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

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.

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.

• Another view would be to look at the inverse transform and to claim that it appears most natural to compose a signal into a sum (or integral) of complex exponentials (with a positive sign in the exponent). But anyway, no significant change would occur if the sign convention was changed. – Matt L. Nov 5 '14 at 14:50
• @MattL. Agreed on both counts. – Jim Clay Nov 5 '14 at 14:55
• @JimClay:Illustration is good.are you saying that since the dot product of vectors include $\cos\theta$,if the rotation is opposite the vectors would add up.Or whether you are saying about cross-product.I couldn't understand what you meant by 'opposite rotation'. – justin Nov 6 '14 at 6:03
• @justin I'm not sure where the $cos\theta$ that you are talking about comes from. Perhaps you are getting that from $e^{j\theta}=cos(\theta) + j*sin(\theta)$? At any rate, the second picture is meant to illustrate the $e^{-j2\pi kn/N}$ in the Fourier transform cross product. It is rotating in a clockwise direction in the complex plane. In other words, each sample is the same phase as the previous sample, minus some constant phase. Low frequencies have small phase differences, high frequencies have large phase differences. – Jim Clay Nov 6 '14 at 11:16
• @JimClay:But in Fourier transform are we really "adding" each signal or "multiplying" them? – justin Nov 6 '14 at 11:22

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 $a$.

• I have updated the post.could you look at it. – justin Nov 5 '14 at 14:13
• @justin: The integrand is $f(t)e^{-st}$. With $s=\sigma+j\omega$ you get $f(t)e^{-\sigma t}e^{-j\omega t}$. For $\sigma>0$ you get exponential damping of $f(t)$ (for $t>0$). Otherwise, you'd get an exponentially increasing factor which could make the integral diverge. – Matt L. Nov 5 '14 at 14:17
• could you say what does $j$,$\omega$ and $\sigma$ represent in the complex s variable for a signal analysis. – justin Nov 5 '14 at 14:19
• @justin: I used $j$ as the imaginary unit (as usual in EE, other folks call it $i$). And since $s=\sigma+i\omega$, $\sigma$ is the real part of $s$, and $\omega$ is the imaginary part of $s$. – Matt L. Nov 5 '14 at 14:22
• could you explain it using signal characterstics instead of theoretical approach. – justin Nov 5 '14 at 14:23

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.

• could you say why $e^{st}$ is an eigen function? – justin Nov 10 '14 at 6:29
• sure, i had already. first, the general input/output equation for an LTI system is the convolution equation: $$y(t) = \int\limits_{-\infty}^{\infty} h(\tau) x(t-\tau) \ d \tau$$ define the input as $$x(t) = e^{s t}$$ then plug that into the convolution equation and see what comes out for $y(t)$. do you need someone to explain how the convolution integral is derived? – robert bristow-johnson Nov 10 '14 at 14:08
• :I would like to know is it only for input $e^{st}$ we would get an output in terms of $e^{st}$ back or is there any other function?More specifically I want to know why an emphasis is given on '$e^{xt}$(x can be complex or real) in most of the transformations like DFT,Laplace tranform,Z transform etc. – justin Nov 24 '14 at 8:57
• i believe that the exponential form, $x(t)=e^{st}$, is the only functional form for an eigenfunction for linear time-invariant (LTI) systems. remember that $s$ only defines, in a general manner, the base of the exponential since $e^{st} = \left(e^s\right)^t = a^t$, so any exponential function of $t$ works. i do not believe that any other general form of function will pass through the convolution integral without getting its form changed. maybe a power series will: $$x(t) = \sum\limits_{n=0}^{\infty} a_n t^n$$ that's about it. – robert bristow-johnson Nov 24 '14 at 18:45