All questions of Signals & Systems for Electrical Engineering (EE) Exam

B

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).

How?...

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.

Properties of the given system:
- Linearity (P)
- Time invariance (Q)
- Causality (R)
- Stability (S)

Explanation:
1. Linearity (P):
A system is said to be linear if it satisfies the superposition principle, which states that the response of the system to the sum of two inputs is equal to the sum of the responses to each individual input. Mathematically, a system is linear if it satisfies the following property:
y[n] = nx[n]

In the given system, y[n] = nx[n], which is a linear relationship between the input x[n] and the output y[n]. Hence, the system satisfies the linearity property.

2. Time invariance (Q):
A system is said to be time-invariant if a time shift in the input signal causes an equal time shift in the output signal. Mathematically, a system is time-invariant if it satisfies the following property:
If y[n] is the output corresponding to x[n], then y[n - k] is the output corresponding to x[n - k], for any integer k.

In the given system, there is no explicit dependence on time. The output y[n] is directly proportional to the input x[n], where the proportionality constant is n. Therefore, the system does not depend on time and satisfies the time-invariance property.

3. Causality (R):
A system is said to be causal if the output at any given time depends only on the present and past values of the input. Mathematically, a system is causal if it satisfies the following property:
If y[n] is the output corresponding to x[n], then y[n] does not depend on x[n + k] for any k > 0.

In the given system, the output y[n] is directly proportional to the input x[n], where the proportionality constant is n. The output at any given time n depends only on the present value of the input x[n] and not on any future values. Hence, the system satisfies the causality property.

4. Stability (S):
A system is said to be stable if bounded inputs produce bounded outputs. Mathematically, a system is stable if it satisfies the following property:
If |x[n]| ≤ M for all n, then |y[n]| ≤ N for all n, where M and N are finite positive constants.

In the given system, the output y[n] is directly proportional to the input x[n], where the proportionality constant is n. As long as the input x[n] is bounded, the output y[n] will also be bounded. Hence, the system satisfies the stability property.

Therefore, the system described by y[n] = nx[n] satisfies the properties of linearity (P), time invariance (Q), causality (R), and stability (S). The correct answer is option C: P, R.

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.

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.

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.

Sampling Frequency and Nyquist Frequency

Sampling Frequency: Sampling frequency is the number of samples taken per second. It is denoted by Fs.

Nyquist Frequency: Nyquist frequency is half of the sampling frequency. It is denoted by Fn = Fs / 2.

Aliasing and Aliased Frequencies

Aliasing: Aliasing is the phenomenon where a signal at a certain frequency appears to be at a different frequency due to undersampling.

Aliased Frequencies: Aliased frequencies are the frequencies at which a signal appears due to undersampling. They are given by the formula:

f_alias = f_signal - k * Fs

where k is an integer.

Solution

Given, the signal frequency is 100 Hz.

Sampling at 140 Hz:
Fn = 140 / 2 = 70 Hz
f_alias = 100 - k * 140 = 100 - 1 * 140 = -40 Hz (negative frequency is same as 140 - 40 = 100 Hz)
Therefore, the aliased frequency is 40 Hz.

Sampling at 90 Hz:
Fn = 90 / 2 = 45 Hz
f_alias = 100 - k * 90 = 100 - 1 * 90 = 10 Hz
Therefore, the aliased frequency is 10 Hz.

Sampling at 30 Hz:
Fn = 30 / 2 = 15 Hz
f_alias = 100 - k * 30 = 100 - 3 * 30 = 10 Hz
Therefore, the aliased frequency is 10 Hz.

Hence, the aliased frequencies correspond to each sampling rate will be 40 Hz, 10 Hz, and 10 Hz respectively. Therefore, option C is the correct answer.

Explanation:
Dirichlet Conditions:
Dirichlet conditions are a set of criteria that a function must satisfy to be expanded in a Fourier series. These conditions are as follows:
1. The function must be single-valued and continuous in the interval for which it is being expanded.
2. The function must have a finite number of maxima and minima in the interval.
3. The function must have a finite number of discontinuities in the interval, which can be of the first kind (jump discontinuities) or the second kind (infinite discontinuities).
4. If the function has discontinuities, then the left and right limits of the function at the discontinuity point must exist and be finite.

f(t) = -f(-t):
The given function f(t) satisfies the odd symmetry property, i.e., f(t) = -f(-t). This means that the function is symmetric about the origin, and its Fourier series will only contain sine terms. This is because the Fourier series of an odd function only contains sine terms, while the Fourier series of an even function only contains cosine terms. Therefore, option A is the correct answer.

Conclusion:
In conclusion, if a function satisfies the odd symmetry property and the Dirichlet conditions, then its Fourier series will only contain sine terms.

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'.