# Online Electronics and Communication Engineering Test - Previous Exam Papers Test 7

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#### Marks : 2/20

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_{c}and C

_{E}can be assumed to be short at signal frequency and the effect of output resistance

*r*

_{0}can be ignored. If C

_{E}is disconnected from the circuit, which one of the following statements is TRUE?

Given circuit after removing C_{E} will behave as current-series feedback.

Overall voltage gain will decrease as feedback signal comes into picture and since it is current-series feedback, input impedance increases.

*z*-axis the field intensity is E = ; which of the following figure shows an exact field distribution?

The given figure fails to show the symmetry with respect to Φ.

The figure shows symmetry with respect to *f*, also the length of arrow is decreasing away from the charge shows that magnitude E is decreasing away from line charge.

But problem with this figure is longest lines must be shown in most crowded region.

Here we use lines of fixed segments but different thickness. But this attempt also makes the region crowded near origin.

This figure shows compromise. A symmetrical distribution of lines (at every 45°) shows azimuthal symmetry and arrowheads show direction.

*h*(

*t*) of a linear time-invariant continuous time system is described by

*h*(

*t*) = exp (a

*t*)

*u*(

*t*) + exp (β

*t*)

*u*(-

*t*), where

*u*(

*t*) denotes the unit step function, and a and β are constants. This system is stable if

*h*(*t*) = *e*^{+at}*u*(*t*) + *e*^{βt}*u*(- *t*)

For *h*(*t*) to be stable*h*(*t*) *dt* < ∞

It will happen when a is negative and β is positive.

Dual of R → G

1 + G(s)H(s) = *s*^{2} + 2*s* + 2

ω_{n} = 2 , 2ε_{e}ω_{n} = 2, ε_{e} = = 0.707.

- The reduction in
*h*_{12}and Miller capacitance result in negligible reverse internal feedback. - A's is large loads cannot be connected.
- As a narrow band amplifier, it provides small range of
*c*_{μ}is reduced. - As a narrow band amplifier it provides small range of frequencies parameters the effect of
*c*_{μ}reduces.

Cascode is a wide band amplifier and it provides wide range of frequencies than CE stage and as is large, large loads can be connected.

*m*' and a denominator polynomial of degree '

*n*', then the integer (

*n*-

*m*) represents the number of:

Asymptotes.

We use Quine - McCluskey method of reduction :

We can observe that minterms 158 and 222 can be grouped together to reduce the variable 'B'.

*= 9). The magnitude of the reflection coefficient is*

_{r}Thus reflection coefficient = = - 0.5

Thus magnitude is 0.5 .

*m*A at a forward potential of 0.8 V and the reverse potential is 0.6 V What are its dynamic forward and reverse resistance at this temperature.

We have,

I = I_{0}Ã¢â‚¬Â¢(*e*^{V/nVT} -1)

**∴**

For forward voltage greater than few teeth's of volts, I >> I_{0}

**∴**

**∴** Forward dynamic resistance,

= 34.3 Ω

Reverse dynamic resistance = = 20 *k*Ω.

*X*(

*t*) with two-sided power spectral density 1 x 10

^{-10}W/Hz is input to a filter whose magnitude squared response is shown below.

The power of the output process Y(

*t*) is given by

Since bandwidth is 10 kHz, thus output power is 10 x 10^{-11} x 10 x 10^{3} = 1 x l0^{-6} W.

*h*(

*t*)

*u*(

*t*) is stable, if __________ (

*n*is even).

h(*t*) =

consider

*ax* = *x*

*ax* = *x*^{2}

*a* = *x*

For *h(t)* to be bounded, exponential term should have a negative power.

i.e. *e ^{-t}* will have response as shown.

Hence should be < 0

**∴** *a* < 0.

_{c(sat)}.

*pn*junction (with applied voltage = V

_{B}) is proportional to

**Assertion (A):** An unbiased p-n junction develops a built-in potential at the junction with the *n*-side positive and the *p*-side negative.

**Reason (R):** The *p-n* junction behaves as a battery and supplies current to a resistance connected across its terminals.