# A discussion about Fourier and Laplace transforms and calculus

Tags:
1. Aug 25, 2015

### khurram usman

i have read many of the answers and explanations about the similarities and differences between laplace and fourier transform.

Laplace can be used to analyze unstable systems.

Fourier is a subset of laplace.

Some signals have fourier but laplace is not defined , for instance cosine or sine from -infinity to +infinity.

i have studied signals and systems and basic control theory in my undergrad. My question is related to the solution of differntial equations using these transforms.

Wherever i have seen it is written that fourier is used for steady state analysis (example : in solution of circuits) whereas for transient response we resort to laplace. What exactly is the thing that enables laplace to incorporate initial conditions (hence unilateral laplace) and solve ODEs ? Similarly what prevents fourier from taking these into account?

Let me explain at least what i understand. when solving for the laplace of the derivative of a function (using integration by parts) we input the initial conditions there and they usually end up appearing as decaying exponentials (at the natural modes/poles of the system) in the final response. The constants are chosen so as to satisfy the initial conditions. On a sidenote, this is also related to linearity and upon initial conditions that are not 0 (rest) the system ends up becoming non-linear because of not satisfying the zeros input zero output property. Now when the fourier is found for the derivative of a function the limits of the integral go from -infinity to + infinity and consequently the initial conditions can't be absorbed. Why cant we define the integral to be from 0 to infinity and incorporate initial conditions just as in laplace? i know that fourier is closely linked to convolution and changing these limits would wrong that but still can this formulation be used for solving an initial value ODE? if yes , kindly provide an example. here is one that i found . i don't understand one step (second step of finding the general solution'u') in it but other than that i can't find a mistake.

http://math.stackexchange.com/quest...ial-value-ode-problem-using-fourier-transform

i know this might be too basic for some or too narrowed down but it has bugged me my whole undergrad and now i really need an answer from an expert. Thank you and regards khurram

2. Aug 26, 2015

### the_wolfman

Fourier and Laplace transforms are two types of integral transforms. A integral transform is of the form:

$F(u) = \int_a^b K(u,x) f(x) dx$

where K is called the kernel. Different integral transforms are defined by both their limits of integration and their particular Kernel.

There's no reason why you can't change the limits of integration for a Fourier transform... but then its no longer a Fourier transform. In fact the Fourier sine and Fourier cosine transforms do just that. They integrate from 0 to infinity and use sin(ux) or cos(ux) respectively for the kernel. They're related to the Fourier transform, but they have their own uniques properties.

Integral transforms are tools that help simplify differential equations. Like all tools you want to pick the right tool for the job. You can "hack" Fourier transforms allowing you to solve initial value problems. But that's like hammering a screw into place. What you really want is a screw driver.

Last edited: Aug 26, 2015
3. Aug 28, 2015

### khurram usman

Thank you very much. I can see what you are trying to convey. So the way i am thinking that changing the limits from 0 to infinity and introducing initial conditions in fourier just like in laplace will be in sync with all the mathematics?
Also if possible then please provide me some example. i have never seen integral transformations in general. i have studied laplace and fourier in my EE undergrad.
Thank you

4. Aug 29, 2015

### jasonRF

Besides the limits of integration, there is another difference between the way EEs use Fourier and lLaplace transforms. Laplace transforms are considered functions of a complex variable $s$, and a given laplace transform has a region of convergence (ROC) of the form $\Re(s)>s_0$. The definitions are:
$$\mathcal{L}\left[y\right] = Y(s) = \int_0^\infty dt \, e^{-s t} y(t)$$
$$\mathcal{L^{-1}}\left[Y\right] = y(t) = \frac{1}{j 2 \pi} \int_{\alpha - j^\infty}^{\alpha - j^\infty} ds \, e^{s t} Y(s)$$
for any $\alpha > s_0$. For example, $\mathcal{L}\left[e^{-t}\right] = 1/(s+1)$ with ROC $\Re(s)>-1$. Another example, $\mathcal{L}\left[e^{2t}\right] = 1/(s-2)$ with ROC $\Re(s)>2$. Finally, for the step function $u(t)$ (u(t)=1 for t>0, u(t)=0 for t<0), $\mathcal{L}\left[u(t)\right] = 1/s$ with ROC $\Re(s)>0$.

In signals and systems, Fourier transforms are considered to be functions of a real variable, but now we allow Fourier transforms to be generalized functions like Dirac deltas and so forth. For example, $\mathcal{F}\left[u(t) e^{-t}\right] = 1/(j\omega+1)$, $\mathcal{F}\left[u(t) e^{2t}\right]$ does not exist, and
$\mathcal{F}\left[u(t)\right] = 1/(j \omega) + \pi \delta(\omega)$. Notice that already the way Fourier is defined for signals and systems analysis cannot be used for unstable systems, or solving ODEs that have solutions that grow exponentially. To allow this, we will need to allow the Fourier transform to be a function of a complex varialbe as well, with a corresponding $ROC$. We will show this below.

Here is a simple example. Solve $y^\prime(t) + y(t) = 0$ with initial condition $y(0)=y_0$. Using Laplace you get,
$$s Y(s) - y_0 + Y(s) = 0$$
so
$$Y(s) = \frac{y_0}{s+1}$$
For the inversion you pick a contour to the right of all of the singularities, or $\Re(s) > -1$, and you get $y(t)=y_0 e^{-t}$.

You can do the same with Fourier. We can define a right-sided Fourier as
$$\hat{Y}_+ (\omega) = \int_0^\infty dt\, e^{-j \omega t} y(t)$$
Notice that in general this will converge for $\Im(\omega)<\gamma$ (imaginary part of $\omega$ less than $\gamma$). For example, $\mathcal{F_+}\left[e^{2t}\right] = 1/(j\omega-2)$ with ROC $\Im(w)<-2$. Anyway, applying this to the differential equation (integrating by parts for the derivative) we get
$$j \omega \hat{Y}_+ (\omega) - y_0 + \hat{Y}_+ (\omega) = 0$$
so
$$\hat{Y}_+ (\omega) = \frac{y_0}{j\omega+1}$$
The inverse transform is of the same form as for the normal Fourier transform, but needs to be on a contour below all of the singularities. In this case, the only singularity is at $\Im(\omega)=1$, so we can take the integral along the real axis:
$$y(t) = \frac{1}{2\pi}\int_{-\infty}^\infty d\omega\, e^{i \omega t}\hat{Y}_+ (\omega)$$
This integral is well defined for this example and we get $y(t)=y_0 e^{-t}$.

What if instead we were solving $y^\prime(t) - 2 y(t) = 0$ with initial condition $y(0)=y_0$. Laplace transforming yields,
$$Y(s) = \frac{y_0}{s-2}$$
For the inversion you pick a contour to the right of all of the singularities, or $\Re(s) > 2$, and you get $y(t)=y_0 e^{2 t}$.

Solving it with the right sided Fourier we get,
$$\hat{Y}_+ (\omega) = \frac{y_0}{j\omega-2}$$
For the inversion contour we need to be below all of the singularities, or $\Im(\omega)<-2$. That is,
$$y(t) = \frac{1}{2\pi}\int_{j \beta -\infty}^{j\beta + \infty} d\omega\, e^{i \omega t}\hat{Y}_+ (\omega)$$
for $\beta < -2$. Once again we get $y(t)=y_0 e^{2 t}$.

What have we done here? We have defined a one-sided Fourier transform that is a function of a complex variable. It is now only different from the unilateral Laplace transform by the $j$ in the exponent and an extra $j$ in the inversion integral. In some sense, we just rotated the Laplace transform by 90 degrees in the complex plane.
jason

5. Aug 29, 2015

### jasonRF

Just wanted to point out that in the first ODE I solved above, we could consider the one-sided Fourier transform as a function of a real variable and we still would get the correct solution. This is because $u(t)e^{-t}$ has a Fourier transform. However, in the second ODE, if we naively considered the transform as a function of a real variable and integrated along the real axis for the inversion integral, we would get that the solution is zero for all t>0. Why? Because the solution ($u(t)e^{2t}$ ) has no Fourier transform unless we allow $\omega$ to be complex. The classical Fourier transform throws away solutions that are not transformable.

In the end, for many problems it makes more sense to use Laplace transforms for this kind of analysis of ODEs and systems. The complex Fourier transform is sometimes used by applied mathematicians, physicists, and electrical engineers working in fields such as electromagnetics, where they may be solving partial differential equations describing diffraction, etc.

jason