Test: Time Scaling, Linear Time Invariant & Casual Systems - Electrical Engineering (EE) MCQ

We have x′(t)=

x′(t) = r(t) – 2r(t − 2) + r(t − 4)

Now,x(t) = x′(2t) compression by 2.

Thus,

x(t) = r(2t) – 2r(2t − 2) + r(2t − 4).

Alternatively, x(t) may also be written as

⇒ {—5, 6, 17, 42, | 38, 30, 8}

y(n) + 5y(n-1) + 6y(n-2) = 5x(n) – 2x(n-1)

Taking the z-transform

Y(z) + 5z^-1 Y(z) + 6z^-2 Y(z) = 5X(z) - 2z^-1X(z)

y(z) [1 + 5z^-1 + 6z^-2] = X(z) [5 – 2z^-1]

Taking inverse Z-transform,

h(n) = 17(-3)ⁿ u(n) – 12(-2)ⁿ u(n)

For an LTI system the output is given by

y(n) = x(n) ⨂ [h₁(n) ⨂ h₂(n)]

given

h₁(n) = δ(n)-5δ(n­-1)

h₂(n) = 5ⁿ u(n)

h₁(n) ⨂ h₂(n) = [δ(n) - 5δ(n-1) ⨂ 5ⁿu(n)

= 5ⁿu(n) – 5 (5^n-1 u(n-1))

= 5ⁿu(n) – 5ⁿu(n-1)

= 5ⁿ [u(n) – u(n-1)]

A = -5

B = 9

Poles of the H(z) is at |z| = 2, |z| = 3. So, these are 3 possible ROC.

Case (1) 2 < |z|<3

Case (2) |z| ≤ 2

Case (3) |z| ≥ 3

Only in case (2) signal is anti-causal and stable for stable ROC must cuts the unity circle so for ROC |z| ≤ 2

h(n)=-5(-2ⁿu(-n-1) + 9(-3ⁿu(-n-1)

h(n)=5×2ⁿu(-n-1) – 9 × 3ⁿu(-n-1)

Note: u(n) – u(n-1) = δ(n)

so h₁(n) ⨂ h₂(n) = 5ⁿδ(n) = δ(n)

y(n) = x(n) ⨂ [h₁(n) ⨂ h₂(n)]

= (1/3)ⁿu(n) ⨂ [δ(n)]

= 

Hence, option B is correct.

Also,

Hence, option C is also correct.

Apply ‘z’ transform on both sides.

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.

Concept:

Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left. 

Time scaling property: A signal is scaled in the time domain with the scaling factor 'a'. 

If a > 1, then the signal is contracted by a factor of 'a' along the time axis. 

if a < 1, then the signal is expanded by a factor of 'a' along the time axis. 

Important Analysis:

Whenever there is a combination of operation care should be taken while following the sequence of operation.

Let, we have x(at - b)

1) Correct approach:

x[a(t - b/a)]

First, time scale the signal by the factor a.

Then, time shift the signal by b/a 

2) Incorrect approach:

x[a(t - b/a)]

First, time shift the signal by b/a.

Then, time scale the signal by a.

or, another approach can be followed

3) Correct approach:

x[at - b]

First, time shift the signal by b.

Then, time scale the signal by a

4) Incorrect approach:

x[at - b]

First, time scale the signal by a

Then, time shift the signal by b.

Application

Given:

x(t)

To plot x(10t - 2)

We can use 3rd approach, 

1. Time shift the signal by 2. 

2. Time scale the signal by 10. 

The waveforms can be shown below:

Concept:

An LTI system is stable if and only if the ROC of the impulse function H(s) includes the jω axis.

For Causal System → ROC is to the right side of the rightmost pole.

For Anti Causal System → ROC is to the left side of the left-most pole.

Observations:

• For a causal system to be stable, the poles must lie on the left half of the complex plane (to include the jω axis)
• A causal system with a pole on the right side cannot be BIBO stable because it's ROC can never include the jω axis. (Statement (II) is therefore correct)
• A BIBO system with a pole in the right half of the complex plane is stable if the system is anti-causal, as this will include the jω axis. (Statement (I) is therefore incorrect)