Question

Given the signal, 
\(X(t) = cos t\), if \(t<0 \)
\(Sin\ t\), if \(t\ge0 \)
The correct statement among the following is?

• A. Periodic with fundamental period 2\(\pi\)
• B. Periodic but with no fundamental period
• C. Non-periodic and discontinuous
• D. Non-periodic but continuous

Hint:

Periodicity and Continuity. A signal is periodic if it repeats exactly after some period T for all time. Piecewise definitions often break periodicity. A signal is continuous at a point if the limit from the left equals the limit from the right equals the function value at that point.

Answer 1

The Correct Option is C

The signal X(t) is defined piecewise: - For \(t<0\), \(X(t) = \cos(t)\)
- For \(t \ge 0\), \(X(t) = \sin(t)\)
A signal is periodic if \(X(t) = X(t+T)\) for all \(t\) and some period \(T\)
While both \(\cos(t)\) and \(\sin(t)\) are individually periodic with period \(2\pi\), the overall signal X(t) is constructed by switching from cosine to sine at \(t=0\)
This "switch" breaks the periodicity required for the *entire* signal definition
For example, consider \(t = -\pi/2\)
\(X(-\pi/2) = \cos(-\pi/2) = 0\)
If the period were \(2\pi\), we'd need \(X(-\pi/2 + 2\pi) = X(3\pi/2)\) to be 0
But \(X(3\pi/2) = \sin(3\pi/2) = -1\)
Since \(X(t) \neq X(t+2\pi)\) for all \(t\), the signal is non-periodic
Now check continuity at \(t=0\)
- Limit from the left: \(\lim_{t \to 0^-} X(t) = \lim_{t \to 0^-} \cos(t) = \cos(0) = 1\)
- Value at \(t=0\): \(X(0) = \sin(0) = 0\)
Since the limit from the left (1) does not equal the value at the point (0), the function is discontinuous at \(t=0\)
Therefore, the signal is non-periodic and discontinuous