TS PGECET 2026 Electronics and Communication (EC) Engineering Question Paper: Download Answer Key with Solutions PDF

Concept:
Use the standard Beta-function identity: \[ \int_{0}^{\pi/2} \sin^{m-1}x \cos^{n-1}x \, dx = \frac{1}{2}B\left(\frac{m}{2},\frac{n}{2}\right) \]

Step 1: Rewrite the integrand.
\[ \sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}} = \sin^{1/2}x \cos^{-1/2}x \]

Thus, \[ I=\int_{0}^{\pi/2}\sin^{1/2}x \cos^{-1/2}x\,dx \]

Step 2: Apply Beta-function formula.

Comparing with \[ \sin^{m-1}x \cos^{n-1}x \]
we get: \[ m-1=\frac12 \Rightarrow m=\frac32 \] \[ n-1=-\frac12 \Rightarrow n=\frac12 \]

Hence, \[ I = \frac12 B\left(\frac34,\frac14\right) \]

Using \[ B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \]
and \[ \Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\pi\sqrt2 \]
with \[ \Gamma(1)=1 \]

we obtain: \[ I = \frac12(\pi\sqrt2) = \frac{\pi}{\sqrt2} \] Quick Tip: Whenever integrals involve powers of \(\sin x\) and \(\cos x\), try converting them into Beta-function form.

Concept:
For a linear differential equation with constant coefficients: \[ ax''+bx'+cx=0 \]
we solve the auxiliary equation: \[ ar^2+br+c=0 \]

Step 1: Form the auxiliary equation.
\[ r^2+4r+5=0 \]
\[ r=\frac{-4\pm\sqrt{16-20}}{2} = -2\pm i \]

Step 2: Write the general solution.

For roots \( \alpha \pm i\beta \), \[ x(t)=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t) \]

Thus, \[ x(t)=e^{-2t}(C_1\cos t+C_2\sin t) \]

Step 3: Apply initial conditions.

Using \(x(0)=1\): \[ 1=e^0(C_1)=C_1 \]
So, \[ C_1=1 \]

Now differentiate: \[ x'(t)=e^{-2t}\left[-2(C_1\cos t+C_2\sin t)-C_1\sin t+C_2\cos t\right] \]

Substitute \(t=0\): \[ 0=x'(0)=-2C_1+C_2 \]

Since \(C_1=1\), \[ C_2=2 \]

Step 4: Final solution.
\[ x(t)=e^{-2t}(\cos t+2\sin t) \] Quick Tip: Complex roots \( \alpha \pm i\beta \) always produce solutions of the form \[ e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t) \] .

Concept:
According to Cauchy’s Integral Theorem, if a function is analytic everywhere inside and on a closed contour, then its contour integral is zero.

Step 1: Identify the singularity:

The integrand is: \[ f(z)=\frac{5z-2}{z-1} \]

The denominator becomes zero at: \[ z=1 \]

Thus, the singularity is at \(z=1\).

Step 2: Analyze the contour:

The contour is: \[ |z|=\frac{1}{2} \]

This is a circle centered at the origin with radius \(\frac12\).

Since: \[ |1|=1>\frac12 \]

the point \(z=1\) lies outside the contour.

Step 3: Apply Cauchy’s Integral Theorem:

Since the singularity is outside the contour, the function is analytic everywhere inside and on the contour.

Therefore, \[ \int_{c}\frac{5z-2}{z-1}dz =0 \] Quick Tip: If all singularities lie outside the closed contour, the contour integral is always zero.

Concept:
A time delay in the time domain corresponds to multiplication by an exponential factor in the Laplace domain.

Step 1: Write the delayed signal:

If the input is delayed by \(T\) seconds: \[ y(t)=x(t-T) \]

Step 2: Apply Laplace Transform:

Using the time-shifting property: \[ \mathcal{L}\{x(t-T)\}=e^{-sT}X(s) \]

Step 3: Determine transfer function:

The transfer function is: \[ H(s)=\frac{Y(s)}{X(s)}=e^{-sT} \] Quick Tip: Delay in time domain always becomes \(e^{-sT}\) in Laplace domain.

Concept:
For a wave incident from air to a dielectric medium, Brewster angle satisfies: \[ \tan \theta_B=\sqrt{\frac{\epsilon_2}{\epsilon_1}} \]

Step 1: Substitute dielectric constants:

Given: \[ \epsilon_1=1,\qquad \epsilon_2=10 \]

Thus, \[ \tan\theta_B=\sqrt{10} \]

Step 2: Find Brewster angle:
\[ \theta_B=\tan^{-1}(\sqrt{10}) \] Quick Tip: For air-to-dielectric transition, Brewster angle is commonly: \[ \theta_B=\tan^{-1}(\sqrt{\epsilon_r}) \]

Concept:
When \(n\) identical amplifier stages are cascaded, the overall bandwidth decreases.

Step 1: Use cascaded bandwidth formula:

For \(n\) identical stages: \[ B_{overall}=B\sqrt{2^{1/n}-1} \]

where: \[ B=bandwidth of each stage \]

Step 2: Match with options:

The obtained expression matches option (C). Quick Tip: Cascading identical stages always reduces bandwidth.

Concept:
Electron concentration equals the number of free electrons per unit volume.

Step 1: Use given atomic concentration:

Given: \[ 1000\ atoms/cm^3 \]

Assuming each atom contributes one free electron: \[ n=1000\ e/cm^3 \]

Step 2: Select correct option:

Hence, the electron concentration is: \[ 1000\,e/cm^3 \] Quick Tip: If one atom contributes one electron, electron concentration equals atomic concentration.

Concept:
The reflection coefficient at normal incidence between two media is: \[ \Gamma=\frac{\eta_2-\eta_1}{\eta_2+\eta_1} \]

For refractive index \(n\), the coefficient from air to glass becomes: \[ \Gamma=\frac{1-n}{1+n} \]

Step 1: Apply air-to-glass condition:

For air: \[ n_1=1 \]

For glass: \[ n_2=n \]

Thus, \[ \Gamma=\frac{1-n}{1+n} \]

Step 2: Match with option:

Hence, option (D) is correct. Quick Tip: Do not confuse reflection coefficient with reflected power coefficient. Power reflection coefficient is \(|\Gamma|^2\).

Concept:
Avalanche breakdown depends on electric field strength and ionization energy.

Step 1: Effect of doping:

Higher doping narrows the depletion region and increases electric field intensity.

Thus, breakdown occurs at lower voltage.

Hence:

\( \text{Breakdown voltage decreases with increase in doping} \)

\( \text{Breakdown voltage decreases with increase in doping} \)

Concept:
A complex limit exists only if the value approaches the same result along every possible path.

Step 1: Express \(z\) in polar form:

Let: \[ z=re^{i\theta} \]

Then: \[ \overline{z}=re^{-i\theta} \]

Step 2: Simplify the function:
\[ f(z)=\frac{re^{i\theta}}{re^{-i\theta}} =e^{i2\theta} \]

Thus, \[ f(z)=\cos 2\theta+i\sin 2\theta \]

Step 3: Check path dependence:

If: \[ \theta=0 \]
then: \[ f(z)=1 \]

If: \[ \theta=\frac{\pi}{2} \]
then: \[ f(z)=-1 \]

Since different paths give different values, the limit does not exist. Quick Tip: For complex limits, if the result depends on \(\theta\) or the path of approach, the limit does not exist.