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The definition of the FT in engineering is:

$$\int_{-\infty}^{\infty}f(x)e^{-j2\pi ft}dt$$

I'm having trouble calculating the FT of a constant, such as $\frac{1}{2}$:

$$\int_{-\infty}^{\infty}\frac{1}{2}\cdot e^{-j2\pi ft}dt =\frac{1}{2} \int_{-\infty}^{\infty}e^{-j2\pi ft}dt $$

$$= \frac{1}{2} \left[\frac{1}{-j2\pi f}e^{-j2\pi ft}\bigg|_{t=-\infty}^{\infty} \right] = \frac{1}{2} \left[\frac{1}{-j2\pi f}(0-\infty) \right] = \infty $$

This is incorrect, as my textbook says $1 \Longleftrightarrow \delta(f)$, which I believe means $\frac{1}{2} \Longleftrightarrow \frac{1}{2}\delta(f)$.

Nate
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2 Answers2

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Your computation is incorrect, because the value of $e^{-j2\pi f t}$ oscillates around the unit circle in the complex plane as $t \to \pm \infty$ (it doesn't approach either $0$ or $\infty$), unless $f = 0$, in which case it is constantly equal to $1$.

Thus, if we average over all values of $t$, we get $0$ if $f \neq 0$ (all the oscillations in the different directions cancel out), while we get $\infty$ if $f = 0$. So we have a function which is zero at all $f \neq 0$ and infinite at $f = 0$, i.e. $\delta(f)$. If we take the initial constant to be $1/2$ instead of $1$, we get $\frac{1}{2} \delta(f)$, as you surmise.

(This is a slightly informal discussion. The correct mathematical formalism for handling $\delta(f)$ is the theory of distributions. It's quite likely that someone else will post an answer, or maybe a link, discussing this more formally correct approach.)

Matt E
  • 123,735
2

The mistake that you are doing is you ignored the 'j' in the exponent. The $e^j$ when raised to infinity is not equal to zero or infinity because

$e^{ix} = \cos(x)+i\sin(x)$,

and the cosine and sine magnitudes never go to beyond 1.

So the $e^{jx}$, as $x$ tends to infinity is not zero or infinity, cause they are just sinusoidal waves with both imaginary and real component.

Now keeping that aside, when you integrate a sinusoidal wave over infinite period you get zero cause the +ve half cycle and -ve half cycle keep cancelling each other for every complete cycle. EXCEPT WHEN THE FREQUENCY IS ZERO that is, if

$\cos (fx)$ has the value of $f=0$

then the value will be a constant (in this case it is 1) over the whole period.

That's the reason why you find the dirac delta function in the answer.

fifaltra
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dushyanth
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