Signal Period Determination

The question asks us to determine the period of the given signal, denoted as X(t). The expression for the signal is:

X(t) = cos (2πt) cos (4πt) – sin (2πt) sin (4πt)

Trigonometric Identity Application

To find the period of this signal, we first need to simplify its expression. The given form of X(t) perfectly matches a well-known trigonometric identity, which is the cosine addition formula.

$$\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$

By comparing the given signal X(t) with this identity, we can identify the values for A and B:

• A = 2πt
• B = 4πt

Now, substitute these values back into the cosine addition formula to simplify X(t):

$$X(t) = \cos(2\pi t + 4\pi t)$$

Combine the terms inside the cosine function:

$$X(t) = \cos(6\pi t)$$

Angular Frequency Calculation

The simplified form of the signal is X(t) = cos(6πt). This is a standard form for a sinusoidal signal, which is generally given as Y(t) = A cos(ωt + φ), where ω represents the angular frequency.

$$\omega = 6\pi \text{ radians/second}$$

Period Calculation

The period T of a sinusoidal signal is related to its angular frequency ω by the formula:

$$T = \frac{2\pi}{\omega}$$

Substitute the calculated angular frequency ω = 6π into this formula:

$$T = \frac{2\pi}{6\pi}$$

Now, simplify the expression to find the period T:

$$T = \frac{1}{3} \text{ sec}$$

Conclusion on Signal Period

The fundamental period of the signal X(t) = cos (2πt) cos (4πt) – sin (2πt) sin (4πt) is found to be 1/3 sec. This demonstrates how trigonometric identities are crucial tools for simplifying complex signals to easily determine their fundamental properties like the period.

The final calculated period is 1/3 sec.