Discrete Time Signals MCQ Quiz - Objective Question with Answer for Discrete Time Signals - Download Free PDF

Explanation:

Identity for Convolution Operation

Definition: The convolution operation is a fundamental mathematical tool widely used in signal processing, systems theory, and various engineering fields. It involves combining two signals to produce a third signal that represents how one signal modifies or overlaps with the other. The identity element for convolution is a special function that, when convolved with any signal, leaves the signal unchanged.

Correct Option: Option 3: Impulse δ(n)

The identity element for the convolution operation is the impulse function, denoted as δ(n). The impulse function has the following properties:

• Definition: δ(n) is a discrete-time signal defined as:
  • δ(n) = 1 for n = 0
  • δ(n) = 0 for n ≠ 0
• Convolution Property: When any signal x(n) is convolved with δ(n), the output is the original signal x(n):

  x(n) * δ(n) = x(n)

Explanation: The impulse function serves as the identity for convolution because of its unique characteristics. During convolution, the impulse function effectively "selects" values of the input signal without altering them. This property is crucial in systems analysis and signal processing, where the impulse function is used to analyze and characterize systems.

Importance:

• The impulse function is used to determine the impulse response of a system, which provides critical insights into the system’s behavior.
• It is the building block for many mathematical operations in signal processing, such as filtering and system identification.

Correct Option Analysis:

The correct option is:

Option 3: Impulse δ(n)

This option is correct because the impulse function is mathematically proven to be the identity element for convolution. When convolved with any signal, it leaves the signal unchanged, reflecting its role as the identity element.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 1

This option is incorrect. While "1" could be considered an identity element in multiplication or addition in certain mathematical contexts, it is not the identity element for convolution. Convolution is not a simple multiplication operation; it involves summing weighted values of overlapping signals. Therefore, "1" does not satisfy the mathematical requirements to be the identity for convolution.

Option 2: 0

This option is incorrect. The function "0" is the zero element for convolution, meaning that convolving any signal with "0" results in "0." It is not the identity element because it does not leave the original signal unchanged when convolved.

Option 4: Unit Step u(n)

This option is incorrect. The unit step function u(n) is a commonly used signal in signal processing, but it does not act as the identity for convolution. Convolving a signal with u(n) modifies the signal by creating cumulative sums of its values. Thus, it does not satisfy the property of leaving the signal unchanged.

Option 5: (No statement provided)

This option is not applicable due to the absence of a statement or definition. It cannot be considered the identity for convolution operation.

Conclusion:

The impulse function δ(n) is the identity element for the convolution operation. Its unique properties make it a fundamental tool in signal processing and systems theory. By understanding the impulse function and its role in convolution, engineers and scientists can effectively analyze and design systems. Evaluating the incorrect options further reinforces the importance of the impulse function as the identity for convolution.

Digital filter:

The response of a digital filter is given below:

The frequency where poles lie is passed whereas the frequencies where zeroes are located are blocked.

For example: H(z) = \({z+1\over z-1}\)

Poles = 1

Zeroes = -1

High frequency is blocked and low frequency is passed, hence it is a low pass filter.

Calculation:

Given, x[n] = an u[n]

\(x (z)= {z \over z-a}\)

Option 1: For poles lying between -1 < a < 0, it is a high-frequency signal.

Option 2: For poles lying between 0 < a < 1, it is a low-frequency signal.

Option 3: For the bandpass signal, poles must be present in the form of complex conjugate pairs.

Option 4: For |a| > 1, the frequency domain representation does not exist. This statement is true.

Hence, option 4 is correct.

Digital filter:

The response of a digital filter is given below:

The frequency where poles lie is passed whereas the frequencies where zeroes are located are blocked.

For example: H(z) = \({z+1\over z-1}\)

Poles = 1

Zeroes = -1

High frequency is blocked and low frequency is passed, hence it is a low pass filter.

Calculation:

Given, x[n] = an u[n]

\(x (z)= {z \over z-a}\)

Option 1: For poles lying between -1 < a < 0, it is a high-frequency signal.

Option 2: For poles lying between 0 < a < 1, it is a low-frequency signal.

Option 3: For the bandpass signal, poles must be present in the form of complex conjugate pairs.

Option 4: For |a| > 1, the frequency domain representation does not exist. This statement is true.

Hence, option 4 is correct.

Calculation: 

The graph of x[n] = αn u(n),

(α > 0)

⇒ it starts from n=0.

 The graph of αx(n - 1),

⇒ it starts from n = 1.

 g(n) is obtained by subtraction of both graphs.

Hence, 

g(n) = δ(n).

Alternate Method

g(n) = x(n) - αx(n - 1) where (α > 0)

⇒ g(n) = αn u(n) - α*αn-1 u(n-1)

⇒ g(n) = αn u(n) - αn u(n-1)

⇒ g(n) = δ(n).

Discrete-time signals are often obtained by sampling continuous-time signals. 

There are three ways to represent discrete time signals.

1) Functional Representation.

Example:

\(x\left[ n \right] = \left\{ {\begin{array}{*{20}{c}} {n\;\;for\;\;0 \le n \le 10}\\ {0\;\;\;\;\;\;\;otherwise} \end{array}} \right.\)

2) Tabular method of representation

Example:

3) Sequence Representation

Example:

\(x\left[ n \right] = \left\{ {\begin{array}{*{20}{c}} {1,}&{5,}&{ - 2,}&3\\ {}& \uparrow &{}&{} \end{array}} \right\}\)

Let a signal p(x) is uniformly distributed between limits – a to + a

Variance, \(\bar \sigma = \mathop \smallint \limits_{ - a}^a {x^2}p\left( x \right)dx\)

\(\begin{array}{l} = \mathop \smallint \limits_{ - a}^a {x^2}.\frac{1}{{2a}}.dx\\ = \frac{1}{{2a}}{\left[ {\frac{{{x^3}}}{3}} \right]^{a}}\\ = \frac{{2{a^3}}}{{6a}} = \frac{{{a^2}}}{3} \end{array}\)

Its mean square value is equal to its variance

\(\begin{array}{l} {P^2}rms = {\sigma _p} = \frac{{{a^2}}}{3}\\ {p_{rms}} = \frac{a}{{\sqrt 3 }} \end{array}\)

Recursive discrete time system:

If a system output y(n) at time n depends on any number of past output value y(n – 1), y(n – 2), …, it is called a recursive system.

This system requires two multiplication, one addition, and one memory location. This is a recursive system which means the output at time n depends on any number of a past output values.

So, a recursive system has feedback output of the system into the input. This feed back loop contains a delay element.

Non-Recursive discrete time system:

Discrete-time signals are often obtained by sampling continuous-time signals. 

There are three ways to represent discrete time signals.

1) Functional Representation.

Example:

\(x\left[ n \right] = \left\{ {\begin{array}{*{20}{c}} {n\;\;for\;\;0 \le n \le 10}\\ {0\;\;\;\;\;\;\;otherwise} \end{array}} \right.\)

2) Tabular method of representation

Example:

3) Sequence Representation

Example:

\(x\left[ n \right] = \left\{ {\begin{array}{*{20}{c}} {1,}&{5,}&{ - 2,}&3\\ {}& \uparrow &{}&{} \end{array}} \right\}\)

Concept:

Linearity: Necessary and sufficient condition to prove the linearity of the system is that the linear system follows the laws of superposition i.e. the response of the system is the sum of the responses obtained from each input considered separately.

y{ax1[t] + bx2[t]} = a y{x1[t]} + b y{x2[t]}

Conditions to check whether the system is linear or not.

• The output should be zero for zero input
• There should not be any non-linear operator present in the system.

Causal system:

If O/P of the system is independent of the future value of input then the system is said to be causal. Causal systems are practical or physically reliable systems.

Analysis:

y(n) = [x(n)]2 

Linearity check:

x₁(n) → x₁(n)²

x₂(n) → x₂(n)²

x₃(n) = [x1(n) + x₂(n)] →  [x1(n) + x2(n)]² = x₁(n)² + x₂(n)² + 2x₁(n) x₂(n)

≠ x1(n)² + x₂(n)²

∴ Non - linear

Causality check:

y(0) = x(0)²

y(1) = x(1)²

y(-1) = x(-1)²

∴ the system is causal.

\(x\left( n \right) = \left\{ { 2,\begin{array}{*{20}{c}} 4\\ \uparrow \end{array},0,3} \right\}\)

The arrow indicates the origin. This can be represented as:

In terms of unit impulse sequences, this can be represented as:

x(n) = 2δ(n + 1) + 4δ(n) + 3δ(n – 2)

Concept:

Discrete time signal x(n) is given as,

x(n) = x1(n) + x2(n)

Time period of x(n) is (N) = LCM (N1, N2)

Time period of x1(n) is given as:

\({N_1} = \frac{{2\pi }}{{{\omega _{01}}}} \)   

Time period of x₂(n) is given as:

\({N_2} = \frac{{2\pi }}{{{\omega _{02}}}} \cdot K\)  

Calculation:

x(n) = 3e^j22nπ – 3e^-j4.3nπ

ω₁ = 22 π, ω₂ = 4.3 π 

\(N_1=\frac{2π}{ω_1} k=\frac{2π}{22π} k, \)

N₁ = 1 for k = 11,

\(N_2=\frac{2π}{ω_2} k=\frac{2π}{4.3π} k, \)

N₂ = 20 for k = 43

Time period (N) = LCM (N₁, N₂)

N = LCM (1, 20)

N = 20

Concept:

The signal u[n] is defined as:

u[n] = 1 for n ≥ 0

u[n] = 0 for n < 0

Also, the shifted version of u[n] is defined as:

u[n - n₀] = 1 for n ≥ n₀

u[n - n₀] = 0 for n < n₀

Now the combination of two discrete time signals can be analysed as:

u[n] - u[n - n₀] = 1 for 0 ≤ n ≤  n₀ - 1

u[n] - u[n - n0] = 0, otherwise

Calculation:

The given signal x[n] can be defined as:

x[n] = u[n] - u[n - 6]

Digital filter:

The response of a digital filter is given below:

The frequency where poles lie is passed whereas the frequencies where zeroes are located are blocked.

For example: H(z) = \({z+1\over z-1}\)

Poles = 1

Zeroes = -1

High frequency is blocked and low frequency is passed, hence it is a low pass filter.

Calculation:

Given, x[n] = an u[n]

\(x (z)= {z \over z-a}\)

Option 1: For poles lying between -1 < a < 0, it is a high-frequency signal.

Option 2: For poles lying between 0 < a < 1, it is a low-frequency signal.

Option 3: For the bandpass signal, poles must be present in the form of complex conjugate pairs.

Option 4: For |a| > 1, the frequency domain representation does not exist. This statement is true.

Hence, option 4 is correct.

\(x\left[ n \right] = \frac{{\sin \left( {\frac{{n\pi }}{6}} \right)}}{{n\pi }}\)

\(h\left[ n \right] = \frac{{\sin \left( {{\omega _c}n} \right)}}{\pi },{\omega _c} > \frac{\pi }{6}\)

output, y[n] = x[n] * h[n]

in frequency domain both will get multiplied

\(y\left[ n \right] = \frac{{\sin \left( {\frac{{n\pi }}{6}} \right)}}{{n\pi }}.\frac{{\sin \left( {{\omega _c}n} \right)}}{\pi }\)

\(= \frac{{\sin \left( {\frac{{n\pi }}{6}} \right)}}{{n\pi }}\;as\;{\omega _c} > \frac{\pi }{6}\)

\(x\left[ n \right] = {e^{j\left( {\frac{{5\pi }}{3}} \right)n}} + {e^{j\left( {\frac{\pi }{4}} \right)n}}\)

x_d[n] = x[4n]

\({x_d}\left[ n \right] = {e^{j\left( {\frac{{20\pi }}{3}} \right)n}} + {e^{j\pi n}}\)

\({n_1} = \frac{{20\pi }}{3},{n_2} = \pi\)

ω₀ = GCD (n₁, n₂)

\(= GCD\left( {\frac{{20\pi }}{3},\frac{{3\pi }}{3}} \right)\)

\(= \frac{\pi }{3}\)