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The signal flow graph given below.
The transfer function=
=
The output of the feedback control system must be a function of:
The response of the control system is the output of the control system that depends upon the transfer function of the system and feedback system and also upon the input of the system.
The unit impulse response of a certain system is found to be e^{8t}. Its transfer function is _______.
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 continuoustime 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.
For the system shown in the figure, Y(s)/X(s) = _________. (Answer in integer )
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)
Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is
Laplace transform of u(t) is given by
Taking Inverse Laplace transform
y(t) = t.u(t)
The sum of the gains of the feedback paths in the signal flow graph shown in fig. is
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.
A control system whose step response is 0.5(1+e^{2t}) is cascaded to another control block whose impulse response is e^{t}. What is the transfer function of the cascaded combination?
To find the transfer function of the cascaded combination, you need to multiply the transfer functions of the individual control blocks.
The transfer function of the first control block is 0.5(1+e^2t), which can be written as 0.5/(s+2).
The transfer function of the second control block is e^t, which can be written as 1/(s+1).
To find the transfer function of the cascaded combination, you need to multiply these two transfer functions:
1/(s+1) * 0.5/(s+2) = 0.5/(s+1)(s+2)
Therefore, the correct answer is 1, "0.5/(s+1)(s+2)."
The overall transfer function C/R of the system shown in fig. will be:
For the signal flow graph shown in fig. an equivalent graph is
While writing the transfer function of this signal flow graph,
e_{2}= t_{a}e_{1} + t_{b}e_{1 }= (t_{a}+ t_{b}) e_{1}
Then, signal flow graph will lokk like this:
The block diagram of a system is shown in fig. The closed loop transfer function of this system is
Consider the block diagram as SFG. There are two feedback loop G_{1}G_{2}H_{1} and G_{2}G_{3}H_{2} and one forward path G_{1}G_{2} G_{3} . So (D) is correct option.
For the system shown in fig. transfer function C(s) R(s) is
Consider the block diagram as a SFG. Two forward path G_{1}G_{2 }and G_{3 }and three loops G_{1}G_{2} H_{2}, G_{2}H_{1}, G_{3} H_{2}
There are no nontouching loop. So (B) is correct.
In the signal flow graph shown in fig. the transfer function is
P_{1} = 5 x 3 x 2 = 30, Δ = 1  (3x  3) = 10
Δ_{1} = 1,
In the signal flow graph shown in fig. the gain C/R is
P_{1} = 2 x 3 x 4 = 24 , P_{2} = 1 x 5 x 1 = 5
L_{1} = 2, L_{2} = 3, L_{3} = 4, L_{4} = 5,
L_{1}L_{3} = 8, Δ = 1 (2  3  4  5) + 8 = 23, Δ_{1} = 1, Δ_{2} = 1  (3) = 4,
The gain C(s)/R(s) of the signal flow graph shown in fig.
X_{1} = 1  H_{1}
X_{2} = (G_{1} + G_{3}) X_{1}  X_{3}H_{1}
X_{3 }= X_{2}G_{2} + G_{4}
The negative feedback closedloop system was subjected to 15V. The system has a forward gain of 2 and a feedback gain of 0.5. Determine the output voltage and the error voltage.
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
For the block diagram shown in fig. transfer function C(s)/R(s) is
Four loops G_{1}G_{4}, G_{1}G_{2}G_{5}, G_{1},G_{2}G_{5}G_{7} and G_{1}G_{2}G_{3}G_{3}G_{7}.
There is no nontouching loop. So (B) is correct.
For the block diagram shown in fig. the numerator of transfer function is
SFG
P_{1} = G_{2}G_{5}G_{6} , P_{2} = G_{3}G_{5}G_{6}, P_{3} = G_{3}G_{6} , P_{4} = G_{4}G_{6}
If any path is deleted, there would not be any loop.
Hence Δ_{1} = Δ_{2} = Δ_{3} = Δ_{4} = 1
For the block diagram shown in fig. the transfer function C(s)/R(s) is
Transfer function
PK = 5 x 2 x 1 = 10
Δ_{K} = 1
Δ = 1  (4) = 5
= 2
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