The condition for Discrete time Fourier transform to exist for function $f(n)$ is given as
$$\sum_{-\infty}^\infty |f(n)| < \infty.$$
In case of continuous Fourier transform the difference is summation is replaced by integration in the above equation
It looks something strange for me because on left side we are taking the signal within 1. infinity limits. Also,2. it is again of summation type and we are expecting right side to be less than infinity. How is it possible?
Let us take two examples
- Suppose there is unit step signal $u(n)$ with constant unity magnitude ranging from o to $\infty$ .then we can clearly see by putting value in the above formula
L.H.S=1+1+1+......upto ${\infty}$ So R.H.S. =$\infty$
So can I say that DTFT of unit step signal doesn't exist?
- Similarly, if I represent the power signal like $cos(wn)$ using the exponential terms and compute it's value from the formula of condition of existence ,it will come out to be $\infty$ .
So can I say that DTFT of power signal like $cos(wn)$ doesn't exist?