If each stage had a gain of 10 dB, and noise figure of 10 dB,then the overall noise figure of a two-stage cascade amplifier will be ___________
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Solution
F = F1 + (F – 1 / ga1) = 10 + 9 / 10 = 10.9
Consider the following op-amp circuit
input Vi and current i are related by
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Solution
i=\(1.\frac{d(v_{i}-v_{1})}{dt}=\frac{dv_{i}}{dt}-\frac{dv_{1}}{dt}\)
\(\frac{(V_{2}-V_{1})}{2}=1.\frac{dv_{i}}{dt}\)
⇒V2=Vi+2\(\frac{dv_{i}}{dt},\frac{v_{1}-v_{i}}{2}=\frac{v_{i}-v_{2}}{1}\)
⇒v1=vi-4\(\frac{dv_{i}}{dt}\)
∴ \(\frac{dv_{1}}{dt}=\frac{dv_{i}}{dt}-4\frac{d^{2}v_{i}}{dt^{2}}\)
∴ i=\(\frac{dv_{i}}{dt}-\frac{dv_{1}}{dt}=\frac{dv_{i}}{dt}-\left [ \frac{dv_{i}}{dt}-4\frac{d^{2}v_{i}}{dt^{2}} \right ]\)
∴i=\(0.25\left ( \int_{0}^{t}\int_{0}^{\lambda }V_{i}(x)dx \right )d\lambda\)
The intermediate frequency of a super heat receiver is 450KHz if it is tuned to 1200 KHz, the image frequency will be___________.
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Solution
Image frequency = 1200 + 2 × 450 = 2100 kHz.
A system has the following transfer function:
\(G(s)=\frac{100(s+5)(s+50)}{s^{4}(s+10)(s^{2}+3s)+10}\)The type and order of the system are respectively
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Solution
\(G(s)=\frac{100(s+5)(s+50)}{s^{4}(s+10)(s^{2}+3s)+10}\)
Since in the denominator we have s4 common, it is a type 4 system. The highest power of s in the denominator is 7 hence it is a 7th-order system.
Consider the signal x(t), shown in fig.let h(t) denotes the impulse response the filter-matched to x(t) with h(t) being non-zero only in the interval 0 to 4 sec. The slope of h(t) in the interval 3 < t < 4 sec is __________.
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Solution
–1
h(t) = x(4 – t)Slope in region t = 3 to 4= – 1
Given Z-transform, X(z)=\(\frac{3z^{2}-\frac{1}{4}z}{z^{2}16}\),\(\left | z \right |\)>4 The time signal x(t) will be
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Solution
x[n] is right sided
\(x(z)=\frac{z-\frac{1}{4}z^{-1}}{1-16z^{-1}}=\frac{\frac{49}{32}}{1+4z^{-1}}+\frac{\frac{47}{32}}{1-4z^{-1}}\)
⇒\(\left [ \frac{49}{32}(-4)^{n}+ \frac{47}{32} 4^{n}\right ]u[n]\)
The table gives the values of a function f (x) corresponding the values of x at the interval of one.
Using the Simpson’s 1⁄3rd. The value of the function between 0 to 6 is _____
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Solution
\(\int_{0}^{6}\)f(x)dx=1⁄3[(1 + 0.027) + 4(0.5 + 0.1+ 0.385) + 2
(0.2 + 0.0588) = 1.3662
Using the Newton-Raphson method, find the first approximation of the function f (x) =xex–2 upto three places of decimals with x0 = 1.
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Solution
f(x) = xex– 2
f(1) = e– 2 = 0.7185
f'(x) = xex + ex
f'(x) = e+ e = 2e =5.4366x1+x0-\(\frac{f(x_{0})}{f'(x_{0})}\)
=1-\(\frac{0.7185}{0.7185}=1-0.1321=0.8679\)
Value of the \(\left | I \right |=\left | \int_{C} \frac{dz}{z^{2}-1} \right |\) if where c is the circle x2+ y2 = 4 is ___________
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Solution
Poles, z2 – 1 = 0
⇒z =± 1
The given circle x2+ y2 = 4 with centre at z = 0 and radius 2 encloses two simple poles at z= 1 and z = –1.
∴\(\int_{c} \frac{dz}{z^{2}-1}=\int_{c1} \frac{dz}{z^{2}-1}+\int_{c2}\frac{dz}{z^{2}-1}\)
=\(\int_{c1} \frac{\frac{1}{z+1}}{z-1}dz+\int_{c2} \frac{\frac{1}{z-1}}{z+1}dz\)
=\(2\pi i\times \left ( \frac{1}{z+1} \right )_{z+1}+2\pi i\left ( \frac{1}{z-1} \right )_{z=-1}\)
=\(2\pi i\times \frac{1}{1+1}+2\pi i\left ( \frac{1}{-1-1} \right )\)
= πi –πi
= 0
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Solution
Let I =\(\int_{0}^{\pi }\)xF(sinx)dx....(i)
I =\(\int_{0}^{\pi }\)(π-x)F[sin(π-x)]dx
I =\(\int_{0}^{\pi }\)(π-x)F(sinx)dx....(ii)
Adding Eqs. (i) and (ii), we get
2I=\(\int_{0}^{\pi }\)(π))F(sinx)dx
∴ I=1⁄2\(\int_{0}^{\pi }\)πF(sinx)dx