f) = |G(f)| e^{j\vartheta}

The Attempt at a Solution


I know that it is a shifted impulse function so would it just be 3δ(f-5)+3δ(f+5)? Any help would be very much appreciated I'm just having a hard time grasping the idea of impulse function having an infinite value but there's an amplitude of 3?
 

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clw said:

Homework Statement


I am confused with the idea that the value of a delta function is infinity but when given a graph why is there an amplitude/magnitude value for the delta function. I have attached a graph. Assuming the phase is zero I'm trying to write a Fourier transform for it.

Homework Equations



G(f) = |G(f)| e^{j\vartheta}

The Attempt at a Solution


I know that it is a shifted impulse function so would it just be 3δ(f-5)+3δ(f+5)? 3?

Your expression is correct.

The delta function has arbitrarily high amplitude but also arbitrarily small spread. The area is what matters.

If you take the inverse Fourier transform of your expression you would get 6cos[2π(5)t]. So what you are displaying is the Fourier transform of a cosine wave with frequency = 5 Hz and amplitute = 6.
 

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