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why fourier transform in AM and FM why not LT

Started by appuindia 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’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.
 

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 your book.
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. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
On Dec 24, 3:32&#2013266080;pm, Jerry Avins <j...@ieee.org> wrote:
> Le Chaud Lapin wrote: > > &#2013266080; &#2013266080;... > > > -Le Chaud Lapin- > > Does Le Chaud Lapin go cold turkey?
Only if the water is leek warm. :p -Le Chaud Lapin-