Test: Transfer Function - 1 - Electronics and Communication Engineering (ECE) MCQ

Explanation: In a feedback control system, the output is determined by a combination of the input signal and the feedback signal. The input signal is the desired output or reference, while the feedback signal is a portion of the actual output that is fed back into the system to compare with the input. This comparison helps the system to adjust its output to minimize the error between the input and the actual output. By considering both the input and feedback signals, the system can continuously adapt and achieve the desired output.

The impulse response is defined as the output of an LTI system due to a unit impulse signal input being applied at time t = 0.

y(t) = h(t) x(t) = h(t) δ(t)

where δ(t) is the unit impulse function and h(t) is the unit impulse response of a continuous-time LTI system.

Calculations:-

Given-

y(t)  = e^-8t 

x(t) = δ(t)

For calculating the transfer function convert the time domain response into Laplace or S domain.

The circuit is redrawn as shown:

[X(s) – Y(s)] G(s) + X(s) = Y(s)

Given G(s) = 2

(X(s) – Y(s)) 2 + X(s) = Y(s)

= 2 X(s) + X(s) = Y(s) + 2Y(s)

= 3 X(s) = 3 Y(s)

Laplace transform of u(t) is given by 

Taking Inverse Laplace transform

y(t) = t.u(t)

• Any feedback path is considered as a valid feedback path if it does not touches any single node twice in its path.
• Feedback paths in the above diagram are af, be, cd, abef, bcde, and abcdef.
• abef is not a valid feedback path because node 'Q' is repeated twice in the path.
• bcde is not a valid feedback path because node 'R' is repeated twice in the path.
• abcdef is not a valid feedback path because node 'Q' and 'R' is repeated twice in the path.

So, af, be and cd are the only valid feedback paths. The sum of the gains of the feedback paths in the signal flow graph is af + be + cd.

TF = H(s) = 1/s,

Input is unit step, R(s) = 1/s

Now output for t > 0 can be calculated as

C(s) = H(s) x R(s)

C(s) = (1/s) x (1/s)

C(s) = 1 / s²

L^-1[C(s)] = t

C(t) = t = ramp function

• P. P₁ = ab, Δ = 1, L = 0 ,T = ab
• Q. P₁ = a, P₂ = 6 , Δ = 1, L = Δ_k = 0,T  = a+b
• R. P₁ = a, L₁ = b, Δ = 1 - b, Δ₁ =1,
• S. P₁ = a, L₁ = ab, Δ = 1 - ab, Δ₁ = 1, 

While writing the transfer function of this signal flow graph,

e₂= t_ae₁ + t_be₁ = (t_a+ t_b) e₁ 

Then, signal flow graph will lokk like this:

Consider the block diagram as SFG. There are two feedback loop -G₁G₂H₁ and -G₂G₃H₂ and one forward path G₁G₂ G₃ . So (D) is correct option.

Consider the block diagram as a SFG. Two forward path G₁G₂ and G₃ and three loops -G₁G₂ H₂, -G₂H₁, -G₃ H₂
There are no nontouching loop. So (B) is correct.

P₁ = 5 x 3 x 2 = 30, Δ = 1 - (3x - 3) = 10
Δ₁ = 1, 

P₁ = 2 x 3 x 4 = 24 , P₂ = 1 x 5 x 1 = 5
L₁ = -2, L₂ = -3, L₃ = -4, L₄ = -5,
L₁L₃ = 8, Δ = 1 -(-2 - 3 - 4 - 5) + 8 = 23, Δ₁ = 1, Δ₂ = 1 - (-3) = 4, 

X₁ = 1 - H₁

X₂ = (G₁ + G₃) X₁ - X₃H₁

X₃ = X₂G₂ + G₄

Given: 
G(s) = 2 
H(s) = 0.5 and R(s) = 10V 
Output voltage:

= (2/1 + 2 x 0.5) x 15 = 15V 
Error voltage:

= (1/1 + 2 x 0.5) x 15 = 7.5V

Four loops -G₁G₄, -G₁G₂G₅, -G₁,G₂G₅G₇ and -G₁G₂G₃G₃G₇.

There is no nontouching loop. So (B) is correct.

SFG

P₁ = G₂G₅G₆ , P₂ = G₃G₅G₆, P₃ = G₃G₆ , P₄ = G₄G₆
If any path is deleted, there would not be any loop.
Hence Δ₁ = Δ₂ = Δ₃ = Δ₄ = 1 

Transfer function

PK = 5 x 2 x 1 = 10
Δ_K = 1
Δ = 1 - (-4) = 5

= 2