Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > or the system, y (t) = x (t - 5) – x (3...
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or the system, y (t) = x (t - 5) – x (3 - t) which of the following holds true?
- a)System is Linear, time-invariant, causal and stable
- b)System is time-invariant, causal and stable
- c)System is Linear, time-invariant and stable
- d)System is Linear, time-invariant and causal
Correct answer is option 'C'. Can you explain this answer?
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or the system, y (t) = x (t - 5) – x (3 - t) which of the follow...
The given system is a time-delay system, where the output y(t) is equal to the input x(t) delayed by 5 units of time.
This means that the output at any given time t is equal to the input value at time t-5.
In other words, the system takes the input signal x(t) and shifts it by 5 units of time to generate the output signal y(t).
This type of system is commonly used in various applications, such as communication systems, control systems, and signal processing. The time delay allows for synchronization and processing of signals in real-time applications.
This means that the output at any given time t is equal to the input value at time t-5.
In other words, the system takes the input signal x(t) and shifts it by 5 units of time to generate the output signal y(t).
This type of system is commonly used in various applications, such as communication systems, control systems, and signal processing. The time delay allows for synchronization and processing of signals in real-time applications.
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or the system, y (t) = x (t - 5) – x (3 - t) which of the follow...
y1 (t) = v (t - 5) – v (3 - t)
y2 (t) = k v (t - 5) – k v (3 - t) = ky1 (t)
Let x1 (t) = v (t), then y1 (t) = v (t - 5) – v (3 - t)
Let x2 (t) = 2w (t), then y2 (t) = w (t - 5)-w (3 - t)
Let x3 (t) = x (t) + w (t)
Then, y3 (t) = y1 (t) + y2 (t)
Hence it is linear.
Again, y1 (t) = v (t - 5) – v (3 - t)
∴ y2 (t) = y1 (t-t0)
Hence, system is time-invariant
If x (t) is bounded, then, x (t-5) and x (3-t) are also bounded, so stable system.
At t = 0, y (0) = x (-5)-x (3), therefore, the response at t=0 depends on the excitation at a later time t = 3.
Therefore Non-Causal.
y2 (t) = k v (t - 5) – k v (3 - t) = ky1 (t)
Let x1 (t) = v (t), then y1 (t) = v (t - 5) – v (3 - t)
Let x2 (t) = 2w (t), then y2 (t) = w (t - 5)-w (3 - t)
Let x3 (t) = x (t) + w (t)
Then, y3 (t) = y1 (t) + y2 (t)
Hence it is linear.
Again, y1 (t) = v (t - 5) – v (3 - t)
∴ y2 (t) = y1 (t-t0)
Hence, system is time-invariant
If x (t) is bounded, then, x (t-5) and x (3-t) are also bounded, so stable system.
At t = 0, y (0) = x (-5)-x (3), therefore, the response at t=0 depends on the excitation at a later time t = 3.
Therefore Non-Causal.
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or the system, y (t) = x (t - 5) – x (3 - t) which of the following holds true?a)System is Linear, time-invariant, causal and stableb)System is time-invariant, causal and stablec)System is Linear, time-invariant and stabled)System is Linear, time-invariant and causalCorrect answer is option 'C'. Can you explain this answer?
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