Question

if a system produces frequency in the output that are not present in the input ,then the system cannot ​

Answer 1

Explanation:

One of the definitive features of LTI systems is that they cannot generate any new frequencies which is not already present in their inputs. Please note that in this context a frequency refers to signals of the type x(t)=ejΩ0t or cos(Ω0t) which are of infinite duration, and are also referred to as eigenfunctions of LTI systems (specifically for the complex exponential only) and whose CT Fourier transforms are expressed by impulse functions in frequency domain as X(Ω)=2πδ(Ω−Ω0) or X(Ω)=πδ(Ω−Ω0)+πδ(Ω+Ω0) repectively.

One way to see why this is so, comes by observing the CTFT, Y(ω), of the output y(t), which is given by the well known relation Y(ω)=H(ω)X(ω) only when the system is LTI (and also stable as a matter of fact so that H(ejω) exists).

(i.e.

y(t)=∫∞−∞x(τ)h(t−τ)dτ⟷Y(ω)=X(ω)H(ω),

holds only when the impulse response h(t) exists and it will exist only when the system is LTI.)

From a little thought, guided by a simple graphical plot, and using the multiplication property above, one can see that the frequency region of support Ry (set of frequencies for which Y(ω) is non-zero), of the output Y(ω) is given by the intersection of the regions of support Rx and Rh of the inputs X(ω) and frequency response H(ω) of the LTI system:

Ry=Rx∩Rh

And from set algebra we know that if A=B∩C then A⊂B and A⊂C . That is, an intersection is always less or equivalent to what are being intersected. Therefore, the region of support for Y(ω) will be less than or at most equal to the support of X(ω). Hence no new frequencies will be observed at the output.

Since this property is a necessary condition for being an LTI system, any system that fails to posses it, therefore, cannot be LTI.