# Question on LSI system

Why second option (1) is not correct ?

• What part of the explanation is unclear? – MBaz Nov 4 '17 at 14:52
• whole explanation.... – fpsshubham Nov 4 '17 at 14:54

Given an impulse response of $$h(t) = \begin{cases} \frac{1}{t^2} &,& \text{ for } t < 1/2 \\ 0 &,& \text{ otherwise } \\ \end{cases}$$

Its associated CTFT (frequency response) is $$H(\Omega) = \int_{-\infty}^{\infty} h(t) e^{-j \Omega t} dt$$

Now look at its value at $\Omega = 0$ which is $H(0)$; $$H(0) = \int_{-\infty}^{\infty} h(t) dt = \int_{-\infty}^{\infty} \frac{1}{t^2} dt = 2$$

Conclusion: since $H(0)$ is $2$ , then $1$ or $1/2$ or any number $x$ less than $2$; i.e., $x < 2$, cannot be an upper bound for $H(\Omega)$ as it already passes that value...