The transmission parameter of two port network
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Solution
The transmission equation are
V1=AV2+ B(– I2)
I1=CV2 +D(– I2)
\(A=\frac{V_{1}}{V_{2}}\mid _{I_{2}=0}=1\)
\(B=-\frac{V_{1}}{V_{2}}\mid _{V_{2}=0}=0\)
\(C=-\frac{I_{1}}{V_{2}}\mid _{I_{2}=0}=\frac{1}{2}\)
\(D=-\frac{I_{1}}{I_{2}}\mid _{V_{2}=0}=\frac{1}{2}\)
The switch in figure shown below has been in position 1 fora long time and is then moved to position 2 at t = 0.
The value of Vc(0+),\(\frac{dv_{c}(t)}{dt}\mid _{t=0}\) are respectively
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Solution
The switch in figure has been in position 1 fora longtime and is then moved to position 2 at t = 0.
VC(0+) = VC(0-) = voltage across resistance 50 kW
VC(0+)=\(6\times \frac{50}{25+50}=4v\)
4 H inductors act like an open circuit at t = 0+
Total current =–10 –4 =–14 mA at t = 0+
thus\(\frac{dv_{c}(t)}{dt}\mid _{t=0}\)=\(\frac{-14mA}{20\mu F}\)=-700V/s
Equivalent inductive reactance of the coupled circuit shown in figure.
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Solution
The dotted convention for the circuit is
L1+L2+ L3– M1– M2+ M3
Zeq=j3 +j5 +j6 + 2j2 – 2j3 + 2j4
=j12 W
In the given figure what will be V01/V02. If the input is existed by the sinusoidal of peak voltage Vm
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Solution
The circuit shown in figure act as voltage multiplication circuit.
The V01= 3Vm(voltage trippler circuit)
V02= 4Vm(voltage quadruple circuit)
\(\frac{v_{01}}{v_{02}}=\frac{3}{4}\)
The values of A, B, C and Dsatisfy the following simultaneous boolean equations is
\(\overline{A}\)+ AB= 0, AB= AC
AB+ AC+ CD= \(\overline{CD}\)
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Solution
Given \(\overline{A}\)+ AB= 0 and AB= AC
Let A=1 and B= 0
then,\(\overline{A}\)+ BA=\(\overline{c1}\) + 1.0
=0+ 0 = 0
And also AB=1.0 = 0
AC=1.0 = 0 then, C = 0
We have also given
=AB+ A\(\overline{c}\)+ CD= 1.0 +1.\(\overline{0}\)+ 0.D
=0+1.1 + 0=0+ 1 + 0 = 1
So, \(\overline{CD}\) should be 1
for this let D = 0
⇒\(\overline{CD}\)=\(\overline{0.0}\)
=0= 1
Hence ,A=1, B= 0, C= 0, D = 0
If f(0) = 2 and f (x) =\(\frac{1}{5-x^{2}}\), then lower and upper bound off (t) estimated by the mean value theorem are
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Solution
From mean value theorem
\(\frac{f(1)-f(0)}{1-0}\)=f'(x),x∈[0, 1]
f (1)=2\(\frac{2x}{(5-x^{2})^{2}}\)
[∵ f(0) = 2]
Lower bound of f(1) = 2 +0 =2
Upper bound of f(1)=2+\(\frac{2}{(5-1)^{2}}=2.125\)
The singular solution of p = tan (px– y) is
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Solution
The given differntial equation is
p = tan (px– y)ory= px – tan-1 p...(1)
This equation is in the form of the Clairaut’s equation.
The general solution of this equation is
y=(x– tan-1 c)...(2)
Differentiating (2) with respect to
we get,0=x-\(\frac{1}{1+c^{2}}\)
⇒x=\(\frac{1}{1+c^{2}}\)
⇒\(\frac{1-x}{x}\)
Eliminating c between (2) and (3) the required singular solution is
\(y=\sqrt{\frac{1-x}{x}}x-tan^{-1}\sqrt{\frac{1-x}{x}}\)
Which of the following is not a solution of the Ordinary Differential Equations (ODE) 2xy’ = 10x3 y25 + y?
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Solution
2xy'= 10x3 y5 + y
Linear equation \(\frac{dy}{dx}\)+py=Q
where, P and Q are f(X).
This can be solved by Integrating Factor(IF)=\(e^{\int pdx}\).
\(ye^{\int pdx}=\int Qe^{\int pdx}dx+c\)
2xy'= 10x3 y5 + y
2x\(\frac{dy}{dx}\)=10x3y5+y
⇒2x\(\frac{dy}{dx}-y=10x^{3}y^{5}\)
\(\frac{1}{y^{5}}\frac{dy}{dx}-\frac{1}{2xy^{4}}=10x^{2}\)
Let t=\(\frac{1}{y^{4}}\)
Differentiating wrt x,
\(\frac{dt}{dx}=\frac{dy}{dx}\frac{1}{y^{5}}\)hence
\(\frac{dt}{dx}+\frac{t}{2x}=10x^{2}\)
IF =\(e^{\int \frac{1}{2x}dx}=e^{in\sqrt{x}}=\sqrt{x}\, since^{In\, x} = x\)
Hence correct solution is
\(y\sqrt{x}=\int 10x^{2}\sqrt{x}+c\)
\(y\sqrt{x}=\frac{20}{7}x^{\frac{7}{2}}+C\)
We can also write this as
\(ye^{\frac{1}{2}}in(x)dx=\int (10x^{2}e^{\int \frac{1}{2x}})dx+c,since^{Inx\, } =x\)
The transfer function of two compensators are given below
\(C_{1}=\frac{10(s+1)}{(s+10)},C_{2}=\frac{s+10}{10(s+1)}\)Which one of the following statement is correct?
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Solution
Given
\(C_{1}=\frac{10(s+1)}{(s+10)},C_{2}=\frac{s+10}{10(s+1)}\)
The z-transform of a signal x[n] is given by 4z-3+ 3z-1 + 2– 6z2 + 2z3. It is applied to a system, with a transfer function H(z) = 3z–1– 2. Let the output to be y[n]. Which of the following is true?
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Solution
Given X(z) = 4z-3+ 3z–1+ 2 – 6z2+ 2z3
H(z) = 3z-1– 2
Y(z) =X(z) H(z)
Y(z) = 12z-4– 8z-3+ 9z-2– 4 – 18z+ 18z2 – 4z3
The Y(z) contain term z, z2, z3.
Hence, y(n) has the term having positive values.
Hence, y(n) is non causal with finite support.