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2 years ago

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Amplifiers are extensively used in the baseband portion of a radio receiver system to condition the baseband signal to produce a

n output signal ready for digital sampling and storage. Some of the key design features of baseband amplifiers include
i. DC gain,
ii. output swing,
iii. power consumption, and
iv. bandwidth.

1 answer:

N76 [4]2 years ago

5 0

Answer:

Please see the attached file for the complete answer.

Explanation:

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To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

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<u></u> $\ T_{c}$ has greater effect.

<u>Explanation</u>:

$\eta_{\max }=1-\frac{T_{c}}{T_{A}}$

$T_{c}\\$ = Temperature of cold reservoir

$T_{H}$ = Temperature of hot reservoir

when $T_{c}$ is decreased by 't',

$\eta_{\text {incre }}$ = $1-\frac{\left(\tau_{c}-t\right)}{T_{H}}$

$=n \ + \frac{t}{T_{n}}$ $-(i)$

when ${T_{H}}$ is increased by 'T'

$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

$\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}$

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2 years ago

An engineer designs a new bus that can drive 30 miles per gallon of fuel. Which of the following was likely one of the client’s

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3 years ago

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For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's

gulaghasi [49]

When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area

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3 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

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3 years ago

A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in

DochEvi [55]

Answer:

$a=5.515\frac{m}{s^{2} }$

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

$1 mile=1609.34 mts.$

$1 hora=3600 segundos$

Performing the operations

$60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}$

Now, we will use the acceleration formula

$a=\frac{v}{t}$

Where v = speed and t = time

Substituting the values of $t=5.2s$

$a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }$

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2 years ago