- GATE Electronics and Communication (EC) Questions Paper With Answer Key Download Pdf [2022]
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GATE - EC 2022 Answers with Explanations
Q. 36
The function f(x) = 8 loge x − x2 + 3 attains its minimum over the interval
[1, e] at x = ______.
(Here loge x is the natural logarithm of x.)
Solution:
Given:
f(x) = 8 ln(x) − x² + 3
Find the minimum of f(x) on the interval [1, e].
Step 1: Find the derivative
f'(x) = \(\frac{d}{dx}\) [8 ln(x) − x² + 3] = \(\frac{8}{x}\) − 2x
Step 2: Set derivative to zero to find critical points
f'(x) = 0 ⇒ \(\frac{8}{x}\) − 2x = 0
⇒ \(\frac{8}{x}\) = 2x
⇒ 8 = 2x²
⇒ x² = 4
⇒ x = ±2
Since the domain is [1, e], and e ≈ 2.718, we consider only x = 2.
Step 3: Evaluate f(x) at critical point and interval endpoints
- At x = 1:
f(1) = 8 ln(1) − 1² + 3 = 0 − 1 + 3 = 2 - At x = 2:
f(2) = 8 ln(2) − 4 + 3 = 8 × 0.693 − 1 = 5.544 − 1 = 4.544 (approx) - At x = e:
f(e) = 8 ln(e) − e² + 3 = 8 × 1 − 7.389 + 3 = 11 − 7.389 = 3.611 (approx)
Step 4: Compare values
f(1) = 2
f(2) ≈ 4.544
f(e) ≈ 3.611
Minimum value of f(x) on [1, e] is at x = 1.
Final answer:
x = 1
Q.52
Option (a)
The cutoff frequency is:
\[ f_{\text{cutoff}} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}} \]
For the cutoff frequency of a TM wave,
\[ \frac{c}{2a} = 1.5 \times 10^{14} \ \text{Hz} \]
Therefore, frequencies above:
\[ 1.5 \times 10^{14} \ \text{Hz} \]
will propagate.
Q.53
Solving the Integral Using Polar Coordinates
We want to evaluate the integral:
$$\iint_D 3(x^2 + y^2)\, dx\, dy$$
In polar coordinates, we use:
- \(x^2 + y^2 = r^2\)
- \(dx\,dy = r\,dr\,d\theta\)
So the integrand becomes:
$$3r^2 \cdot r = 3r^3$$
1. Describe the Region \(D\) in Polar Coordinates
The triangular region lies between the lines:
- \(y = x \Rightarrow \theta = \pi/4\)
- \(y = -x \Rightarrow \theta = -\pi/4\)
The right boundary is \(x = 4\):
$$r \cos\theta = 4 \quad \Rightarrow \quad r = \frac{4}{\cos\theta}$$
Thus the region in polar form is:
$$-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}, \quad 0 \le r \le \frac{4}{\cos\theta}$$
2. Set Up the Integral
$$\int_{\theta=-\pi/4}^{\pi/4} \int_{0}^{4/\cos\theta} 3r^3 \, dr \, d\theta$$
3. Integrate with Respect to \(r\)
$$\int_{0}^{4/\cos\theta} 3r^3 \, dr = \frac{3}{4}\left(\frac{4}{\cos\theta}\right)^4 = \frac{192}{\cos^4\theta}$$
4. Integrate with Respect to \(\theta\)
$$192 \int_{-\pi/4}^{\pi/4} \sec^4\theta \, d\theta$$
Use the substitution \(u = \tan\theta\), \(du = \sec^2\theta \, d\theta\):
$$192 \int_{-1}^{1} (u^2 + 1)\, du$$
$$\int_{-1}^{1}(u^2 + 1)\, du = \left[\frac{u^3}{3} + u\right]_{-1}^{1} = \frac{8}{3}$$
$$192 \cdot \frac{8}{3} = 512$$
Final Answer:
512
Q.54
Calculation:
First, find Vth and Rth from the graph.
From the graph
V2 = −(1 / 4m)·I2 + 5
When I2 = 0
V2 = Vosc = Vth = 5
Replace the network with its Thevenin's equivalent and replace the variable resistance with the load:
I2 = (10 − 5) / 1250 = 4 mA
Q.55
DFT Problem Solution
Given the sequence:
x̄ = [1, 0, 0, 0, 2, 0, 0, 0]
The problem asks to compute:
y = DFT(DFT(x̄))
and to find y[0].
Step 1 — First DFT
The DFT is defined as:
X(k) = Σ x[n] · exp(-j·2Ï€nk/8), n = 0..7
The only non-zero samples are x[0] = 1 and x[4] = 2, so:
X(k) = 1 + 2·exp(-jÏ€k) = 1 + 2(-1)k
Thus:
X = [3, -1, 3, -1, 3, -1, 3, -1]
Step 2 — Second DFT
We need:
y[0] = Σ X[k], k = 0..7
Summing the values:
- Four values are 3 → sum = 12
- Four values are -1 → sum = -4
y[0] = 12 - 4 = 8