A square matrix B is skew symmetric if
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Solution
B-1 = BT
P=\(p=\frac{1}{2\sqrt{2}}\begin{bmatrix} 2\, -2 & \\ 2\, 2 & \end{bmatrix}\)
and \(\overrightarrow{x}=[x_{1},x_{2}],\left | \overrightarrow{x} \right |=x_{1}\widehat{i}+x_{2}\widehat{j}\)
Length \(\left | \overrightarrow{x}\right |=\sqrt{x_{1}^{2}+x_{2}^{2}}\)
Now,(px)=\(\frac{1}{2\sqrt{2}}\begin{bmatrix} 2\, -2 & \\ 2\, 2 & \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}\)
(px)=\(\frac{1}{2\sqrt{2}}\begin{bmatrix} 2x_{1}\, -2x_{2} & \\ 2x_{1}\, +2x_{2} & \end{bmatrix}\)
=\(\frac{1}{\sqrt{2}}\left [ \left ( x_{1}-x_{2} \right )\widehat{i}(x_{1}+x_{2})\widehat{j} \right ]\)
\(\left | p\overrightarrow{x} \right |=\frac{1}{\sqrt{2}}\sqrt{(x_{1}-x_{2})^{2}+(x_{1}+x_{2})^{2}}\)
=\(\frac{1}{\sqrt{2}}\sqrt{x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}+x_{1}^{2}+x_{2}^{2}+2x_{1}x_{2}}\)
\(=\frac{1}{\sqrt{2}}.\sqrt{2}\sqrt{x_{1}^{2}+x_{2}^{2}}\)
=\(\sqrt{x_{1}^{2}+x_{2}^{2}}\)
=\(\left | \overrightarrow{x} \right |\)
\(\left | p\overrightarrow{x} \right |=\left | \overrightarrow{x} \right |for\,all\, \overrightarrow{a}\)
The differential equation \(\frac{dx}{dt}=\frac{1-x}{\tau }\) is discretised using Euler’s numerical integration method with a time step ΔT > 0. What is the maximum permissible value of ΔT to ensure stability of the solution of the corresponding discrete time equation?
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Solution
The maximum permissible value of ΔT = 2τ
Equation ex– 1 = 0 is required to be solved using Newton’s method with an initial guess x2= –1. Then, after one step of Newton’s method, estimate x21 of the solution will be given by ______________
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Solution
Let f(x) = ex– 1 = 0
⇒f'(x)=ex
Now, by Newton-Raphson method,
xk+1=xk-\(\frac{f(x_{1})}{f'(x_{1})}\)
xk+1=xk-\(\frac{(e^{x_{k-1}})}{(e^{x_{k}})}=\frac{(e^{x_{k}}.x_{k})-e^{xk+1}}{e^{x_{k}}}\)... (i)
But at (k = 0), x0 = –1
x1=\(\frac{(x_{0}.e^{x_{0}})-e^{x_{0}}+1}{e^{x_{0}}}=\frac{(-1.e^{-1})-e^{-1}+1}{e^{-1}}\)
=\(\frac{\left ( \frac{-1}{e}-\frac{1}{e}+1 \right )}{\frac{1}{e}}=\frac{(-2)+e}{1}=x_{1}= 0.71828.\)
A differential equation \(\frac{dx}{dt}\)= e–2t u(t) has to be solvedu sing trapezoidal rule of integration with a step size h = 0.01s. Function u(t)indicates a unit step function. If x (0–) = 0, then value of x at t = 0.01 s will be given by
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Solution
The given differential equation
\(\frac{dx}{dt}=^{-2t}\)u(t)
⇒x =\(\int e^{-2t}\)ut dt(After integrating)
Let x =\(\int\)f(t) dt⇒[f(t) = e–2t u(t)]
where, u(t) = unit step function
that means v(t)= 1
Given, step size h =0.01 s
Also,x(0) = 0
Now, f(0) = e–0.1 =1 and f(0.01) = e–2(0.01).1 = e–0.02
Then, by trapezoidal rule,
x =\(\int\)f(t)dt=h⁄2[(f(0)+f(0.01)]
=\(\frac{0.01}{2}[1+e^{-0.02}]=\frac{0.01}{2}(1.98)\)
x = 0.0099.
f(x, y) is a continuous function defined over (x, y)∈[0,1]×[0, 1]. Given the two constraints, x > y2 and y > x2, the volume under f(x, y) is
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Solution
Given that,
f(x, y) is a continuous function.
and (x, y)∈[0, 1] × [0, 1]
Then, the intersection point of the curves x = y2 and
y = x2 is
y = y4 ⇒ y(y3– 1) = 0y(y– 1) (y2 + 1 + y) = 0
⇒y = 0 and 1
⇒x =0 and 1
So, points are (0, 0) and(1, 1)
So, the volume under f(x, y) is
\(\int_{y=0}^{y=1}\int_{x=y^{2}}^{x=\sqrt{y}}f(x,y)dxdy\)
An ammeter has a range of 0 to 50 A. The instrument gave the following readings.
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Solution
The non-linearity of the instrument in terms of full scale reading in percent is________.
Full scale deflection of ammeter is 0 to 50 A,
Non linearity=\(=\frac{Maximum\, deviation\, of\, output}{Full\, scale\, deflection}\times 100\)
=\(\frac{45-40}{50}\)×100=10%
A discrete signal y[n] is related to another discrete signal x[n] given as, y[n]= x[n/2] when n is even. For odd values of n holds the value of earlier samples. Express z-transform of y[n] in terms of z-transform of x[n].
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Solution
(i)y[n]=\(\left [ \frac{n}{2} \right ]\) when n is even.
(ii)For odd values of n, y [n] holds the value of earlier samples.
Y(z) = y(0) + y(1) z-1+ y(2)z-2+ y(3)z-3+ .......∞
Y(z) = x(0) + x(0) z-1+ x(1)z-2+ x(1)z-3+ x(2) z-4+ x(2)z-5 +x(3)z-6 + x(3)z-7+ .....∞
Y(z) = [x(0) + x(1)z-2+ x(2)z–4+ x(3)z–6+ ....... ∞]+ z-1[x(0) +x(1)z-2+ x(2)z-4+ x(3)z-6+ ......∞]
Y(z) = [x(0) + x(1)(z2)-1+ x(2)(z2)-2x(3)(z2)–3+ ....... ∞]+ z-1[x(0) + x(1)(z-2)-1+ x(2)(z2)-2+ ......∞]
Y(z) = X(z2) + z-1X(z2)
Y(z) = (1 + z-1) X (z2)
The forced response for the capacitor voltage vf(t) is
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Solution
\(\frac{d^{2}v(t)}{dt^{2}}+\frac{R}{L}\frac{dv(t)}{dt}+\frac{1}{LC}v(t)=\frac{V_{s}(t)}{LC}\)
\(\frac{d^{2}v(t)}{dt^{2}}+70\frac{dv(t)}{dt}\)+12000v(t)=12000 Vs(v)
Trying vf= A + Bt
0 +70B + 12000 (A +Bt) = 12000 (0.2t)
⇒B = 0.2
A=-70\(\times \frac{0.2}{12000}=-1.17-10^{-3}\)
Vf= 0.2t – 1.17×10-3 V
In the following lattice network the value of RL for the maximum power transfer to it is
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Solution
The circuit is as shown below
RTH= 7 ||5 + 6 || 9 = 6.52 Ω
For maximum power transfer
RL= RTH= 6.52 Ω
In the circuit of the fig., the value of the voltage source E is_________.
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Solution
Going from 10 V to 0 V
10 + 5 + E+ 1 = 0 or E =– 16 V