Question

The signal \( \sin(\sqrt{2\pi t}) \) is

• A. periodic with period \( T = \sqrt{2\pi} \)
• B. not periodic
• C. periodic with period \( T = 2\pi \)
• D. periodic with period \( T = 4\pi^2 \)

Hint:

For a function to be periodic, its argument must repeat at regular intervals. If the argument involves non-linear terms like \( \sqrt{t} \), the function is typically not periodic.

Answer 1

The Correct Option is B

The function in question is \( \sin(\sqrt{2\pi t}) \). To determine if it is periodic, we need to investigate whether there exists a period \( T \) such that: \[ \sin(\sqrt{2\pi(t + T)}) = \sin(\sqrt{2\pi t}). \] For a function to be periodic, it must repeat itself after a certain interval \( T \). However, in this case, the argument of the sine function involves \( \sqrt{2\pi t} \), which is a non-linear function of \( t \). This makes the function \( \sin(\sqrt{2\pi t}) \) non-periodic because no single value of \( T \) will make the argument \( \sqrt{2\pi t} \) repeat at regular intervals. Therefore, the function \( \sin(\sqrt{2\pi t}) \) is not periodic. Final Answer: not periodic