This set of Signals & Systems Questions & Answers for Exams focuses on “Properties of Systems – 5”.
1. For the system y (t) = x2(t), which of the following holds true?
a) Invertible
b) Non-Invertible
c) Invertible as well as Non-Invertible in its respective domains
d) Cannot be determined
View Answer
Explanation: When we pass the signal x (t) through the system y (t), the system squares the input. Hence, inverse system should take square root, i.e.
W (t) = \(\sqrt{y (t)} = \sqrt{x^2 (t)}\) = ± x (t)
Thus, two outputs are possible, + x (t) or – x (t)
This means there is no unique output for unique input. Hence, this system is Non-Invertible.
2. For the system \( y (t) = ∑_{k=-∞}^n x(k)\), which of the following holds true?
a) Invertible
b) Non-Invertible
c) Invertible as well as Non-Invertible in its respective domains
d) Cannot be determined
View Answer
Explanation: \( y (t) = ∑_{k=-∞}^n x(k)\)
Or, y (n) = x (n) + x (n-1) + x (n-2) + …….
Or, y (n+1) = x (n+1) + x (n) + (terms after this)
Or, y (n+1) = x (n+1) + y (n)
Or, y (n+2) = x (n+2) + y (n+1)
Or, y (n) = x (n) + y (n-1)
This is an alternate form of given system equation.
∴ x (n) = y (n) – y (n-1)
Taking z-transform on both sides,
X (z) = Y (z) = (1-z-1)
\(H (z) = \frac{Y(z)}{X(z)} = \frac{1}{1-z-1}\)
Or, \(\frac{w(z)}{y(z)} \)= H-1(z) = (1-z-1)
∴ w (n) = y (n) – y (n-1)
Hence, the system is invertible.
3. For the system, y (t) = u{x (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 causal and stable
d) System is stable
View Answer
Explanation: Let x1(t) = v (t), then y1 (t) = u {v (t)}
Let x2(t) = k v (t), then y2 (t) = u {k v (t)} = k y1 (t)
Hence, non-linear
y1 (t) = u {v (t)}
y2 (t) = u {v (t-t0)} = y1 (t-t0)
Hence, time-invariant
Since the response at any time depends only on the excitation at time t=t0, and not on any further values, hence causal.
4. For 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
View Answer
Explanation: 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.
5. For the system, y (t) = x (\(\frac{t}{2}\)), which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is Linear and time-invariant
c) System is Linear and causal
d) System is Linear and stable
View Answer
Explanation: y1 (t) = v (\(\frac{t}{2}\))
And y2 (t) = k v (\(\frac{t}{2}\)) = k y1 (t)
If x3 = v (t) + w (t),
Then, y3 (t) = v (\(\frac{t}{2}\)) + w (\(\frac{t}{2}\))
= y1 (t) + y2 (t)
Since, it satisfies both law of homogeneity and additivity, it is also linear.
Again, y1 (t) = v (\(\frac{t}{2}\)), y2 (\(\frac{t}{2}\)-t0) ≠ y (t-t0) = v (\(\frac{t-t_0}{2})\)
∴ The system is time variant.
If x (t) is bounded then y (t) is bounded, so a stable system.
At time t=-2, y (-2) = x (-1), therefore, the response at time t=-2 depends on the excitation at a later time, t=-1, so a Non-causal system.
6. For the system, y (t) = cos 2πt x (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, causal and stable
d) System is Linear, time-invariant and stable
View Answer
Explanation: y2 (t) k cos 2πt v (t) = k y1 (t)
Again, x3 (t) = v (t) + w (t)
So, y3 (t) = cos 2πt [v (t) + w (t)] = y1 (t) + y2 (t)
Since, the system is both homogeneous and additive, therefore it is linear.
Again, y1 (t) = cos 2πt v (t)
And y2 (t) = cos 2πt (t-t0) ≠ y (t-t0)
= cos [2π (t-t0)] v (t-t0)
∴ The system is time variant.
The response at any time t=t0, depends only on the excitation at that time and not on the excitation at any later time, so Causal system.
If x (t) is bounded then y (t) is also bounded, so a stable system.
7. For the system, y (t) = |x (t)|, which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is Linear, time-invariant and causal
c) System is Linear, time-invariant and stable
d) System is Linear, causal and stable
View Answer
Explanation: y1 (t) = |v (t)|, y2 (t) = |k v (t)|= |k|y1 (t)
If k is negative, |k| y1 (t) ≠k y1 (t)
Since it is not homogeneous, so non-linear system.
Again, y1 (t) = |v (t)|, y2 (t) = |y (t-t0)| = y1 (t-t0)
∴ System is time invariant.
The response at any time t=t0, depends only on the excitation at that time and not on the excitation at any later time, so causal system.
If x (t) is bounded then y (t) is also bounded, so stable system.
8. For the system, \(t\frac{dy (t)}{dt}\) – 8 y (t) = x (t), which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is Linear, time-invariant and causal
c) System is time-invariant, causal and stable
d) System is Linear, causal and stable
View Answer
Explanation: All options are linear. Hence linearity is not required to be checked.
Let x1 (t) = v (t), then \(t\frac{dy_1 (t)}{dt}\) – 8 y1 (t) = v (t)
Let x2 (t) = v (t-t0)
Then, \(t\frac{dy_2 (t)}{dt}\) – 8 y2 (t) = v (t-t0)
The first equation can be written as (t-t0) \(t\frac{dy (t-t_0)}{dt}\) – 8 y (t-t0) = x (t-t0)
This equation is not satisfied if y2 (t) = y1 (t-t0). Therefore y2 (t) ≠ y1 (t-t0)
∴ System is time variant.
The response at any time t=t0, depends only on the excitation at that time and not on the excitation at any later time, so Causal system.
The response will increase without bound as time increases, so unstable system.
9. For the system, \(y (t) = \int_{-∞}^{t+3} x(t) \,dt\), which of the following holds true?
a) System is Linear, time-invariant and causal
b) System is time-invariant and causal
c) System is Linear and time-invariant
d) System is Linear and stable
View Answer
Explanation: \(y_1 (t) = \int_{-∞}^{t+3} v(t) \,dt\)
And, \(y_2 (t) = \int_{-∞}^{t+3} kv(t) \,dt\)
= \(k \int_{-∞}^{t+3} x(t) \,dt\) = k y1 (t)
Now, x3 (t) = v (t) + w (t)
And, y3 (t) = \(\int_{-∞}^{t+3}\) [v(t) + w(t)]dt
= \(\int_{-∞}^{t+3}\)v(t) dt + \(\int_{-∞}^{t+3}\)w(t) dt
= y1 (t) + y2 (t)
Since, it is both homogeneous and additive, so linear system.
Again, y1 (t) = \(\int_{-∞}^{t+3}\) v(t) dt
And, y2 (t) = \(\int_{-∞}^{t+3}\) v(t-t0) dt
= y1 (t-t0)
∴ System is time invariant.
The response at any time t=t0, depends partially on the excitation at time t0 < t < (t0 + 3), which are in future, so non-causal system.
The response will increase without bound as time increases, so unstable system.
10. The impulse response of a continuous time LTI system is \(H (t) = (2e^{-2t} -e^{\frac{t-100}{100}}) \,u (t)\). The system is ____________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
View Answer
Explanation: For t<0, h (t) = 0.
Therefore from the definition of causality, we can infer that the system is Causal.
Now, \(\int_{-∞}^∞ |h(t)| \,dt = ∞\)
∴ From the definition of stability, we can infer that the system is unstable.
Hence, the given system is causal but not stable.
11. The impulse response of a continuous time LTI system is H (t) = e-|t|. The system is ___________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
View Answer
Explanation: For t<0,
H (t) ≠ 0
Therefore the system is not causal
Again, \(\int_{-∞}^∞ |h(t)| \,dt\) = \(\frac{1}{3}\) < ∞
∴ The system is stable.
12. The impulse response of a continuous time LTI system is H (t) = e-t u (3-t). The system is __________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
View Answer
Explanation: For t<0, h (t) ≠ 0
Therefore the system is not causal.
Again, \(\int_{-∞}^∞ |h(t)| \,dt = \int_{-∞}^∞ e-t \,u(3-t) \,dt = ∞ \)
∴ System is unstable.
13. The impulse response of a continuous time LTI system is H (t) = e-t u (t-2). The system is __________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
View Answer
Explanation: Since, h (t) = 0 for t<0, so the system is causal.
Again, \(\int_{-∞}^∞ |h(t)| \,dt = \int_{-∞}^∞ e-t \,u(t-2) \,dt\) < ∞
∴ The system is stable.
14. The continuous time convolution integral y(t) = cos πt [u (t+1) – u (t-1) * u(t)] is __________
a) \(\frac{sinπt}{π}\) [u (t+1) – u(t-1)]
b) \(\frac{sinπt}{π}\) u(t-1)
c) \(\frac{sinπt}{π}\) u(t+1)
d) \(\frac{sinπt}{π}\) u(t)
View Answer
Explanation: For t<-1, y (t) = 0
For t<1, y (t) = \(\int_{-1}^t cosπt \,dt = \frac{sinπt}{π}\)
For t>1, y (t) = \(\int_{-1}^t\) cos πt dt = 0
∴ y (t) = \(\frac{sinπt}{π}\) [u (t+1) – u(t-1)].
15. The continuous time convolution integral y(t) = e-3tu(t) * u(t+3) is ___________
a) \(\frac{1}{3}\)[1 – e-3(t+3)] u(t+3)
b) \(\frac{1}{3}\)[1 – e-3(t+3)] u(t)
c) \(\frac{1}{3}\)[1 – e-3t] u(t)
d) \(\frac{1}{3}\)[1 – e-3t] u(t+3)
View Answer
Explanation: For t+3<0 or t<-3, y(t)=0
For t≥-3, y (t) = \(\int_{-∞}^3 e^{-3t} \,dt\)
= \(\frac{1}{3}\)[1 – e-3(t+3)]
∴ y(t) = \(\frac{1}{3}\)[1 – e-3(t+3)] u(t+3).
Sanfoundry Global Education & Learning Series – Signals & Systems.
To practice all exam questions on Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
