dbt in taking Fourier transform of signum and step functions

[bhupala]

signum fourier

i have a doubt in calculating the FT of a step function

let me eloborate

d/dt(u(t)) = δ(t) ---->(1)

as we know.

similarly

d/dt(signum(t)) = 2 δ(t) ---->(2)

Taking FT of the eqns (1) and (2) we get

jωFT[u(t)] = 1 ---->(3)

2jωFT[signum(t)] = 2 ---->(4)

From (3) we can get that FT[u(t)] = 1/jω which is not correct so how to resolve it?

thanx in advance

sri hari

 

[athar_s]

Re: dbt in taking Fourier transform of signum and step funct

i dont know the solution to ur problem but i know of book hat may help u.signals and system:Continious and Discrete second edition Rodger E. Ziemer

 

[claudiocamera]

Re: dbt in taking Fourier transform of signum and step funct

i have a doubt in calculating the FT of a step function

let me eloborate

d/dt(u(t)) = δ(t) ---->(1)

as we know.

similarly

d/dt(signum(t)) = 2 δ(t) ---->(2)

Taking FT of the eqns (1) and (2) we get

jωFT[u(t)] = 1 ---->(3)

2jωFT[signum(t)] = 2 ---->(4)

From (3) we can get that FT[u(t)] = 1/jω which is not correct so how to resolve it?

thanx in advance

sri hari

It is not correct because u(t) is not defined at t=0 . what its value at t=0 ? 0 or 1 ? So you should not use the diferentiation property in singular point.

Write Signum(t) as lim(a->0 ) { exp(-at)*u(t) - exp(at)*u(-t)]

Write u(t) as 1/2[ 1 + Signum(t)]

Revise your equation 2.

Revise your equation 4.

Now apply the properties and you will get the result.

 

[steve10]

Re: dbt in taking Fourier transform of signum and step funct

Let me take u(t) as an example. According to your post,

du(t)/dt=δ(t).

Applying FT, you get

jFT[u(t)]=1,

which means

FT[u(t)]=1/(jω).

Now you claim it is NOT correct. So, why is it not correct?
What you got is perfectly right. The next thing you do is to take the inverse transform:

u(t) = (1/(2π))∫_{-∞}^{∞}((e^{jtω})/(jω))dω
= (1/(2π))∫_{-∞}^{∞}((cos(tω))/(jω))dω+(1/(2π))∫_{-∞}^{∞}((sin(tω))/ω)dω.

The first integral is zero according to the integral of Cauchy Principal Value. Therefore,

u(t)=(1/(2π))∫_{-∞}^{∞}((sin(tω))/ω)dω.

Notice the famous integral

∫_{-∞}^{∞}((sin(ω))/ω)dω=π.

You obtain,

u(t)=-(1/2), if t<0, (1/2) if t>0,

which might differ from what you want (Heaviside function) by a constant 1/2. This is understandable, because you have been actually solving a differential equation, which usually produces the solution with an arbitrary constant. Just look:

d(u(t)+C))/dt=δ(t),

which still produces the same solution.

 

[bhupala]

Re: dbt in taking Fourier transform of signum and step funct

then what abt signum function it is also it is also singular at t=0 at t=0 what is its value? 0,-1,1?

thnx

sri hari

Added after 3 minutes:

Mr steve u r awesum thank you very much. It was a very good explanation which my teacher never gave. Thanks once again

sri hari

 

[claudiocamera]

Let start with the definition of u(t):

u(t) = 0 para t<0
u(t) = 1 para t>0

Lets take the Fourier Transform of u(t) according to step proposed in my last message:

FT[u(t)]= 1/(jω) + πδ(ω)

So FT[u(t)] is not 1/(jω) as exposed in your first doubt.

u(t) is not -(1/2), if t<0, (1/2) if t>0, in every book of this world you will see the definition that:
u(t) = 0 para t<0
u(t) = 1 para t>0

In circuit analysis u(t) is a pure DC signal with unitary amplitude that is switched on in t=0 which means that it is 0 in t=0- and 1 in t>=0+, and undefined in t=0 thats why you have to introduce an impulse i w=0 in its frequency response.

The function used in the given proof is actually (1/2) signum(t) whose FT is 1/(jω) . See (1/2) signum(t)= -(1/2), if t<0, (1/2) if t>0
since signum(t)= -1 if t<0, 1 if t>0, isnt it ?


That's why your teacher had never given you this explanation.

About your doubt: signum function value at t=0 ? 0,-1,1? I dont know , nobody knows, it is a singular point that doesnt make any problem when you do:

Signum(t) = lim(a->0 ) { exp(-at)*u(t) - exp(at)*u(-t)]
u(t) = 1/2[ 1 + Signum(t)]