Ambiguous Results for two Fourier transform techniques

[PainterGuy]

TL;DR Summary

I have been trying to trace an error in my result without any success. I was trying to compare the application of two different forms of Fourier transform formulae.

Hi,

I tried to apply different forms of Fourier transform, exponential and trigonometric forms, to the same function, f(t)=a⋅e^-(bt)⋅u(t). The result reached using exponential form is correct.

Please notice that while appling the trigonometric form of Fourier transform, the factor of 1/π was ignored.

At the very end, the difference lies between "+a²" and "-ia²". I believe that I should have gotten "-ia²" instead of +a²" for A(α)+B(α) result. The error might be somewhere in the calculation for B(α), or the reason might be ignoring the factor of 1/π.

Could you please help me with tracing down that error? Thank you!

 

[AVBs2Systems]

Hi
Painter

The Fourier Transform of a function is defined as:

$$
x(j\omega) = \displaystyle \int_{-\infty}^{\infty} f(t) \cdot e^{-j(\omega t)} \,\,\,\, \text{dt}
$$

The trigonometric Fourier series for a function is defined as:

$$
f(t) = \dfrac{a_{0}}{2} + \displaystyle \sum_{k=1}^{k \to \infty} \bigg(a_{k} cos(k \omega_{0} t) + b_{k} \sin(\omega_{0} t) \bigg)
$$
Whereas the exponential Fourier series (represents the same thing as trigonometric and is just a restatement of it) is defined as:
$$
f(t) = \dfrac{a_{0}}{2} + \displaystyle \sum_{k=1}^{k \to \infty} \bigg(a_{k} cos(k \omega_{0} t) + b_{k} \sin(\omega_{0} t) \bigg) = \displaystyle \sum_{k \to -\infty}^{k \to \infty} c_{k} e^{j(k \omega_{0}) }
$$
The Fourier transform of:
$$A \cdot e^{-bt} \cdot \epsilon(t) \iff \frac{ A }{ b + j \omega }
$$

I think, you might be confusing the Fourier transform and the Fourier series.
They don't mean the same thing, but you already know they are quite related..
\begin{array}{ |c|c| c | }

\hline \textbf{Transform}& \textbf{On signal in time domain} & \textbf{Spectrum produced} \\

\hline \text{Fourier transform} & \text{ Continuous, aperiodic} & \text{Continuous, aperiodic but discrete if signal was periodic}\\

\hline \text{Fourier series} & \text{Continuous, periodic} & \text{Discrete aperiodic sequences $ a_{k}, b_{k}, c_{k}$ } \\

\hline \text{Discrete time Fourier transform} & \text{Discrete, aperiodic} & \text{Continuous, periodic with the sampling angular velocity} \\

\hline \text{Discrete Fourier transform} & \text{Discrete, periodic} & \text{Discrete, periodic}

\end{array}

I hope I could help.

 

[RPinPA]

Before going through your calculations in detail, I'll note one thing about the sinusoidal vs complex Fourier representations. The exponential version is defined for both positive and negative frequency. If you are transforming a real signal, then the transform has conjugate symmetry: $F(-\omega) = F^*(\omega)$. The inverse transform combines both positive and negative frequencies to get back a real signal.

The trigonometric version is defined only in terms of positive frequencies. The inverse transform sums only over positive frequencies. Everything is kept real. The two transformed functions are each real, and they contain in some sense the same information as the complex Fourier transform.

Something like that may be at the heart of what you're seeing.

OK, now I'll go through the analysis. One question, what is $u(t)$? Looks like it may be a unit step, equal to 1 for $t \geq 0$ and 0 for $t < 0$?

 

[RPinPA]

Actually before looking at your specific results, I'll look at the general transforms and show how they're related. I will also ignore factors of $1/\pi$, etc so whatever I say will be off by some multiplicative constant.

I will use these identities:
$$\cos(x) = \frac {e^{ix}+e^{-ix}} {2} \\
\sin(x) = \frac {e^{ix}-e^{-ix}} {2i} $$
So
$$A(\alpha) = \int_{-\infty}^{\infty} f(x) \cos(\alpha x) dx \\
= \frac{1}{2}\left [\int_{-\infty}^{\infty}f(x)e^{i\alpha x}dx + \int_{-\infty}^{\infty}f(x)e^{-i\alpha x}dx \right ] \\
= \frac{1}{2} \left [ F(-\alpha) + F(\alpha) \right ]$$

And similarly
$$B(\alpha) = \int_{-\infty}^{\infty} f(x) \sin(\alpha x) dx \\
= \frac{1}{2i}\left [\int_{-\infty}^{\infty}f(x)e^{i\alpha x}dx - \int_{-\infty}^{\infty}f(x)e^{-i\alpha x}dx \right ] \\
= \frac{1}{2i} \left [ F(-\alpha) - F(\alpha) \right ]$$

This is the relationship you should expect (up to the constants I ignored) between the two types of transform.

 

[RPinPA]

So let's check. You have $F(\alpha) = \frac {\alpha} {b + i\alpha}$. So that gives $F(-\alpha) + F(\alpha)$ = $\left ( \frac {-\alpha} {b - i\alpha} \right ) + \left ( \frac {\alpha} {b + i\alpha} \right )$ = $\frac {-\alpha(b + i\alpha) + \alpha(b - i\alpha)} {b^2 + \alpha^2}$ = $\frac {-\alpha b -i \alpha^2 + \alpha b - i\alpha^2}{b^2 + \alpha^2}$ = $\frac {-2i\alpha^2} {b^2 + \alpha^2}$

Hmm. I see a problem already. This should be real and it's not. And that's because the expression for $F(-\alpha)$ is not the conjugate of the expression for $F(\alpha)$ as it should be. So I'll need to go through your complex integral in more detail to understand what's going on.

 

[PainterGuy]

Thank you for the help. I understand that it's not easy to track down an error in someone else's work.

One question, what is u(t)u(t)u(t)? Looks like it may be a unit step, equal to 1 for t≥0t≥0t \geq 0 and 0 for t<0t<0t < 0?

Yes, it's a unit step function.

I think, you might be confusing the Fourier transform and the Fourier series.
They don't mean the same thing, but you already know they are quite related..

Thank you but I'm not confusing the both. This thread is actually an extension of another thread; the excerpt of which is quoted below. I was trying to make a general comparison between exponential and trigonometric forms of Fourier transform, and as a self-study was doing this problem to reach some conclusion. I hesitated to post in the other thread in order not to overwhelm the people who were helping me.

I was trying to find that why the exponential form of Fourier transform is more popular compared to its trigonometric equivalent. I was able to find an answer, https://math.stackexchange.com/ques...er-series-versus-trigonometric-fourier-series, which gives the reasons in terms of exponential form of Fourier series and trigonometric form of Fourier series. It has more to do with cleanliness, compactness, and easy manipulation of exponential form compared to the trigonometric and it wouldn't be wrong to say that both are equally applicable mathematically.

 

[AVBs2Systems]

Thank you for the help. I understand that it's not easy to track down an error in someone else's work.
Yes, it's a unit step function.
Thank you but I'm not confusing the both. This thread is actually an extension of another thread; the excerpt of which is quoted below. I was trying to make a general comparison between exponential and trigonometric forms of Fourier transform, and as a self-study was doing this problem to reach some conclusion. I hesitated to post in the other thread in order not to overwhelm the people who were helping me.

Yes, I apologise for not being more thorough.

So:

$$
\Re\big( X(j \omega) \big) = X(j \omega)_{Even} \,\,\,\,\, j \cdot \Im\big(X(j\omega) \big) = X(j \omega)_{odd}
$$
Here:
$$
X(j \omega) = \dfrac{A}{b + j\omega)}= \dfrac{A \cdot(b - j\omega) }{b^{2} + \omega^2} = ´\dfrac{Ab}{b^{2} + \omega^2} - j \cdot \dfrac{A \omega}{ b^{2} + \omega^2}
$$
Hence:
$$
X(j \omega)_{Even} = \dfrac{Ab}{b^{2} + \omega^2}
$$
And:
$$
X(j \omega)_{odd} = - j \cdot \dfrac{A \omega}{ b^{2} + \omega^2}
$$
Using the theorem that neither odd nor even functions can be decomposed into even and odd parts
We can say that:
$$
X(j \omega ) = \big( X(j \omega)_{odd} + X(j\omega)_{even} \big)
$$

So now, if I understood correctly, you want to strictly evaluate the real and imaginary parts first, and then see if they match what we got above?

Now I got for the real part, evaluating it directly:

$$
\displaystyle \int_{0}^{\infty} A \cdot e^{-b t} \cdot \cos(\omega t) \,\,\,\, \text{d}t = \displaystyle \int_{0}^{\infty} \frac{A}{2} \cdot e^{-b t} \cdot \big( e^{j(\omega t) } + e^{-j(\omega t) } \big) \text{d}t= \frac{A}{2} \cdot \bigg[ \frac{1}{b + j \omega} + \frac{1}{ b - j \omega } \bigg] = \frac{Ab}{b^{2} + \omega^{2} }
$$
This matches what we get when we decompose the original spectrum.

Now, I would do the same integral with the sine, and multiply the result with j.
$$j \cdot \Im(\displaystyle \int_{0}^{\infty} A e^{-bt}e^{-j(\omega t)} \,\,\,\, \text{d}t ) $$ $$ =j \cdot \displaystyle \int_{0}^{\infty} A \cdot e^{-b t} \cdot \sin(\omega t) \,\,\,\, \text{d}t = \displaystyle \int_{0}^{\infty} \frac{A}{2j} \cdot e^{-b t} \cdot \big( e^{j(\omega t) } - e^{-j(\omega t) } \big) \text{d}t= \frac{A}{2j} \cdot \bigg[ \frac{1}{b + j \omega} - \frac{1}{ b - j \omega } \bigg]
$$
Then we get:
$$
j \cdot \frac{A}{2 j} \cdot \frac{ -2 j \omega }{b^{2} + \omega^{2} } = - j \cdot \frac{A \omega}{b^{2} + \omega^{2} }
$$
They match.
So to recap:

$$
X(j\omega)_{Even} = \Re \big( \displaystyle \int_{-\infty}^{\infty} f(t) \cdot e^{- j(\omega t) } \,\, \text{d}t \big) = \displaystyle \int_{-\infty}^{\infty} f(t) \cos(\omega t) \,\,\,\, \text{d}t
$$

$$
X(j\omega)_{Odd} = j \cdot \Im \big( \displaystyle \int_{-\infty}^{\infty} f(t) \cdot e^{- j(\omega t) } \,\, \text{d}t \big)= j \cdot \displaystyle \int_{-\infty}^{\infty} f(t) \sin(\omega t) \,\,\,\, \text{d}t
$$

 

[PainterGuy]

Hi,

First of all, I just realized that there was some errors in my calculations from post #1; the main error occurred in calculation of B(α) where I had mistaken letter "a" for alpha "α" and hence ended with α². The same thing happened at the very end where I said "Converting result from exponential form into relevant form".

I have fixed the errors below.

Now let's rephrase the question.

You can see that the difference lies between "aα" and "-iaα". I believe that I should have gotten "-iaα" instead of +aα" for A(α)+B(α) result. The difference is also shown below.

Now moving on to trace the error and fix it.

@AVBs2Systems, thank you for your attempt to help me. I didn't follow your post closely but intend to do it later. First, I wanted to trace the error.

Thanks.