GATE Electronics and Communication Online Test Series 1

G(s) = k(s+3)/s^4(s+2)
We can write as
G(s) = k(s+3)/(s^5+2s^4+ks+1)
The system is always unstable because s³ and s² are missing here.

Option (1) is correct.
X (s) = 2 – 1 / (s+2) (s+3)
= 2 – 1 / (s+2) + 1 / (s+3)
X (t) = 2(t) + (e-3t – e-2t) u (t)

Option (2) is correct.
Sum of poles = 0-1-2 = -3
Sum of zeros = 0
Therefore, no. of poles – no.of zeros = N. of values for which response is infinite = 3 poles and none zeros. Hence,

Option (1) is correct.
The function has poles at z = 1, 3/4 . Thus, the final value theorem applies.

= (z – 1)

According to the circuit of MUXs: Z₁ = 1 when Z₀ = 0 otherwise Z₁ is equal to 0. Z₀ = 0 when  b = 1, at this time a = 0. So, Z₁ = a’ . b. c .
This is the correct answer. Also one cannot represent the operation of MUX in simple binary expression unless the conditions are met.

Option (3) is correct.
2x² + x + 1 = 64 + 5 x 8 + 2
x = 7

I_d = I_s 

V₁ – V₂ = V_t In  = 0.0259 In 10 = 59.6 mV ≈ 59mV

We know
I_c = 
I_C 
So, if W_B increases by a factor of 3, then I_C is decreased by a factor of 3.

0.5 = 
V_GS =  + 0.8 = 2.38 V

4i₂ + 6i₃ – 2i₁ = 0
i₁ + i₂ = 2
i₂ = 5 + i₃
i₁ = -5/6 A

The circuit is as shown below.

T_TH = 7 || 5 + 6 || 9 = 6.52 Ω
For maximum power transfer
R_L = T_TH = 6.52 Ω

=  = 0

For E_y = 0, 2y sin 2x = 0
y = 0
sin 2x = 0
2x = 0, π,3π
x = 0, 3π/2

I_t = I_c  = 0.68

Initial 4 x 10 = 40 kHz
Now if deviation is d, then Δ BW = 2d = 2 x 2 = 4 kHz
New Bandwidth BW = 40 + 4 = 44 kHz

p =  , q =  and n = 3
Thus, mean (np) = 1/3×3 = 1

f` (c) = 
c = log (e – 1)

x : 0 0.1 0.2 0.3 0.4
Euler method gives:
y_n+1 = y_n + h (x_n, y_n) …………. (1)
n = 0 in (1) gives:
y₁ = |y₀ + h f (x₀, y₀)|
Here x₀ = 0,
y₀ = 1,
h = 0.1
y₁ = 1 + 0.1 f (0, 1) = 1 + 0 = 1
n = 0 in (1) gives y₂ = y₁ + hf (x₁, y₁)
= 1 + 0.1 f (0.1, 1) = 1 + 0.1 (0.1) = 1 + 0.01
Thus y₂ = y_{(0, 2)} = 1.01
n = 2 in (1) gives:
y₃ = y₂ + hf (|x₂, y₂) = 1.01 + 0.1 f (0.2, 1.01)
y₃ = y(0, 3) = 1.01 + 0.0202 = 1.0302
n = 2 in (1) gives:
y₃ = y₂ + hf (x₂, y₂) = 1.01 + 0.1 f (0.2, 1.01)
y₃ = y_{(0, 3)} = 1.01 + 0.0202 = 1.0302
n = 3 in (1) gives:
y₄ = y₃ + hf (x₃, y₃) = 1.0302 + 0.1 f (0.3, 1.0302)
= 1.0302 + 0.03090
Hence y₄ = y_{(0, 4)} = 1.0611

dy/dx – y tan x = y² sec x
Or y^-2 dy/dx – y^-1 tan x = sec x
Put y^-1 = v to get – y² dy/dx = dv/dx
Substituting this in the given equation, we get
-dv/dx -v tan x = sec x or dv/dx + (tan x). v = – sec x
I. F = 
v. sec x = –   + c = – tan x + c
1/y = = – sin x + cos x
Or    y-1 = – sin x + c2 cos x

From the given solution            Ht2 = 54.8Ux – 22.76 Uy + 10.34 Uz           Hn2 = -1.93 Ux – 2.9Uy + 3.86 Uz           H2   = Ht2 + Hn2           H2    = 52.87Ux -25.66 Uy + 14.2 Uz