# why fourier transform in AM and FM why not LT

Started by December 22, 2007
```Hello friends,
I am little confused that why do we always apply fourier transform for
amplitude modulation, frequency modulation why not laplace transform.
Second question is that when I studied control system then everywhere
Laplace Transform is used.
So please tell me why we use LT in controls and why FT in communication.

```
```The Fourier Transform is a decomposition into sinusoids. The Laplace
Transform starts with the Fourier Transform and adds something more- the
ability to use exponentials as well as sinusoids in the decomposition.
For many problems this has no advantage.  For instance, the Laplace
Transform of a rectangular pulse doesn&rsquo;t tell you anything of interest.

However, there is a special case where the Laplace Transform is very
useful.  Say you attach a vibration sensor to an automobile, drive it over
a bumpy road, and acquire a signal for a few minutes.  This signal is not
arbitrary; it is constrained by the differential equations that govern the
vehicle&rsquo;s motion (i.e., the mass, springs, and shocks).  We could also
repeat this example using an electric circuit of inductors, capacitors,
and resistors, which is also controlled by differential equations.  While
the Fourier Transform examines the signals, the Laplace Transform is
capable of examining the differential equations that underlie the signals.

In general, control theory is based on differential equations and its main
tool is therefore the Laplace Transform.  Likewise, communications and
other signal processing problems usually do not involving differential
equations, are therefore handled by the Fourier Transform.

Regards,
Steve Smith

Steve.Smith1@SpectrumSDI.com

```
```On Dec 22, 3:50 pm, "SteveSmith" <Steve.Smi...@SpectrumSDI.com> wrote:
> The Fourier Transform is a decomposition into sinusoids. The Laplace
> Transform starts with the Fourier Transform and adds something more- the
> ability to use exponentials as well as sinusoids in the decomposition.
> For many problems this has no advantage.  For instance, the Laplace
> Transform of a rectangular pulse doesn't tell you anything of interest.
>
> However, there is a special case where the Laplace Transform is very
> useful.  Say you attach a vibration sensor to an automobile, drive it over
> a bumpy road, and acquire a signal for a few minutes.  This signal is not
> arbitrary; it is constrained by the differential equations that govern the
> vehicle's motion (i.e., the mass, springs, and shocks).  We could also
> repeat this example using an electric circuit of inductors, capacitors,
> and resistors, which is also controlled by differential equations.  While
> the Fourier Transform examines the signals, the Laplace Transform is
> capable of examining the differential equations that underlie the signals.

the continuous-time Fourier Transform is a degenerate case of the
(double-sided) Laplace Trasform where s = sigma + j*w and the real
part, sigma, is set to zero.  the L.T. can be thought of as a
generalization of the F.T. where the imaginary quantity jw is replaced
by a more general complex quantity "s".  both can be used for
differential equations, but there are some notational conventions that
are simpler with L.T. and setting up the diff eqs with known initial
conditions at t=0 is easier with the single-sided L.T.

likewise, the discrete-time Fourier Transform (DTFT) is the
corresponding degenerate case of the (double-sided) Z Transform where
z = |z|*e^(j*omega) and the magnitude, |z|, is set to one.  the Z.T.
can be thought of as a generalization of the DTFT where the unit
magnitude quantity e^(j*omega) is replaced by a more general complex
quantity "z" that can have any |z|.  both can be used for difference
equations, but there are some notational conventions that are simpler
with Z.T. and setting up the difference equations with known initial
conditions at n=0 is easier with the single-sided Z.T.

and the Z Transform (single or double-sided) is related to the Laplace
Transform (single or double-sided, respectively) by the sampling
function.

if
+inf
x(t)   =   SUM{  x[n] * sinc((t-n*T)/T)  }
n=-inf

(note that x[n] = x(n*T), and the FT{ x(t) } is bandlimited to 1/(2T))

then the continuous-time representation of the sampled function is

+inf
xs(t)  =   SUM{  x[n] * delta((t-n*T)/T)  }
n=-inf

or

+inf
xs(t)  =   SUM{  x[n] * T*delta(t-n*T)  }
n=-inf

where delta(t) is the unit impulse function (and i'm not in the mood
to debate the properties of the unit impulse function as the Dirac
delta function or the legitimacy of leaving the impulse functions
"naked" outside of an integral, i am asserting to the pure
mathematicians, that the above expression for xs(t) is meaningful, but
i don't wanna argue the case now).

with that T scaling, the continuous-time Fourier Transforms of x(t)
and xs(t) are identical for -pi/T < w < +pi/T

anyway, the Z.T. and L.T. are related as:

ZT{ x[n] } = LT{ (SUM x[n] * delta(t-n*T) }

or

ZT{ x[n] } = LT{ (1/T)*xs(t) }

and, likewise, the F.T. and DTFT are related as

DTFT{ x[n] } = (1/T) * FT{ xs(t) }

or

DTFT{ x[n] } = (1/T) * FT{ x(t) }

for  -pi/T < w < +pi/T  and where omega = w*T + integer*pi, and
usually we pick that integer to be zero so -pi < omega < +pi.

> In general, control theory is based on differential equations and its main
> tool is therefore the Laplace Transform.  Likewise, communications and
> other signal processing problems usually do not involving differential
> equations, are therefore handled by the Fourier Transform.

i would agree, but because of words: "main tool is" and "handled by".

F.T. and L.T. are enough alike that either can be used virtually
interchangably with the correct substitutions and restrictions.

r b-j

Steve, when you have time, please look at that post about fast
convolution and the cost function.  i really would like to know what
your assumptions or restrictions or even if "rules of thumb" were (FFT
"N" as  function of FIR "L") in making that plot of Figure 18-3 in

```
```Fourier transform captures only the steady state phenomena.

Laplace captures both, steady state and transient phenomena.

S  = Sigma + j * Omega.

The sigma component brings in the transient capture possible
in laplace transform. If you make Sigma = 0, then laplace
transform and fourier transform are one and the same.

Hence Laplace Transform is Superset of Fourier Transform.

It is kind of interesting that Laplace was teacher of Fourier.

Many a times we are not interested in transients but want
to understand steady state behaviour, and hence fourier
transform is used in such applications.

Regards
Bharat

>Hello friends,
>I am little confused that why do we always apply fourier transform for
>amplitude modulation, frequency modulation why not laplace transform.
>Second question is that when I studied control system then everywhere
>Laplace Transform is used.
>So please tell me why we use LT in controls and why FT in communication.
>
>
>
>
```
```On Dec 23, 8:21 am, "appuindia" <mail2apoo...@gmail.com> wrote:
> Hello friends,
> I am little confused that why do we always apply fourier transform for
> amplitude modulation, frequency modulation why not laplace transform.
> Second question is that when I studied control system then everywhere
> Laplace Transform is used.
> So please tell me why we use LT in controls and why FT in communication.

In control theory you need transient behaviour for step response of
systems (say).
In coms it is nomally signals (not so much the system) that we look at
and hence the Fourier TF.

Hardy
```
```On Dec 23, 10:11 pm, HardySpicer <gyansor...@gmail.com> wrote:
> On Dec 23, 8:21 am, "appuindia" <mail2apoo...@gmail.com> wrote:
>
> > Hello friends,
> > I am little confused that why do we always apply fourier transform for
> > amplitude modulation, frequency modulation why not laplace transform.
> > Second question is that when I studied control system then everywhere
> > Laplace Transform is used.
> > So please tell me why we use LT in controls and why FT in communication.
>
> In control theory you need transient behaviour for step response of
> systems (say).
> In coms it is nomally signals (not so much the system) that we look at
> and hence the Fourier TF.

You can also imagine what would happen if you wanted to know the zero-
state response of an LTI system to a real exponential signal:

x(t) = Ae^(alpha*t), where alpha is a positive real number, for all t
> 0.

Such systems can be man-made or natural, the latter being necessarily
modeled to be LTI because obviously we cannot live long enough write
down the true answer for t=infinity, nor produce a signal that is
infinite, and in any case, long before infinity, the LTI model would
break down.

The input x(t) can also be man-made. It is conceivable to generate an
electrical signal having value 0 at t=0 that grows exponetially as
e^(alpha*t), but you cannot find the Fourier transform of x(t) because
Fourier transforms explicitly restrict choices of s to be j-omega, and
for any omega you choose, j-omega woud not lie in the region of
available choices for s that will allow a value for X(s) to be
obtained, the so-called region of convergence.

Also, x(t)  would not be able to "see itself" in the inverse-Fourier
transform:

x(t) = Ae^(alpha*t) cannnot equal S|{w}[X(jw)*e^(jwt)dw]

x(t) being able to see itself in the inverse-Fourier transform is
necessary if the Fourier transform is to be used under the principle
of eigenfunctions against H(s).

-Le Chaud Lapin-
```
```Le Chaud Lapin wrote:

...

> -Le Chaud Lapin-

Does Le Chaud Lapin go cold turkey?

Jerry
--
Engineering is the art of making what you want from things you can get.
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```
```On Dec 24, 3:32&#4294967295;pm, Jerry Avins <j...@ieee.org> wrote:
> Le Chaud Lapin wrote:
>
> &#4294967295; &#4294967295;...
>
> > -Le Chaud Lapin-
>
> Does Le Chaud Lapin go cold turkey?

Only if the water is leek warm. :p

-Le Chaud Lapin-

```