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

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.

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) = e^+atu(t) + e^βtu(- t)

For h(t) to be stableh(t) dt < ∞

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

Dual of R → G

1 + G(s)H(s) = s² + 2s + 2

ω_n = 2 , 2ε_eω_n = 2, ε_e = = 0.707.

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.

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'.

Thus reflection coefficient = = - 0.5

Thus magnitude is 0.5 .

We have,

I = I₀•(e^V/nV_T -1)

∴

For forward voltage greater than few teeth's of volts, I >> I₀

∴

∴ Forward dynamic resistance,

= 34.3 Ω

Reverse dynamic resistance = = 20 kΩ.

Since bandwidth is 10 kHz, thus output power is 10 x 10^-11 x 10 x 10³ = 1 x l0^-6 W.

h(t) =

consider

ax = x

ax = x²

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.