Question

2. Give an example of a signal O) that is Fourier Transformable but whose BLT does not exist. You must show what the CTFT of the signal is and why for that signal there is no valid BLT. You cannot use as an example sinusoidal signals (already mentioned in class) or periodic signals because those can be expressed as sums of sinusoids

Answer 1

The best example of a signal which is fourier transformable but doesnt have BLT is-

f(t) = 1

First let us calculate fourier transform of f(t)=1

For a very common impulse function : F[δ(t)]= 1

Applying duality property:

F(1) = 2π δ(-ω) = 2π δ(ω)

as impulse is an even function.

Note: We could have gone by the integration definition, but that would have taken more steps and the idea is to find the transform. Another thing is Fourier transform exists for power signals as approximation to energy signals. Since f(t) =1 is a power signal fourier transform will exist.

Coming to BLT, we know that Laplace transform will exist only for that function which can give a convergence condition. As f(t) =1 cannot give a definite convergence, BLT does not exist.

We cannot find the bilateral laplace transform of 1 as the signal will not converge at any point and laplace transform will not exist.

But we can find the unilateral laplace transform of 1 only if we define a signal such that its value is 1 and

Limits from 0 to infinity which is basically a unit step signal.

So its laplace transform is as follows:

$L_u(1) = (-1/s) [e^{-\infty} - e^0] = 1/s$

where L_u is unilateral laplace transform.