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11) The feedback system shown below oscillates at 2 rad/s when
(A) K = 2 and a = 0.75 (B) K = 3 and a = 0.75
(C) K = 4 and a = 0.5 (D) K = 2 and a = 0.5
(C) K = 4 and a = 0.5 (D) K = 2 and a = 0.5
$$Open loop Transfer Function, G(s) = \frac{K(S+1)}{s^{3}+as^{2}+2s+1}$$
$$Closed loop Transfer Function with unity feedback, GH = \frac{G(s)}{1+G(s)}$$
$$GH = \frac{\frac{K(S+1)}{s^{3}+as^{2}+2s+1}}{1+\frac{K(S+1)}{s^{3}+as^{2}+2s+1}} $$
$$GH = \frac{K(S+1)}{s^{3}+as^{2}+(K+2)s+(k+1)}$$
Drawing Routh Hurwitz table
$$s^{3} 1 k+2$$
$$s^{2} a k+1$$
$$s^{1} \frac{ak+2a-k-1}{a} 0 $$
For oscillation there should be an all zero row in the Routh table,
$$ \frac{ak+2a-k-1}{a} = 0 $$
Simplifying
$$a = \frac{k+1}{k+2} ----(1)$$
We can either solve this equation by substituting choices given in question(easy method).
Otherwise write auxiliary equation,ie the second row of the Routh table
$$ as^{2}+(K+1) = 0 $$
$$ a(jw)^{2}+(K+1) = 0 $$
$$ a(j2)^{2}+(K+1) = 0 $$
$$ -4a+K+1 = 0 ----(2) $$
Use (1) and (2) to solve for getting k=2 and a= 0.75
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12) The input x(t) and output y(t) of a system are related$$\int_{-\infty }^{t}x(\tau)cos(3 \tau))d \tau$$ The system is
(A) time-invariant and stable (B) stable and not time-invariant
(C) time-invariant and not stable (D) not time-invariant and not stable
(C) time-invariant and not stable (D) not time-invariant and not stable
solution
Time invariance
$$x_{1}(t) => y_{1}(t) = \int_{-\infty }^{t}x_{1}(\tau)cos(3 \tau))d \tau$$
$$x_{2}(t)=x_{1}(t - t_{0}) => y_{2}(t) = \int_{-\infty }^{t}x_{2}(\tau)cos(3 \tau))d \tau$$
$$= \int_{-\infty }^{t}x_{1}(\tau - t_{0} )cos(3 \tau)d \tau$$
$$= \int_{-\infty }^{t-t_{0}}x_{1}(\tau)cos(3 (\tau+t_{0})))d \tau$$
$$\neq y(t - t_{0})$$
Because
$$y(t - t_{0})= \int_{-\infty }^{t-t_{0}}x_{1}(\tau)cos(3 \tau))d \tau$$
Therefore it is not time invariant
stability
For stable systems, bounded input should give bounded output.
consider x(t) as bounded(or of finite value), the other input term cosine is also bounded.
But at output, for each instant of t, for example say t=1, we have to integrate from -infinity to t=1, thus making the output value unbounded due to the presence of infinity.
So system is unstable
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13) The region of convergence (ROC) of Z-transform in the Z-plane of the function $$\frac{1}{3}^{|n|}-\frac{1}{2}^{n}u[n]$$ is
(A) (1/3)<|z|<3 (B) (1/3)<|z|< (1/2)
(C) (1/2)<|z|<3 (D) (1/3)<|z|
Solution
Given
$$\frac{1}{3}^{|n|}-\frac{1}{2}^{n}u[n]$$
Z-transform is given by,
$$X(Z)=\sum_{-\infty }^{\infty } x[n]z^{-n}$$
$$X(Z)=\sum_{-\infty }^{\infty }(\frac{1}{3}^{|n|}-\frac{1}{2}^{n}u[n])z^{-n}$$
Splitting summation range as -infinity->zero and zero->infinity.
At -infinity->zero, |n|= -n
At zero->infinity, |n|= +n
$$X(Z)=\sum_{-\infty }^{0 }(\frac{1}{3})^{-n}z^{-n}+\sum_{0 }^{\infty }(\frac{1}{3})^{n}z^{-n}-\sum_{0 }^{\infty }(\frac{1}{2})^{n}z^{-n}$$
$$X(Z)=\sum_{-\infty }^{0 }(\frac{1}{3}z)^{-n}+\sum_{0 }^{\infty }(\frac{1}{3}z^{-1})^{n}-\sum_{0 }^{\infty }(\frac{1}{2}z^{-1})^{n}$$
put m=-n for first summation part,
$$X(Z)=\sum_{0 }^{\infty }(\frac{1}{3}z)^{m}+\sum_{0 }^{\infty }(\frac{1}{3}z^{-1})^{n}-\sum_{0 }^{\infty }(\frac{1}{2}z^{-1})^{n}$$
Using the identity
$$\sum_{-\infty }^{\infty }a^{n}=\frac{1}{1-a} when: |a|<1$$
$$ X(Z)=\frac{1}{1-\frac{1}{3}z}+\frac{1}{1-\frac{1}{3}z^{-1}}-\frac{1}{1-\frac{1}{2}z^{-1}}$$
conditions are
$$\left |\frac{1}{3}z \right |<1 ---(1) $$
$$\left |\frac{1}{3}z^{-1} \right |<1 ---(2) $$
$$\left |\frac{1}{3}z^{-1} \right |<1 ---(3) $$
(1) gives |z|<3
(2) gives |z|> (1/3)
(2) gives |z|> (1/2)
Final ROC will be intersection of (1)(2) and (3)
or,
(1/2) < |z| < 3
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14) If x =√-1 , then the value of xx is
(A) e-π/2 (B) eπ/2 (C) x (D) 1
Solution
13) The region of convergence (ROC) of Z-transform in the Z-plane of the function $$\frac{1}{3}^{|n|}-\frac{1}{2}^{n}u[n]$$ is
(A) (1/3)<|z|<3 (B) (1/3)<|z|< (1/2)
(C) (1/2)<|z|<3 (D) (1/3)<|z|
Solution
Given
$$\frac{1}{3}^{|n|}-\frac{1}{2}^{n}u[n]$$
Z-transform is given by,
$$X(Z)=\sum_{-\infty }^{\infty } x[n]z^{-n}$$
$$X(Z)=\sum_{-\infty }^{\infty }(\frac{1}{3}^{|n|}-\frac{1}{2}^{n}u[n])z^{-n}$$
Splitting summation range as -infinity->zero and zero->infinity.
At -infinity->zero, |n|= -n
At zero->infinity, |n|= +n
$$X(Z)=\sum_{-\infty }^{0 }(\frac{1}{3})^{-n}z^{-n}+\sum_{0 }^{\infty }(\frac{1}{3})^{n}z^{-n}-\sum_{0 }^{\infty }(\frac{1}{2})^{n}z^{-n}$$
$$X(Z)=\sum_{-\infty }^{0 }(\frac{1}{3}z)^{-n}+\sum_{0 }^{\infty }(\frac{1}{3}z^{-1})^{n}-\sum_{0 }^{\infty }(\frac{1}{2}z^{-1})^{n}$$
put m=-n for first summation part,
$$X(Z)=\sum_{0 }^{\infty }(\frac{1}{3}z)^{m}+\sum_{0 }^{\infty }(\frac{1}{3}z^{-1})^{n}-\sum_{0 }^{\infty }(\frac{1}{2}z^{-1})^{n}$$
Using the identity
$$\sum_{-\infty }^{\infty }a^{n}=\frac{1}{1-a} when: |a|<1$$
$$ X(Z)=\frac{1}{1-\frac{1}{3}z}+\frac{1}{1-\frac{1}{3}z^{-1}}-\frac{1}{1-\frac{1}{2}z^{-1}}$$
conditions are
$$\left |\frac{1}{3}z \right |<1 ---(1) $$
$$\left |\frac{1}{3}z^{-1} \right |<1 ---(2) $$
$$\left |\frac{1}{3}z^{-1} \right |<1 ---(3) $$
(1) gives |z|<3
(2) gives |z|> (1/3)
(2) gives |z|> (1/2)
Final ROC will be intersection of (1)(2) and (3)
or,
(1/2) < |z| < 3
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14) If x =√-1 , then the value of xx is
(A) e-π/2 (B) eπ/2 (C) x (D) 1
Solution
ejπ/2 = cos(π/2) + j sin(π/2)
= 0+ j
=
j = √-1
= x
xx = (ejπ/2)x
= (ejπ/2)j
= (ej*j*π/2)
= e-π/2 ( because j2 = -1)
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Solution
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16) The circuit shown is a
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15) A system with transfer function
$$\frac{(s^{2}+9)(s+2)}{(s+1)(s+3)(s+4)} $$
is excited by sin(ωt) . The steady-state output of the system is zero at
(A) w = 1 rad/s (B) w = 2 rad/s
(C) w = 3 rad/s (D) w = 4 rad/s
(C) w = 3 rad/s (D) w = 4 rad/s
For a stable LTI system with transfer function(H(jω), steady-state response to a sinusoidal input sin(ωt) is given by asin(ωt+φ), where a=|H(jω)| and φ=angle(H(jω))
a sin(ωt+φ)=0 (given condition)
or a=0(taking magnitude only)
The system given here is stable since there is no poles on right side of Jω axis, we can use above given relation here
$$H(jw)=\frac{((jw)^{2}+9)(jw+2)}{(jw+1)(jw+3)(jw+4)} $$
$$H(jw)=\frac{(-w^{2}+9)(jw+2)}{(jw+1)(jw+3)(jw+4)} $$
As per condition given in question,steady state response = 0
$$|H(jw)|=\frac{|(-w^{2}+9)||(jw+2)|}{|(jw+1)||(jw+3)||(jw+4)|} =0$$
We are equating numerator to zero, since denominator cannot be zero
$$|(-w^{2}+9)||(jw+2)|=0$$
$$(-w^{2}+9)\sqrt{w^{2}+2^{2}}=0$$
$$(-w^{2}+9)=0$$
or
$$\sqrt{w^{2}+2^{2}}=0$$
which gives ω=3 (rest all are negative values-thus avoided)
16) The circuit shown is a
(A) low pass filter with f3dB =1/(R1+ R2
)C rad/s
(B) high pass
filter with f3dB
=1/R1C
rad/s
(C) low pass
filter with f3dB
=1/R1C
rad/s
(D) high pass
filter with f3dB
=1/(R1+
R2 )C rad/s
Solution
The circuit given is a practical differentiator and R2
has no influence in cutoff frequency
Hence it is a high pass
filter with f3dB
=1/R1C
rad/s
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17) The state transition diagram for the logic circuit shown is
Solution
Start with Q = 0 condition
1)When Q = 0 and A=0
Qbar =1
X0 of mux is selected
So Qbar will goto Y, from there to D ; D=Qbar=1
On next clock D-flipflop output will be Q=1
2)When Q = 0 and A=1
Qbar =0
X1 of mux is selected
So Q will goto Y, from there to D ; D=Q=0
On next clock D-flipflop output will be Q=0
3)When Q = 1 and A=0
Qbar =0
X0 of mux is selected
So Qbar will goto Y, from there to D ; D=Qbar=0
On next clock D-flipflop output will be Q=0
4)When Q = 1 and A=1
Qbar =0
X1 of mux is selected
So Q will goto Y, from there to D ; D=Q=1
On next clock D-flipflop output will be 1 ie Q=1
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17) The state transition diagram for the logic circuit shown is
Solution
Start with Q = 0 condition
1)When Q = 0 and A=0
Qbar =1
X0 of mux is selected
So Qbar will goto Y, from there to D ; D=Qbar=1
On next clock D-flipflop output will be Q=1
2)When Q = 0 and A=1
Qbar =0
X1 of mux is selected
So Q will goto Y, from there to D ; D=Q=0
On next clock D-flipflop output will be Q=0
3)When Q = 1 and A=0
Qbar =0
X0 of mux is selected
So Qbar will goto Y, from there to D ; D=Qbar=0
On next clock D-flipflop output will be Q=0
4)When Q = 1 and A=1
Qbar =0
X1 of mux is selected
So Q will goto Y, from there to D ; D=Q=1
On next clock D-flipflop output will be 1 ie Q=1
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