Signals and Systems MCQ Quiz - Objective Question with Answer for Signals and Systems - Download Free PDF

Explanation:

Minimum Clock Frequency for Proper Sampling:

The problem revolves around determining the minimum clock frequency required for proper sampling of the inputs in a four-channel data acquisition system. The inputs are provided as:

• A = 3cos(2π8000t)
• B = 5cos(2π4000t)
• C = 6sin(2π2000t)

To ensure proper sampling, the Nyquist criterion must be satisfied. The Nyquist criterion states that the sampling frequency should be at least twice the highest frequency present in the signal. This is essential to avoid aliasing, which can lead to distortion and incorrect data acquisition.

Step 1: Determine the Highest Frequency

For each input signal, the frequency can be extracted from the coefficient of t in the trigonometric function:

• A = 3cos(2π8000t): The frequency is 8000 Hz.
• B = 5cos(2π4000t): The frequency is 4000 Hz.
• C = 6sin(2π2000t): The frequency is 2000 Hz.

The highest frequency among the signals is 8000 Hz.

Step 2: Apply the Nyquist Criterion

According to the Nyquist criterion, the sampling frequency must be at least twice the highest frequency. Hence:

Minimum Sampling Frequency = 2 × Highest Frequency = 2 × 8000 Hz = 16000 Hz

Step 3: Consider the Four-Channel System

In a four-channel system, the clock frequency determines how quickly the multiplexer switches between the input channels. To properly sample all four channels, the sampling frequency for each channel must be maintained at the minimum required value.

Since the system sequentially switches between four channels, the clock frequency must be 4 × Minimum Sampling Frequency to ensure each channel is sampled at least at the Nyquist rate. Therefore:

Clock Frequency = 4 × 16000 Hz = 64000 Hz

Concept:

An even function satisfies the condition: \(f(-x) = f(x)\).

An odd function satisfies: \(f(-x) = -f(x)\).

Checking the given functions:

1) sin(x): \(\sin(-x) = -\sin(x)\) → odd function 

2) cos(x): \(\cos(-x) = \cos(x)\) → even function 

3) tan(x): \(\tan(-x) = -\tan(x)\) → odd function 

4) x3: \((-x)^3 = -x^3\) → odd function 

Answer:

Hence, the correct option is 2

Concept:

For a linear time-invariant (LTI) discrete-time system, stability is determined by the location of the poles (roots of the characteristic equation). The system is:

• Stable if all poles lie inside the unit circle (|z| < 1).
• Unstable if any pole lies outside the unit circle (|z| > 1).
• Marginally stable if poles lie on the unit circle and are non-repeating.

Given:

Difference equation:

y(n) = 1.8y(n - 1) - 0.72y (n - 2) + x (n) + 0.5x (n - 1)

Characteristic equation (from homogeneous part):

\(H(z): z^2 - 1.8z + 0.72 = 0 \)

Calculation:

Use the quadratic formula to find roots:

 \(\frac{1.8 \pm \sqrt{(1.8)^2 - 4 \cdot 0.72}}{2}\)

Since one root is greater than 1, the system is unstable.

Explanation:

Discrete Fourier Transform (DFT) and Symmetry Properties

Introduction: The Discrete Fourier Transform (DFT) is a mathematical technique used to analyze the frequency content of discrete signals. When a signal is real-valued, its DFT exhibits specific symmetry properties that simplify computations and provide insights into the relationship between various frequency components. In this problem, we are tasked with finding the value of X(6) of an 8-point DFT, given the values of X(0), X(1), X(2), and X(3). The key to solving this lies in exploiting the symmetry properties of the DFT.

Symmetry Properties of DFT for Real-Valued Sequences:

• Conjugate Symmetry: If the input sequence is real-valued, the DFT coefficients satisfy the following property:

  X(N-k) = conjugate(X(k)) for k = 1, 2, ..., N-1.

  Here, N is the number of points in the DFT (in this case, N = 8).

• Special Cases:
  • X(0) is always real because it represents the sum of all input values (DC component).
  • If N is even, X(N/2) (middle frequency) is also real.

In this problem, the input sequence is real-valued, so we can apply the conjugate symmetry property to find X(6).

Given:

• X(0) = 5
• X(1) = 1 + j
• X(2) = 3 + j
• X(3) = 2 + 3j

We need to determine the value of X(6). Using the conjugate symmetry property of the DFT:

X(N-k) = conjugate(X(k))

For N = 8, the relationship for X(6) becomes:

X(6) = conjugate(X(2))

From the given data:

X(2) = 3 + j

Taking the complex conjugate:

conjugate(X(2)) = 3 - j

Therefore:

X(6) = 3 - j

Correct Answer: Option 2

Explanation:

Minimum Phase LTI System

Definition: A Linear Time-Invariant (LTI) system is considered to be a minimum phase system if all its poles and zeros are located within the unit circle in the Z-plane. The unit circle is defined as the set of points in the complex plane where the magnitude of the complex number is equal to 1. For a minimum phase system, both stability and minimum phase conditions are satisfied.

Key Characteristics of a Minimum Phase System:

• All poles of the system must lie strictly inside the unit circle. This ensures the system is stable.
• All zeros of the system must also lie strictly inside the unit circle. This ensures the system has the minimum phase property.
• The system's phase response is minimized, which is crucial in many control and signal processing applications where phase linearity or minimal phase shift is required.

Explanation of the Correct Option:

The correct option is:

Option 4: All poles and zeros are inside the unit circle.

This is the correct condition for a minimum phase LTI system. For a system to be classified as minimum phase, it is not sufficient for just the poles or just the zeros to be inside the unit circle—both must satisfy this condition. This ensures that the system is stable and has the minimum phase property.

Definition:

Z transform is defined as

\(X\left( z \right) = \mathop \sum \limits_{ - \infty }^\infty x\left[ n \right]{z^{ - n}}\)

Properties:

Differentiation in z domain:

If X(z) is a z transform of x(n), then the z transform of n x(n) is,

\(nx\left( n \right) \leftrightarrow - z\frac{d}{{dz}}\left( {X\left( z \right)} \right)\)

Time-shifting:

If X(z) is a z transform of x(n), then the z transform of x(n – n₀) is,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

Application:

Let x(n) = 1

\(X\left( z \right) = \mathop \sum \limits_0^\infty {z^{ - n}} = \frac{1}{{1 - {z^{ - 1}}}} = \frac{z}{{z - 1}}\)

Now, by applying the property of differentiation in the z domain,

\(x\left( n \right) = n \leftrightarrow - z\frac{d}{{dz}}\left( {\frac{z}{{z - 1}}} \right) = \frac{z}{{{{\left( {z - 1} \right)}^2}}}\)

Now, by applying the property of differentiation in the z domain,

\(x\left( n \right) = {n^2} \leftrightarrow - z\frac{d}{{dz}}\left( {\frac{z}{{{{\left( {z - 1} \right)}^2}}}} \right) = \frac{{z\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}}\)

Now, by applying the property of time-shifting,

\(x\left( n \right) = {\left( {n + 1} \right)^2} \leftrightarrow z.\frac{{z\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}} = \frac{{{z^2}\left( {z + 1} \right)}}{{{{\left( {z - 1} \right)}^3}}}\)

Concept:

Shifting property of impulse function

\(\rm\displaystyle\int_{-\infty}^{\infty} x(t) \delta (t -a ) dt = x(a)\)

Scaling property of impulse function

\(\delta (at) = \frac{1}{|a|)} \delta (t)\)

Explanation:

Let:

\(\rm I =\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(2t - 2) dt\)

\( I = \rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta[2(t - 1)] dt\)

Using scaling property of impulse function in the above equation, we'll get:

\( I =\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \frac{1}{|2|}\delta(t - 1) dt\)

Applying Shifting property of impulse function to the above equation, we'll get:

\( I =\frac{1}{2}\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(t - 1) dt\)

\(I = \frac{1}{2}. \left. e^{-t} \right|_{t = 1}\)

\(\frac{1}{{2e}}\)

Concept:

Minimum sampling rate to avoid aliasing:

f_s = 2f_m = (Nyquist rate)

Calculation:

Given that, ω_m = 400 π

f_m = 200 Hz = maximum frequency of signal

Sampling frequency f_s = 2 × 200 = 400 Hz

Concept:

Bilateral Laplace transform:

\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = \;\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)

Unilateral Laplace transform:

\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = \;\mathop \smallint \limits_0^\infty x\left( t \right){e^{ - st}}dt\)

Some important Laplace transforms:

Calculation:

\(\sin \omega t. u(t)\leftrightarrow \frac{\omega }{{{s^2} + {\omega ^2}}}\)

By applying frequency differentiation property,

\({e^{ - at}}\sin \omega t. u(t) \leftrightarrow \frac{\omega }{{{{\left( {s + a} \right)}^2} + {\omega ^2}}}\)

Modulation:

• The process in which the characteristics of carrier signal is varied in accordance with baseband message signal to make bandpass signal.

Demultiplexing :

• The process or technique of transmitting multiple analog or digital input signals or data streams over a single channel is called multiplexing.
• The reverse of the multiplexing process. Demultiplexing is a process reconverting a signal containing multiple analog or digital signal streams back into the original signals.

Sampling :

• The process of conversion of continuous-time signals into discrete time signals.

Quantization :

• The process of mapping of continuous-time signals into discrete time signal.
• Hence correct option is "3"

Concept:

Final value theorem:

A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression

Final value theorem states that the final value of a system can be calculated by

\(x\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} sX\left( s \right)\)

 Where X(s) is the Laplace transform of the function.

For the final value theorem to be applicable system should be stable in steady-state and for that real part of poles should lie in the left side of s plane.

Initial value theorem:

\(x\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} x\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sX\left( s \right)\)

It is applicable only when the number of poles of X(s) is more than the number of zeros of X(s).

Calculation:

Given that, \(X\left( s \right) = \frac{{4s + 1}}{{{s^2} + 6s + 3}}\)

Initial value,

 \(x\left( 0 \right) = \mathop {\lim }\limits_{s \to \infty } s\frac{{4s + 1}}{{{s^2} + 6s + 3}}\\=\mathop {\lim }\limits_{\frac{1}{s} \to 0 } \frac{{4 + \frac{1}{s}}}{{{1} + \frac{6}{s} + \frac{3}{s^2}}} = 4\)

Given, the signal

V (t) = 30 sin 100t + 10 cos 300 t + 6 sin (500t+π/4)

So, we have

ω₁ = 100 rads

ω₂ = 300 rads

ω₃ = 500 rads

∴ The respective time periods are

\(\begin{array}{l} {T_1} = \frac{{2\pi }}{{{\omega _1}}} = \frac{{2\pi }}{{100}}sec\\ {T_1} = \frac{{2\pi }}{{{\omega _2}}} = \frac{{2\pi }}{{300}}sec\\ {T_3} = \frac{{2\pi }}{{500}}sec \end{array}\)

So, the fundamental time period of the signal is

\(LCM\left( {{T_1},{T_2},{T_3}} \right) = \frac{{LCM\left( {2\pi ,2\pi ,2\pi } \right)}}{{HCF\left( {100,\ 300,\ 500} \right)}}\)

as \({T_0} = \frac{{2\pi }}{{100}}\)

∴ The fundamental frequency, \({\omega _0} = \frac{{2\pi }}{{{T_0}}} = 100\ rad/s\)

Concept:

Final value theorem:

It states that:

\(x\left( \infty \right) = \mathop {\lim }\limits_{z \to 1} \left( {1 - {z^1}} \right)X\left( z \right)\)

Conditions:

1. It is valid only for causal systems. 

2. Pole of (1 – z-1) X(z) must lie inside the unit circle.

Calculation:

The final value theorem for z-transform is:

\( = \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\frac{{\left( {1 - {z^{ - 1}}} \right){z^{ - 1}}\left( {1 - {z^{ - 4}}} \right)}}{{{{\left( {1 - {z^{ - 1}}} \right)}^2}}}\)

\(= \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\;\frac{{\left( {{z^2} - 1} \right)\left( {{z^2} + 1} \right)}}{{{z^4}\left( {z - 1} \right)}}\)

\( = \mathop {\lim }\limits_{z \to 1} \frac{1}{4}\;\frac{{\left( {z + 1} \right)\left( {{z^2} + 1} \right)}}{{{z^4}}} \)

= 1/4 × 1 × 2 × 2 = 1

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

\(X(\omega) = \left. {\frac{{{e^{j\left( {10 - \omega } \right)t}}}}{{j\left( {10 - \omega } \right)}}} \right|_{ - 1}^1 = \frac{{2\sin \left( {\omega - 10} \right)}}{{\left( {\omega - 10} \right)}} \)

Concept:

Initial value theorem:

The initial value theorem is one of the basic properties of the Laplace transform used to find the response of the system at the initial state (t = 0) in the Laplace domain. Mathematically it is given by

\({\bf{f}}\left( {{0^ + }} \right) = \mathop {\lim }\limits_{t \to 0} {\bf{f}}\left( {\bf{t}} \right) = \mathop {\lim }\limits_{s \to \infty \;} s\;F\left( s \right)\)

Where

f(t) is system function

F(s) is Laplace transform of system function f(t)

f(0+) is the initial value of the system

NOTE:

• For the final value theorem to be applicable system should be stable in a steady-state and for that real part of the poles should lie on the left side of the s plane.
• In the given problem exactly the final value theorem is not applied but just X(0+) is calculated.
   

Calculation:

Given that, 

\(X(s) = \frac{{3{s} + 5}}{{{s^2} + 10s + 21}}\)

\({\bf{x}}\left( {{0^ + }} \right) = \mathop {\lim }\limits_{x \to 0} {\bf{x}}\left( {\bf{t}} \right) = \mathop {\lim }\limits_{s \to \infty \;} s\;X\left( s \right)\)

\({\bf{x}}\left( {{0^ + }} \right) = \mathop {\lim }\limits_{x \to 0} {\bf{f}}\left( {\bf{t}} \right) = \mathop {\lim }\limits_{s \to \infty \;} s.\;\frac{{3{s} + 5}}{{{s^2} + 10s + 21}}\)

\(= \mathop {\lim }\limits_{s \to \infty \;} \;\frac{{3{s^2} + 5s}}{{{s^2} + 10s + 21}}\)

= 3