Which of the following operation of the discrete time system is NOT memory less?
Unit delay
In the realm of discrete time systems, understanding the concept of 'memory' is crucial. A system is considered memoryless if its output at any specific time instant depends exclusively on the input at that exact same time instant. This means that for a discrete time system, if the output \( y[n] \) at time \( n \) is determined solely by the input \( x[n] \) at time \( n \), then the system operates without memory. Conversely, if the output relies on past input values (such as \( x[n-1], x[n-2], \dots \)) or future input values (like \( x[n+1], x[n+2], \dots \)), then the system is said to possess memory.
Let's thoroughly examine each listed operation to ascertain whether it exhibits the characteristic of memory or operates as a memoryless system.
A signal multiplier performs a multiplication operation between two input signals to produce an output. The output at any given time instant is the product of the input signals at that same instant. For example, if we consider two input signals, \( x_1[n] \) and \( x_2[n] \), the output \( y[n] \) would be mathematically expressed as:
\( y[n] = x_1[n] \cdot x_2[n] \)
In this operation, the calculation of \( y[n] \) at time \( n \) solely depends on the values of \( x_1[n] \) and \( x_2[n] \) at the same time instant \( n \). There is no dependency on past or future input values. Therefore, a signal multiplier is classified as a memoryless discrete time system operation.
A constant multiplier scales an input signal by a fixed numerical constant. If \( x[n] \) represents the input signal and \( K \) is a constant value, the output \( y[n] \) is defined by:
\( y[n] = K \cdot x[n] \)
Here, the output \( y[n] \) at time \( n \) is entirely determined by the input \( x[n] \) at the identical time instant \( n \). No historical or subsequent input values are required to compute the current output. Consequently, a constant multiplier is also considered a memoryless discrete time system operation.
A unit delay system operation shifts the input signal by one unit of time. This means that the current output at time instant \( n \) is equivalent to the input signal's value at the immediate previous time instant \( n-1 \). The mathematical representation of this operation is:
\( y[n] = x[n-1] \)
For the system to generate the output \( y[n] \), it must retrieve or "remember" the value of the input \( x \) from the previous time instant \( n-1 \). Since the output \( y[n] \) explicitly relies on a past value of the input \( x[n-1] \), this system operation inherently possesses memory. Thus, a unit delay is categorically NOT memoryless; it is a system that clearly exhibits memory.
An adder combines two input signals by summing their values at each time instant. If \( x_1[n] \) and \( x_2[n] \) are the input signals, the resultant output \( y[n] \) is given by:
\( y[n] = x_1[n] + x_2[n] \)
The output \( y[n] \) at time \( n \) is exclusively determined by the values of the input signals \( x_1[n] \) and \( x_2[n] \) at the same time \( n \). No past or future input values are involved in this calculation. Therefore, an adder is also a memoryless discrete time system operation.
Based on the detailed analysis, operations such as the signal multiplier, constant multiplier, and adder are all memoryless because their output at any time \( n \) is solely dependent on the input at that identical time \( n \). In stark contrast, the unit delay operation explicitly utilizes a past input value to compute the current output, thereby classifying it as a system with memory.
Hence, among the given choices, the operation of the discrete time system that is NOT memoryless is the Unit delay.
All energy signals will have an average Power of
Arrange the following system types in ascending order of their intersection with 0 dB axis.
A. Type N system
B. Type 3 system
C. Type 2 system
D. Type 4 system
Choose the correct answer from the options given below
Given below are two statements
Statement I: The scaling theorem relates scale changes in frequency domain to the consequent changes in scale in the time domain.
Statement II: The scaling theorem is not applicable to an impulse signal.
In light of the above statements, choose the correct answer from the options given below
The period of the signal X(t) = cos (2 πt) cos (4 πt) – sin (2πt) sin (4πt) is
The time period of a 50 kHz sine wave is