Here is the highly condensed, step-by-step layout of how you should write your answers on the physical exam script to secure maximum marks within the 20-minute limit.
Question 1 [06 Marks]
1. Symmetry Definitions: [1.00, 1.117]
- Even Signal: \(x(-t) = x(t)\) for all \(t\)
- Odd Signal: \(x(-t) = -x(t)\) for all \(t\)
2. Proof using the Delta Scaling Property: [1.00, 2.2] The scaling property of the Dirac delta function is: \[\delta(at) = \frac{1}{|a|} \delta(t)\]
Substitute \(a = -1\): [1.00] \[\delta(-t) = \frac{1}{|-1|} \delta(t) = \delta(t)\]
3. Conclusion: [1.00] Since \(x(-t) = \delta(-t) = \delta(t) = x(t)\), the signal satisfies even symmetry.
\[\mathbf{\text{Therefore, } x(t) = \delta(t) \text{ is an Even Signal.}}\]
Question 2 [07 Marks]
Given: \(x(t) = \cos(6\pi t) + \sin(9\pi t)\)
1. Calculate Individual Periods (\(T_1, T_2\)): [1.00]
- For \(x_1(t) = \cos(6\pi t) \implies \omega_1 = 6\pi \text{ rad/s}\) [1.00] \[T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{6\pi} = \mathbf{\frac{1}{3}\text{ s}}\]
- For \(x_2(t) = \sin(9\pi t) \implies \omega_2 = 9\pi \text{ rad/s}\) [1.00] \[T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{9\pi} = \mathbf{\frac{2}{9}\text{ s}}\]
2. Rationality Test: [1.00, 1.118] \[\frac{T_1}{T_2} = \frac{1/3}{2/9} = \frac{1}{3} \times \frac{9}{2} = \mathbf{\frac{3}{2}} \in \mathbb{Q} \quad \mathbf{\text{(Rational fraction)}}\] Since the ratio of the periods is a rational number, the signal is periodic. [1.00]
3. Calculate Fundamental Period (\(T_0\)): [1.00, 1.118] \[T_0 = \text{LCM}\left(T_1, T_2\right) = \text{LCM}\left(\frac{1}{3}, \frac{2}{9}\right)\]
Using the fractional LCM formula: [1.00, 1.118] \[T_0 = \frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{\text{LCM}(1, 2)}{\text{HCF}(3, 9)} = \mathbf{\frac{2}{3}\text{ s}}\]
\[\mathbf{\text{Answer: The signal is Periodic with a fundamental period of } T_0 = \frac{2}{3}\text{ seconds.}}\]
Question 3 [07 Marks]
Given: \(x(t) = \cos(t)\) (Periodic with fundamental period \(T_0 = 2\pi\text{ s}\)) [6, 1.136]
1. Calculate Total Energy (\(E\)): [1.00, 1.136] \[E = \int_{-\infty}^{\infty} |x(t)|^2 , dt = \int_{-\infty}^{\infty} \cos^2(t) , dt\] Since the integrand \(\cos^2(t) \ge 0\) is a non-zero periodic function integrated over infinite boundaries, the integral diverges: \[\mathbf{E = \infty \text{ Joules}}\]
2. Calculate Average Power (\(P\)): [1.00, 1.136] For a periodic signal, average power is calculated over exactly one period: [1.00, 1.136] \[P = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 , dt = \frac{1}{2\pi} \int_{0}^{2\pi} \cos^2(t) , dt\] Here is the highly condensed, step-by-step layout of how you should write your answers on the physical exam script to secure maximum marks within the 20-minute limit.
Question 1 [06 Marks]
1. Symmetry Definitions: [1.00, 1.117]
- Even Signal: \(x(-t) = x(t)\) for all \(t\)
- Odd Signal: \(x(-t) = -x(t)\) for all \(t\)
2. Proof using the Delta Scaling Property: [1.00, 2.2] The scaling property of the Dirac delta function is: \[\delta(at) = \frac{1}{|a|} \delta(t)\]
Substitute \(a = -1\): [1.00] \[\delta(-t) = \frac{1}{|-1|} \delta(t) = \delta(t)\]
3. Conclusion: [1.00] Since \(x(-t) = \delta(-t) = \delta(t) = x(t)\), the signal satisfies even symmetry.
\[\mathbf{\text{Therefore, } x(t) = \delta(t) \text{ is an Even Signal.}}\]
Question 2 [07 Marks]
Given: \(x(t) = \cos(6\pi t) + \sin(9\pi t)\)
1. Calculate Individual Periods (\(T_1, T_2\)): [1.00]
- For \(x_1(t) = \cos(6\pi t) \implies \omega_1 = 6\pi \text{ rad/s}\) [1.00] \[T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{6\pi} = \mathbf{\frac{1}{3}\text{ s}}\]
- For \(x_2(t) = \sin(9\pi t) \implies \omega_2 = 9\pi \text{ rad/s}\) [1.00] \[T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{9\pi} = \mathbf{\frac{2}{9}\text{ s}}\]
2. Rationality Test: [1.00, 1.118] \[\frac{T_1}{T_2} = \frac{1/3}{2/9} = \frac{1}{3} \times \frac{9}{2} = \mathbf{\frac{3}{2}} \in \mathbb{Q} \quad \mathbf{\text{(Rational fraction)}}\] Since the ratio of the periods is a rational number, the signal is periodic. [1.00]
3. Calculate Fundamental Period (\(T_0\)): [1.00, 1.118] \[T_0 = \text{LCM}\left(T_1, T_2\right) = \text{LCM}\left(\frac{1}{3}, \frac{2}{9}\right)\]
Using the fractional LCM formula: [1.00, 1.118] \[T_0 = \frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{\text{LCM}(1, 2)}{\text{HCF}(3, 9)} = \mathbf{\frac{2}{3}\text{ s}}\]
\[\mathbf{\text{Answer: The signal is Periodic with a fundamental period of } T_0 = \frac{2}{3}\text{ seconds.}}\]
Question 3 [07 Marks]
Given: \(x(t) = \cos(t)\) (Periodic with fundamental period \(T_0 = 2\pi\text{ s}\)) [6, 1.136]
1. Calculate Total Energy (\(E\)): [1.00, 1.136] \[E = \int_{-\infty}^{\infty} |x(t)|^2 , dt = \int_{-\infty}^{\infty} \cos^2(t) , dt\] Since the integrand \(\cos^2(t) \ge 0\) is a non-zero periodic function integrated over infinite boundaries, the integral diverges: \[\mathbf{E = \infty \text{ Joules}}\]
2. Calculate Average Power (\(P\)): [1.00, 1.136] For a periodic signal, average power is calculated over exactly one period: [1.00, 1.136] \[P = \frac{1}{T_0} \int_{0}^{T_0} |x(t)|^2 , dt = \frac{1}{2\pi} \int_{0}^{2\pi} \cos^2(t) , dt\]
Apply trigonometric identity \(\cos^2(t) = \frac{1 + \cos(2t)}{2}\): [6, 1.00] \[P = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1 + \cos(2t)}{2} , dt\] \[P = \frac{1}{4\pi} \left[ t + \frac{\sin(2t)}{2} \right]_{0}^{2\pi}\] \[P = \frac{1}{4\pi} \left[ \left(2\pi + \frac{\sin(4\pi)}{2}\right) - (0 + 0) \right]\] \[P = \frac{2\pi}{4\pi} = \mathbf{0.5\text{ Watts}}\]
3. Conclusion: [1.00, 1.136] Since total energy is infinite (\(E = \infty\)) and average power is finite and non-zero (\(P = 0.5\text{ W}\)):
\[\mathbf{\text{Therefore, } x(t) = \cos(t) \text{ is a Power Signal.}}\] Apply trigonometric identity \(\cos^2(t) = \frac{1 + \cos(2t)}{2}\): [6, 1.00] \[P = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1 + \cos(2t)}{2} , dt\] \[P = \frac{1}{4\pi} \left[ t + \frac{\sin(2t)}{2} \right]_{0}^{2\pi}\] \[P = \frac{1}{4\pi} \left[ \left(2\pi + \frac{\sin(4\pi)}{2}\right) - (0 + 0) \right]\] \[P = \frac{2\pi}{4\pi} = \mathbf{0.5\text{ Watts}}\]
3. Conclusion: [1.00, 1.136] Since total energy is infinite (\(E = \infty\)) and average power is finite and non-zero (\(P = 0.5\text{ W}\)):
\[\mathbf{\text{Therefore, } x(t) = \cos(t) \text{ is a Power Signal.}}\]