All questions of Signals & Systems for Electronics and Communication Engineering (ECE) Exam

Explanation:

When a function f(t) u(t) is shifted to the right side by t0, it means that the function starts at t0 instead of t=0. Therefore, we need to shift the entire function to the right by t0.

Option A: f(t - t0) u(t)
This option shifts the function to the right by t0, but it still starts at t=0. Therefore, it is incorrect.

Option B: f(t) u(t -t0)
This option delays the function by t0, but it does not shift it to the right. Therefore, it is incorrect.

Option C: f(t - t0) u(t - t0)
This option shifts the entire function to the right by t0 and also delays it by t0. Therefore, it is the correct answer.

Option D: f(t + t0) u(t + t0)
This option shifts the entire function to the left by t0 instead of to the right. Therefore, it is incorrect.

In summary, the correct option for expressing a function f(t) u(t) shifted to the right side by t0 is f(t - t0) u(t - t0).

The Sampling Theorem
The sampling theorem states that in order to accurately reconstruct a continuous-time signal from its samples, the sampling rate must be at least twice the highest frequency component of the signal. This is known as the Nyquist rate.

In the case of telephonic communications, the highest frequency component of a speech signal is about 3.1 kHz. Therefore, to accurately capture and transmit the speech signal, the sampling rate must be at least twice this frequency, i.e., 2 * 3.1 kHz = 6.2 kHz.

Available Options
Let's evaluate the available options for the suitable value of the sampling rate:

a) 1 kHz: This sampling rate is below the Nyquist rate, as it is less than 6.2 kHz. Therefore, it is not suitable for accurately capturing the speech signal.

b) 2 kHz: Similar to option a), this sampling rate is also below the Nyquist rate and cannot accurately capture the speech signal.

c) 4 kHz: Again, this sampling rate is below the Nyquist rate and insufficient for accurately capturing the speech signal.

d) 8 kHz: This sampling rate is twice the highest frequency component of the speech signal (3.1 kHz * 2 = 6.2 kHz). Therefore, it satisfies the Nyquist rate and is suitable for accurately capturing the speech signal.

Conclusion
To accurately capture the highest frequency component of a speech signal for telephonic communications, a suitable value for the sampling rate is 8 kHz (option d). This sampling rate ensures that the signal is adequately sampled, allowing for faithful reconstruction during transmission and reception.

Answer is 0.5A

Answer:

Frequency content of the continuous-time signal: 10 MHz, 50 MHz, and 70 MHz.

Sampling frequency: 56 MHz.

Low-pass filter cutoff frequency: 15 MHz.

Output frequency content after passing through the low-pass filter: 10 MHz, 6 MHz, and 14 MHz.

Explanation:

The Nyquist-Shannon sampling theorem states that a continuous-time signal can be perfectly reconstructed from its samples if the sampling frequency is greater than twice the highest frequency component of the signal. In this case, the highest frequency component of the signal is 70 MHz. Therefore, the sampling frequency of 56 MHz is sufficient to capture all the frequency content of the signal.

However, since the sampling frequency is less than twice the frequency of the highest component of the signal, aliasing can occur. Aliasing is the phenomenon where high-frequency components of the signal are folded back into the frequency band of the lower-frequency components.

To prevent aliasing, a low-pass filter is used to remove the high-frequency components of the signal before sampling. In this case, a low-pass filter with a cutoff frequency of 15 MHz is used.

The low-pass filter removes all frequency components above 15 MHz, including the 50 MHz and 70 MHz components of the original signal. The only frequency component that remains is the 10 MHz component.

However, since the original signal had frequency components at 10 MHz, 50 MHz, and 70 MHz, the low-pass filter output will also contain frequency components due to aliasing. These alias frequencies are given by:

- 2 × 56 MHz – 70 MHz = 42 MHz (alias of 70 MHz)
- 2 × 56 MHz – 50 MHz = 62 MHz (alias of 50 MHz)

Therefore, the frequency content of the output of the filter will be 10 MHz (due to the original 10 MHz component) and 6 MHz and 14 MHz (due to aliasing of the original 50 MHz and 70 MHz components, respectively). The correct answer is option 'C'.

T) is a mathematical function that describes how a signal or time series is correlated with itself at different time lags. It is defined as the expected value of the product of two time-shifted samples of a signal or time series, given by:

Rx(t) = E[X(t) X(t + τ)]

where X(t) is the signal or time series at time t, τ is the time lag, and E[·] denotes the expected value operator.

The auto-correlation function is a useful tool for analyzing the properties of time series data, such as their periodicity, stationarity, and spectral content. It can be used to estimate the power spectrum of a signal, which describes the distribution of its energy over different frequencies.

The auto-correlation function is related to the Fourier transform of the signal through the Wiener-Khinchin theorem, which states that the power spectral density of a stationary signal is the Fourier transform of its auto-correlation function.

In practice, the auto-correlation function is often estimated from a finite sample of data using statistical methods such as the sample auto-correlation function or the maximum likelihood estimator. These estimators are commonly used in time series analysis, signal processing, and statistical inference.

Explanation:

Linear systems satisfy the properties of superposition and homogeneity, whereas time-invariant systems have the same response for a given input, regardless of when the input is applied.

Let's analyze the given system:

y(t) = tx(t)

1. Superposition:

The system is not additive since it doesn't satisfy the superposition property. If we apply two inputs, say x1(t) and x2(t), then the output will be:

y1(t) = tx1(t) and y2(t) = tx2(t)

y1(t) + y2(t) = tx1(t) + tx2(t)

y1(t) + y2(t) ≠ t(x1(t) + x2(t))

Thus, the system is not linear.

2. Homogeneity:

The system satisfies the homogeneity property since applying a scalar value to the input results in a scaled output.

y(at) = atx(t) = a(tx(t))

Thus, the system is homogeneous.

3. Time-invariance:

The system is clearly time-varying because the output depends on the value of time t, which means the system's response changes with time.

Therefore, the system is linear and time-varying, which matches with option B.

Conclusion:

The system described by y(t) = tx(t) is linear and time-varying.

Kindly help

Solution : b)

Sampling and Aliasing

Sampling is the process of converting a continuous-time signal into a discrete-time signal by taking samples at regular intervals. The sampling rate determines how frequently the samples are taken, and it is usually expressed in samples per second or Hertz (Hz). Aliasing is a phenomenon that occurs when the sampling rate is not sufficient to accurately represent the original signal.

Aliasing Effect

The Nyquist-Shannon sampling theorem states that a signal must be sampled at a rate greater than or equal to twice the highest frequency component of the signal in order to avoid aliasing. If the sampling rate is not sufficient, the higher frequency components of the original signal will fold back into the lower frequency range, resulting in a distorted and unfaithful representation of the original signal.

Analysis of Sampling Rates

In this question, the message signal is a single tone with a frequency of 4 kHz. We are given three sampling rates: 9 kHz, 7 kHz, and 5 kHz. Let's analyze each sampling rate to determine if aliasing will occur.

1. Sampling rate of 9 kHz:
- The Nyquist-Shannon sampling rate for the 4 kHz signal is 8 kHz (twice the signal frequency).
- Since the sampling rate of 9 kHz is greater than the Nyquist rate, there will be no aliasing. The reconstructed signal will accurately represent the original signal.

2. Sampling rate of 7 kHz:
- The Nyquist-Shannon sampling rate for the 4 kHz signal is 8 kHz (twice the signal frequency).
- Since the sampling rate of 7 kHz is less than the Nyquist rate, aliasing will occur. The higher frequency components of the original signal will fold back into the lower frequency range, causing distortion in the reconstructed signal.

3. Sampling rate of 5 kHz:
- The Nyquist-Shannon sampling rate for the 4 kHz signal is 8 kHz (twice the signal frequency).
- Since the sampling rate of 5 kHz is less than the Nyquist rate, aliasing will occur. The higher frequency components of the original signal will fold back into the lower frequency range, causing distortion in the reconstructed signal.

Conclusion

Based on the analysis, we can conclude that aliasing will occur in the reconstructed signal when the signal is sampled at a rate of 7 kHz or 5 kHz. The sampling rate of 9 kHz is sufficient to avoid aliasing, and the reconstructed signal will accurately represent the original signal. Therefore, the correct answer is option 'D' - aliasing will occur when the signal is sampled at both 7 kHz and 5 kHz.

Understanding the Transfer Function
The transfer function given is H(s) = 1/s^3. This represents a system with a third-order pole at the origin, indicating a specific response to inputs.
Impulse Response Calculation
To find the impulse response h(t), we need to take the inverse Laplace transform of H(s). The inverse transform of H(s) = 1/s^n is known to be:
- h(t) = (1/n!)*u(t) for n = 1, 2, 3, ...
For n = 3:
- h(t) = (1/3!)*u(t) = (1/6)*u(t).
This indicates that the impulse response h(t) is a scaled unit step function.
Connection with Convolution
The impulse response can also be expressed in terms of convolutions of unit step functions.
- Convolution of u(t) with itself n times yields the function that corresponds to the nth integral of the unit step function.
Thus, the impulse response can be expressed as:
- h(t) = u(t) ⊗ u(t) ⊗ u(t).
This is the convolution of three unit step functions, which is mathematically equivalent to the impulse response we derived.
Options Analysis
Now, analyzing the options given:
- Option A: u(t) ⊗ u(t) ⊗ u(t) - This is indeed the correct representation of the impulse response, as it corresponds to the convolution of three unit step functions.
- Option B: u(t) ⊗ u(t) ⊗ u(t) ⊗ u(t) - This would represent a fourth convolution and is not applicable here.
- Option C: u(t) × u(t) × u(t) - This denotes multiplication, which does not apply to our impulse response.
- Option D: u(t) ⊗ u(t) - This is only a second convolution and does not match the required order.
Thus, the correct answer is option A, as it accurately represents the impulse response derived from the transfer function.

Πt) + 2cos(10πt)

1. What is the frequency of the first cosine term?

The frequency of the first cosine term is 6 Hz, which can be calculated as 6π/2π = 3 cycles per second.

2. What is the frequency of the second cosine term?

The frequency of the second cosine term is 10 Hz, which can be calculated as 10π/2π = 5 cycles per second.

3. What is the amplitude of the first cosine term?

The amplitude of the first cosine term is 1, as it has a coefficient of 1 in front of the cosine function.

4. What is the amplitude of the second cosine term?

The amplitude of the second cosine term is 2, as it has a coefficient of 2 in front of the cosine function.

5. What is the total amplitude of the signal x(t)?

The total amplitude of the signal x(t) is the sum of the amplitudes of each cosine term, which is 1 + 2 = 3.

Unilateral Laplace Transform for Constant Coefficient Differential Equations with Non-zero Initial Condition

Unilateral Laplace Transform is a mathematical tool used to solve linear constant coefficient differential equations. It is particularly useful in solving differential equations with non-zero initial conditions. Let us discuss in detail the application of Unilateral Laplace Transform for constant coefficient differential equations with a non-zero initial condition.

Constant Coefficient Differential Equations

A constant coefficient differential equation is a differential equation in which the coefficients are constant. These types of differential equations are commonly found in physics and engineering. Examples of constant coefficient differential equations include the following:

- d^2y/dt^2 + 3dy/dt + 2y = 0
- d^2y/dt^2 + 2dy/dt + 2y = f(t)

Unilateral Laplace Transform

Unilateral Laplace Transform is a mathematical tool used to solve linear constant coefficient differential equations. It is defined as follows:

L{f(t)} = F(s) = ∫0∞ e^-st f(t) dt

where L denotes the Laplace transform, f(t) is the function to be transformed, s is the complex variable, and F(s) is the Laplace transform of f(t). The Laplace transform is a powerful tool that can be used to solve differential equations.

Application of Unilateral Laplace Transform for Constant Coefficient Differential Equations with Non-zero Initial Condition

The Unilateral Laplace Transform can be used to solve constant coefficient differential equations with non-zero initial conditions. The general process involves the following steps:

- Take the Laplace transform of both sides of the differential equation.
- Solve for the Laplace transform of the dependent variable.
- Take the inverse Laplace transform to find the solution to the original differential equation.

For example, consider the following differential equation:

d^2y/dt^2 + 2dy/dt + 2y = 0, y(0) = 1, dy/dt(0) = 0

Taking the Laplace transform of both sides of the differential equation, we get:

s^2Y(s) - s + 2sY(s) + 2Y(s) = Y(0) + sdy/dt(0) + 2y(0)

Simplifying this equation, we get:

Y(s) = (s+1)/(s^2 + 2s + 2)

Taking the inverse Laplace transform, we get:

y(t) = e^-t(cos(t) - sin(t))

Thus, we have solved the differential equation with non-zero initial conditions using the Unilateral Laplace Transform.

Conclusion

In conclusion, the Unilateral Laplace Transform is a powerful tool that can be used to solve constant coefficient differential equations with non-zero initial conditions. By taking the Laplace transform of the differential equation, solving for the Laplace transform of the dependent variable, and taking the inverse Laplace transform, we can find the solution to the original differential equation.