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Signals and Systems Questions and Answers

1.

Determine the fundamental period of the signal \(x(t) = \sin 50\pi t\).

Exam Year(s)

DSSSB AE - 2023,

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.

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2.

Discrete time sinusoidal signals are identical when their frequencies are separated by the integer multiple of ______.

Exam Year(s)

DSSSB AE - 2023,

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.

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3.

Identify whether the given statements are true or false.

1. Amplitude of antisymmetric signal is always zero at origin.

2. For the discrete time signal, the condition for periodicity is \(x(n) = x(n + N)\).

Exam Year(s)

DSSSB AE - 2023,

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.

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4.

Is the given discrete time signal periodic? If periodic, what is the fundamental period?

\(\cos(3\pi n)\)

Exam Year(s)

DSSSB AE - 2023,

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.

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5.

What is the average power of the given signal?

Graph of a square wave signal: amplitude +1.0 for the first interval T/2, then -1.0 until T, alternating between +1 and -1 with the marks T/2, T and 2T on the time axis

Exam Year(s)

DSSSB AE - 2023,

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.

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6.

Determine whether the given signal is a power or energy signal? What is its power or energy level?

\(X(n) = (1/2)^n u(n)\)

Exam Year(s)

DSSSB AE - 2023,

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.

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7.

A function is said to be even if :

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

A function is even if \(f(t) = f(-t)\).

An even function is symmetrical about the vertical axis: its value at \(-t\) equals its value at \(t\), as for \(\cos t\) or \(t^{2}\). The Fourier series of an even periodic function contains only cosine terms and a constant.

\(f(t) = -f(-t)\) defines an odd function, such as \(\sin t\). \(f(t) = -f(t - T_{0}/2)\) is the condition for half-wave symmetry, under which only odd harmonics appear. \(f(t) = -f(t)\) holds only for the zero function.

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8.

Differentiation of unit impulse function results ............... function.

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

Differentiating the unit impulse gives the unit doublet.

The singularity functions form a chain in which each is the derivative of the one before it: the unit ramp differentiates to the unit step, the unit step to the unit impulse \(\delta(t)\), and the unit impulse to the unit doublet \(\delta'(t)\), a pair of opposite impulses infinitely close together.

So the unit ramp and unit step come before the impulse in the chain, being its integrals rather than its derivative, and the impulse itself is what is being differentiated, not the result.

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