Classification of Signals Questions and Answers

Answer: B

For \(x(t) = \sin \omega_0 t\) the fundamental period is \(T = 2\pi / \omega_0\). Here \(\omega_0 = 50\pi\) rad/s.

\[ T = \frac{2\pi}{50\pi} = \frac{1}{25} = 0.04\ \text{s} \]

The frequency is 25 Hz. The value 0.02 s would be the period of a 50 Hz signal, which is what you get if you take \(\omega_0 = 100\pi\), and 0.06 s and 0.08 s have no basis. The answer is 0.04 sec, option B.

Answer: A

A discrete time sinusoid \(x(n) = \cos(\omega n)\) satisfies

\[ \begin{aligned} \cos((\omega + 2\pi k)n) \\ &= \cos(\omega n + 2\pi k n) = \cos(\omega n) \end{aligned} \]

because \(kn\) is an integer. So two sinusoids whose frequencies differ by an integer multiple of \(2\pi\) are the same signal, which is why the frequency of a discrete time signal is only meaningful in an interval of length \(2\pi\). A difference of \(\pi\), \(\pi/2\) or \(\pi/3\) gives different sequences (for example, a difference of \(\pi\) changes the sign of every other sample). The answer is \(2\pi\), option A.

Answer: A

Both statements are true. An antisymmetric (odd) signal satisfies \(x(-t) = -x(t)\); putting \(t = 0\) gives \(x(0) = -x(0)\), so \(x(0) = 0\), which means its amplitude is zero at the origin. For a discrete time signal, periodicity means that there is a smallest positive integer \(N\) with \(x(n) = x(n + N)\) for all \(n\), which is the usual definition.

So the pair true and true is correct, and the options calling either statement false are wrong. The answer is option A.

Answer: A

A discrete time sinusoid \(\cos(\omega_0 n)\) is periodic only if \(\omega_0 / 2\pi\) is a rational number, and the fundamental period is the smallest integer \(N\) with \(\omega_0 N = 2\pi k\) for some integer \(k\).

\[ \frac{\omega_0}{2\pi} = \frac{3\pi}{2\pi} = \frac{3}{2} \]

This is a ratio of integers, so the signal is periodic, and the smallest \(N\) is 2, since \(3\pi \times 2 = 6\pi = 2\pi \times 3\). The signal is \(\cos(3\pi n) = (-1)^n\), which repeats every 2 samples. The answer is periodic with fundamental period 2, option A.

Answer: B

The signal is a square wave that takes only the values +1 and −1 (it is never zero). The average power of a signal is the mean of its squared value over a period.

\[ \begin{aligned} P = \frac{1}{T}\int_0^T x^2(t)\,dt \\ &= \frac{1}{T}\int_0^T (\pm 1)^2\,dt = 1 \end{aligned} \]

Because \((+1)^2 = (-1)^2 = 1\) at every instant, the mean square is 1, whichever the proportion of time spent at each level. The mean value (the dc component) may be zero, but the power is not: a power of 0 would need a zero signal. Powers of 2 and 4 would need amplitudes of about 1.41 and 2. The answer is 1, option B.

Answer: C

The signal \(x(n) = (1/2)^n u(n)\) decays to zero, so its total energy is finite and its average power is zero. It is an energy signal.

\[ \begin{aligned} E \\ &= \sum_{n=0}^{\infty} \left|\left(\tfrac12\right)^n\right|^2 \\ &= \sum_{n=0}^{\infty} \left(\tfrac14\right)^n \\ &= \frac{1}{1 - 1/4} = \frac{4}{3} \end{aligned} \]

The sum is a geometric series with ratio 1/4. The options calling the signal a power signal are wrong, since a power signal has finite nonzero power and infinite energy. The value 3/4 is the reciprocal of the energy. The answer is energy signal, 4/3, option C.