Signals & Systems Questions and Answers Part-8

1. Determine the nature of the system: y[n] = x[n]x[n – 1] with unit impulse function as an input.
a) Dynamic, output always zero, non-invertible
b) Static, output always zero, non-invertible
c) Dynamic, output always 1, invertible
d) Static, output always 1, invertible

Explanation: Since the system depends on present and past values, therefore, it is not memory less(dynamic).
Now, input is a unit impulse function. Unit impulse function = 1 at n = 0, otherwise it is equal to 0.
For, y[0] = x[0]x[-1] = 1 × 0 = 0
For, y[1] = x[1]x[0] = 0 × 1 = 0
For, y[2] = x[2]x[1] = 0 × 0 = 0
For any time, output is always zero.
Since, the output is always same, the system is non-invertible

2. Determine the nature of the system: y(t)= t2 x(t-1)
a) Linear, time invariant
b) Linear, time variant
c) Non-linear, time invariant
d) Non-linear, time variant

Explanation: For linearity:
For input x1(t): y1 (t)= t2 x1 (t-1)
For input x2(t): y2 (t)= t2 x2 (t-1)
⇒ ay1 (t)+by2 (t)= t2 [x1 (t-1)+ bx2 (t-1)] Equation 1
For input x3(t): y3 (t)= t2 x3 (t-1) = t2 [ax1 (t-1)+ bx2 (t-1)] Equation 2
∴ The system is linear.
For time invariancy: Shift in input:
⇒y(t,T)= t2 x(t-1-T)
Shift in output: y(t- T)= (t-T)2 x(t-1-T)
∵ The shift in output is not equal to the shift in input, therefore, the system is time variant.

3. y[n]=rn x[n] is ________ system.
a) LTI
b) Time varying
c) Linear and time invariant
d) Causal and time invariant

Explanation: The input-output relationship of the given system shows it does not satisfy the condition of time-invariant system. Hence it is time varying system.

4. A system is said to be linear if _______
a) It satisfies only the principle of superposition theorem
b) It satisfies only amplitude scaling
c) It satisfies both amplitude scaling and principle of superposition theorem
d) It satisfies amplitude scaling but not the principle of superposition theorem

Explanation: By the definition of linearity a system is said to be linear if it satisfies the condition y1(t) + y2(t) = ax1(t) + bx2(t).

5. If the input-output relationship is given by y(t) = 2x(t) ddx x(t). What kind of system it represents?
a) Linear system
b) Non linear system
c) LTI system
d) Linear but time-invariant system

Explanation: The given input-output relationship of the system does not satisfy the principle of superposition theorem hence it is an example for non linear system.

6. 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

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.

7. 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

Explanation: $$y (t) = ∑_{k=-∞}^n x(k)$$
y (n) = x (n) + x (n-1) + x (n-2) + …….
y (n+1) = x (n+1) + x (n) + (terms after this)
y (n+1) = x (n+1) + y (n)
y (n+2) = x (n+2) + y (n+1)
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.

8. 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

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.

9. 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

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.

10. 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

Explanation: y1 (t) = v ($$\frac{t}{2}$$)
And y2 (t) = k v ($$\frac{t}{2}$$) = k y1 (t)
Then, y3 (t) = v ($$\frac{t}{2}$$) + w ($$\frac{t}{2}$$)
Again, y1 (t) = v ($$\frac{t}{2}$$), y2 ($$\frac{t}{2}$$-t0) ≠ y (t-t0) = v ($$\frac{t-t_0}{2})$$