A discrete signal y[n] is related to another discrete signal x[n] given as, y[n] = × [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]=x\(\left [ \frac{n}{2}\right ]\) when n is even.
(ii)For odd values of n, y [n] holds the value of earlier samples.
z-transform of y[n] in terms of z-transform of x[n].
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-4x(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)
In the voltage doubler circuit shown in the figure, the switch S is closed at t= 0. Assuming diodes D1 and D2 to be ideal,load resistance to be infinite and initial capacitor voltage to be zero, the steady voltage capacitor C1 and C2 will be
The shift register is shown in figure is initially loaded with the bit pattern 1010.Subsequently the shift register is clocked and with each clock pulse, the pulse gets shifted by one bit position to right with each shift, the bit at the serial is pushed to the left most position (msb). After how many clock pulses will the contact of the shit register become 1010 again?
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Solution
We can write
Y2=Q1 ⊕ Q0
Y21=Q2 ⊕ Y2
Three devices A, Band C have to be connected to a 8085 microprocessor. Device 4 has the highest priority and device C has the lowest priority. In this context which of the following is correct assignment of interrupt inputs?
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Solution
The priority from highest to lowest is TRAP, RST 7.5,RST 6.5,RST 5.5,INTR
TRAPS are usually reserved for unusual event such as power supply failure.
IN/OUT device are usually using other four interrupt with priority required.
Hence,A uses RST 7.5, Buses RST 6.5 and Cuses RST 5.5
Which of the following matching is correct?
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Solution
H(s) =\(\frac{s}{s+1}\)
The transfer function has pole at –1 and zero at 0.Hence, h(t) =d(t) – e-t u(t)
H(s) =\(\frac{s}{(s+1)^{2}}\)
h(t)=te-t u(t)
H(s)=\(\frac{1}{s^{2}+s+1}\)
s2+ 2ξωns+ ω\(_{n}^{2}\)=0
2ξωn=1
ξ=1⁄2<1
Underdamp
H(s)=\(\frac{1}{s^{2}+1}\)
h(t)=sin t
The system is controllable for what values of a,b and c______ ?
\(\begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{13} \end{bmatrix}=\begin{bmatrix} 0\, 1\, 0 & & \\ 0\, 0\, 1 & & \\ -a\, -b\, -c & & \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}+\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}u\)-
Solution
\(\begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{13} \end{bmatrix}=\begin{bmatrix} 0\, 1\, 0 & & \\ 0\, 0\, 1 & & \\ -a\, -b\, -c & & \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}+\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}u\)
Controll ability matrix of system is Qc= [B AB A2B... An–1B]
AB=\(\begin{bmatrix} 0\\ 1\\ -c \end{bmatrix}\)
A2B=\(\begin{bmatrix} 0\, 0\, 1 & & \\ -a\, -b\, -c & & \\ ac\, -a+cb\, -b+c^{2} & & \end{bmatrix}\)
\(\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}=\begin{bmatrix} 1\\ -c\\ -b+c^{2} \end{bmatrix}\)
∵ Q2=\(\begin{bmatrix} 0 \,0\, 1 & & \\ 0\, 1\, -e & & \\ 1\, -e\, e^{2}-b & & \end{bmatrix}\)
\(\left | Q_{e} \right |=-1\)
So for any value of a, band c system is completely controllable.
Consider the system described by
\(\begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2} \end{bmatrix}=\begin{bmatrix} 1\, 1 & \\ -2\, -1 & \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}+\begin{bmatrix} 0\\ 1 \end{bmatrix}U\) \(y=[10]\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}\)The system is
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Solution
\(\begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2} \end{bmatrix}=\begin{bmatrix} 1\, 1 & \\ -2\, -1 & \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}+\begin{bmatrix} 0\\ 1 \end{bmatrix}U\)
\(y=[10]\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}\)
A=\(\begin{bmatrix} 1\, 1 & \\ -2\, -1 & \end{bmatrix}\)
B=\(\begin{bmatrix} 0\\ 1 \end{bmatrix}\)
C=[1,0]
Matrix[B AB]=\(\begin{bmatrix} 0\, +1 & \\ 1\, -1 & \end{bmatrix}\)
The rank of matrix is 2, the system is completely state controllable
Consider the matrix [CT ATCT] =\(\begin{bmatrix} 1\, 1 & \\ 0\, 1 & \end{bmatrix}\)
Rank of matrix [CT AT CT] is 2.
Hence, the system is completely observable.
A network is shown below
The voltage across AF in volt is _________
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Solution
The circulating current in the loop ABCD is
\(\frac{10}{2+3}=\frac{10}{5}=2A\)
Voltage across AC = 10V
Drop across EF = 2× 5= 10V
Drop across AF = –10 = –5 V
The image impedance of the network looking form the input port in Ω is _________.
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Solution
We can conveniently express the image impedance in terms of ABCD parameter
Image impedance z0i=\(\sqrt{\frac{AB}{CD}}\)
But for a symmetrical ABCD network
A = DSo, z0=\(\sqrt{\frac{B}{C}}\)
B = z1+ z2+\(\frac{z_{1}z_{2}}{z_{3}}\)
B =100 + 100+\(\frac{100\times 100}{400}=225\Omega\)
c=\(\frac{1}{z_{3}}=\frac{1}{400}\Omega\)
z0=\(\sqrt{\frac{B}{C}}=\sqrt{\frac{225}{1/400}}=300\Omega\)
The low frequency gain of the low pass filter in dB shown in the given figure is __________.
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Solution
R = 100 kΩ
R1= 1kΩ
C = 0.01 μF\(A=\left | \frac{V_{2}(s)}{V_{1}(s)}\right |=\left | \frac{R}{[R(sC+1)]R_{1}} \right |\)
A=\(\left | \frac{100}{j10^{-3}\omega +1} \right |\)
\(\left | A \right |_{db}=20 log 100\cong 40dB\)