Why is a negative exponent present in Fourier and Laplace transform?

Question

could anyone explain why there is a need of negative exponent in fourier and laplace transform.I looked through the web but I couldn't get anything.Does anything happen if a positive exponent is placed in these transforms.

While looking through http://1drv.ms/1tbV45S it says that if $s>0$ it becomes a rapidly decreasing function while if $s<0$ it becomes an rapidly increasing functin of t.I couldn't understand that.Can anyone illustrate this.

Answer 1

Matt is correct that the sign is convention. I think that there is a reason for it beyond that though.

If we look at complex frequencies in the complex plane, they look like a constant vectors that rotate in one direction or another. Positive frequencies rotate counter-clockwise, negative frequencies rotate clockwise, and "0 Hz" frequencies don't rotate at all.

The Fourier transform has a negative sign to intentionally rotate in the opposite direction as the frequencies that they are "looking" for.

The reason for the opposite rotation is that when the two frequency vectors are multiplied, their phases will repeatedly cancel out, so when the results are summed together there will be a massive vector due to all of the individual vectors lining up.

$$ X(f) = \sum\limits_{n=0}^{N-1}x(n)e^{-j2\pi kn/N} $$

This is how the Fourier transform "looks" for frequencies. If the two frequencies are the same or "close" (how close they need to be depends on the length of the DFT) they will line up well and cause a massive response in the summation. I have showed how this works for the discrete Fourier transform (DFT), but the exact same reasoning applies to the continuous transform.

Hopefully this explains why the Fourier transform would want the vectors rotating in the opposite direction. To be perfectly honest I don't know the Laplace transform well enough to give solid reasoning for its negative sign. Since the two transforms are very closely related though (the Laplace transform being a generalization of the Fourier transform), I assume that it is for similar reasons.

Answer 2

For the Fourier transform the sign of the exponent is pure convention. Note that for the inverse transform you have a positive sign in the exponent. You could also define the Laplace transform with a positive sign in the exponent. In any case, you want exponential damping of the time domain function to be transformed, so the real part of the complex exponent should be negative. If you changed $s$ to $-s$ then the region of convergence of the unilateral Laplace transform would change from $\Re\{s\}>a$ to $\Re\{s\}<a$ for some real-valued constant $a$.

Answer 3

The negative exponent in the forward transform is necessary and inevitable, because inner product axioms for complex vectors or functions without conjugation are inconsistent.

For example, the inner product of a complex vector with itself would not be real and non-negative without conjugation.

Answer 4

i would just say that the original convention is to represent complex sinusoids with a positive exponent. so a voltage "phasor" would be

$$ v(t) = V e^{j \omega t} $$

($V$ is a complex constant, and $|V|$ represents the magnitude of the phasor and $\arg\{V\}$ represents the phase of the phasor.) i s'pose we could define the convention as

$$ v(t) = V e^{-j \omega t} $$

but my question would be "why bother?"

why a complex exponential? because $e^{s t}$ is an eigenfunction (essentially the eigenfunction) of linear time-invariant (LTI) systems, which are what we apply Fourier and Laplace transforms to. when $e^{s t}$ goes into an LTI system, something times $e^{s t}$ comes out.

LTI systems can be completely described by, or have their input/output relationship completely described by their impulse response $h(t)$. that description is convolution:

$$ y(t) = \int\limits_{-\infty}^{\infty} h(\tau) x(t-\tau) \ d \tau $$

if the input is

$$ x(t) = e^{s t}$$

the output is

$$ \begin{align} y(t) & = \int\limits_{-\infty}^{\infty} h(\tau) x(t-\tau) \ d \tau \\ & = \int\limits_{-\infty}^{\infty} h(\tau) e^{s (t-\tau)} \ d \tau \\ & = \int\limits_{-\infty}^{\infty} h(\tau) e^{-s \tau} \ d \tau \ \ e^{s t} \\ & = H(s) \ e^{s t} \\ & = H(s) \ x(t) \\ \end{align} $$

so $x(t)=e^{s t}$ is an eigenfunction and the eigenvalue, the thing that scales the eigenfunction in an LTI system is $H(s)$ and directly related to $h(t)$.

then the rest is all about Fourier. so Fourier generalizes a little, first with a periodic $x(t)$ that Fourier posits that can be represented with sinusoids all having the same period as $x(t)$.

$$ x(t+T) = x(t) \quad \forall t $$

$$ x(t) = \sum\limits_{k=-\infty}^{\infty} X[k] \ e^{j \frac{2 \pi k}{T} t} $$

it's still the original convention: define the signal as a phasor $e^{j \omega t}$. the positive exponent remains. $X[k]$ are the "Fourier coefficients".

so we know that the output is

$$ \begin{align} y(t) & = \sum\limits_{k=-\infty}^{\infty} H\left(j \frac{2 \pi k}{T} \right) X[k] \ e^{j \frac{2 \pi k}{T} t} \\ & = \sum\limits_{k=-\infty}^{\infty} Y[k] \ e^{j \frac{2 \pi k}{T} t} \\ \end{align} $$

another periodic function, having the same period, but with different Fourier coefficients.

so, positive $\omega$ in the exponent.

so what are those Fourier coefficients?

$$ \begin{align} \int\limits_{0}^{T} x(t) e^{-j \frac{2 \pi m}{T} t} \ dt & = \int\limits_{0}^{T} x(t) e^{-j \frac{2 \pi m}{T} t} \ dt \\ & = \int\limits_{0}^{T} \sum\limits_{k=-\infty}^{\infty} X[k] e^{j \frac{2 \pi k}{T} t} e^{-j \frac{2 \pi m}{T} t} \ dt \\ & = \int\limits_{0}^{T} \sum\limits_{k=-\infty}^{\infty} X[k] e^{j \frac{2 \pi (k-m)}{T} t} \ dt \\ & = \sum\limits_{k=-\infty}^{\infty} X[k] \int\limits_{0}^{T} e^{j \frac{2 \pi (k-m)}{T} t} \ dt \\ \end{align} $$

for every $k$ in the sum where $k \ne m$, the integral is zero so the term in the summation is zero.

$$ \int\limits_{0}^{T} e^{j \frac{2 \pi (k-m)}{T} t} \ dt = \begin{cases} 0, & \text{for } k \ne m \\ T, & \text{for } k = m \end{cases} $$

for the single non-zero term, when $k=m$, we have

$$ \int\limits_{0}^{T} x(t) e^{-j \frac{2 \pi m}{T} t} \ dt = X[m] T $$

so

$$ X[m] = \frac{1}{T} \ \int\limits_{0}^{T} x(t) e^{-j \frac{2 \pi m}{T} t} \ dt $$

that is where the negative exponent comes from. we need that exponent to be negative so that only the $m^{\text{th}}$ term in the summation survives (when $k=m$ and $e^{j \frac{2 \pi (k-m)}{T} t}=1$), thus isolating a single $X[m]$ so we know what it is. otherwise it would be the $-m^{\text{th}}$ term surviving and we would have to change the convention in our original definition of $x(t)$.

this remains essentially the case as the Fourier series representation is generalized to non-periodic $x(t)$, where the summation becomes an integral. because we define our signal as a sort of integral summation of these exponential (with positive exponents) eigenfunctions:

$$ x(t) = \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} X(j \omega) e^{j \omega t} \ d \omega $$

again, to get those Fourier "coefficients", we need a negative exponent:

$$ X(j \omega) = \int\limits_{-\infty}^{\infty} x(t) e^{-j \omega t} dt $$

Laplace generalizes further by allowing that purely imaginary value $j \omega$ to be a more general complex value, $s = \sigma + j \omega$. but that does not change the sign convention.