Discrete Time Signals MCQ Quiz in मल्याळम - Objective Question with Answer for Discrete Time Signals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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).

at t = 2, t = 3, the value falls by 1 point each, hence u(t – 2) & u(t – 3)

at t = 4, the value falls by 2 points

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.

Concept:

Time Shifting Property:

If \(x\left( n \right) \overset{DTFT}{\longleftrightarrow } x\left( {{e^{j\omega }}} \right)\) 

Then, \(x\left[ {n + T} \right] \overset{DTFT}{\longleftrightarrow } {e^{j\omega T}}x\left( {{e^{j\omega }}} \right)\)

\(x\left[ {n - T} \right] \overset{DTFT}{\longleftrightarrow } {e^{ - j\omega T}}x\left( {{e^{j\omega }}} \right)\)

Application:

Given

\(y\left( n \right) = \frac{1}{2}x\left( {n + T} \right) + x\left( n \right) + \frac{1}{2}x\left( {n - T} \right)\) 

Taking the discrete-time Fourier transform of the above equation, we get:

\(y\left( {{e^{j\omega }}} \right) = \frac{1}{2}{e^{j\omega T}}x\left( {{e^{j\omega }}} \right) + x\left( {{e^{j\omega }}} \right) + \frac{1}{2}{e^{ - j\omega T}}x\left( {{e^{j\omega }}} \right)\;\) 

Frequency response:

\(H\left( {{e^{j\omega }}} \right) = \frac{{Y\left( {{e^{j\omega }}} \right)}}{{X\left( {{e^{j\omega }}} \right)}}\) 

\(H\left( {{e^{j\omega }}} \right) = \frac{1}{2}\left( {{e^{j\omega T}} + {e^{ - j\omega T}}} \right) + 1\) 

\(H\left( {{e^{j\omega }}} \right) = 1 + \cos \omega T\) 

\(\sin \theta = \frac{{{e^{j\theta }} - {e^{ - j\theta }}}}{{2j}}\)

\(= {e^{ - 2t}}\left[ {\frac{{{e^{j\left( {6t + \frac{\pi }{6}} \right)}} - {e^{ - j\left( {6t + \frac{\pi }{6}} \right)}}}}{{2j}}} \right] + {e^{jt}}\)

\(\Rightarrow \frac{1}{{2j}}\left[ {{e^{\left( { - 2 + j6} \right)t}}{e^{\left( { - 2t + \frac{\pi }{6}} \right)}} - {e^{\left( { - 3 - j6} \right)t}}{e^{\left( { - 2t - \frac{\pi }{6}} \right)}}} \right] + {e^{jt}}\)

Complex frequencies are

Conjugate odd symmetric part in

\(\begin{array}{l} {f_{oc}}\left( n \right) = \frac{{f\left( n \right) - {f^ \star }\left( { - n} \right)}}{2}\\ f\left( { - n} \right) = \left\{ {3 + 5j,1 - 2j,2 + 4j} \right\}\\ {f^ \star }\left( { - n} \right) = \left\{ {3 - 5j,1 + 2j,2 - 4j} \right\}\\ \therefore {f_{oc}}\left( { - 1} \right) = \frac{{2 + 4j - 3 + 5j}}{2}\\ = \frac{{ - 1 + 9j}}{2}\\ {f_{\left( {oc} \right)}}\left( o \right) = \frac{{1 - 2j - 1 - 2j}}{2} \end{array}\)

= - 2j

\(\begin{array}{l} {f_{\left( {oc} \right)}}\left( 1 \right) = \frac{{3 + 5j - 2 + 4j}}{2}\\ = \frac{{1 + 9j}}{2}\\ {f_{\left( {oc} \right)}}\left( n \right) = \left\{ {\frac{{ - 1 + 9J}}{2},-2J,\frac{{1 + 9J}}{2}} \right\} \end{array}\)

Periodic↔discrete

It says that a signal that is periodic in one domain is discrete in another and

Continuous ↔ aperiodic

Signal that is continuous in one domain is periodic in another

\(\rm{y[n]}\) is a discrete time signal

\(\rm{\Rightarrow FT[y[n]]}\) is periodic irrespective of sampling time \(\rm{T.}\) Only \(\rm T\) decides whether \(\rm{x[n]}\) can be recovered back from \(\rm{y[n]}\).

Now, nothing has been said about \(\rm{x[n]}\)'s periodicity so we cannot tell that \(\rm{y[n]}\) will be periodic or not and consequently we cannot tell that \(\rm{FT[y[n]]}\) will be continuous or discrete.