Unit 5 - Practice Quiz

The Laplace transform of the unit impulse function is defined as . Due to the sifting property of the impulse function, the integral evaluates to , which is 1.

Right-sided signals have a Region of Convergence that is a right half-plane, located to the right of the rightmost pole in the s-plane. For , the pole is at , and the ROC is .

The Laplace transform is a linear operator. This means the transform of a weighted sum of signals is equal to the weighted sum of their individual Laplace transforms.

For a system to be stable, its impulse response must be absolutely integrable. This condition requires all poles of the transfer function to have negative real parts, meaning they must lie in the left-half of the s-plane.

The Laplace transform converts a time-domain signal into an s-domain function . The inverse Laplace transform performs the reverse operation, converting back to .

By convention in signal processing, poles (values of 's' for which the transfer function goes to infinity) are marked with a cross ('x') on the s-plane.

The Laplace transform of is given by the integral , with the ROC being .

The transfer function is a fundamental property of an LTI system and is defined as the Laplace transform of the system's impulse response, .

This is the time-differentiation property. Differentiating a signal in the time domain corresponds to multiplying its Laplace transform by 's' in the s-domain (assuming zero initial conditions).

The Laplace transform integral converges only for values of 's' within the ROC. At the location of a pole, the transform's magnitude is infinite, so the integral does not converge. Therefore, the ROC can never contain any poles.

The Fourier Transform is a special case of the Laplace Transform evaluated along the imaginary axis (). The Laplace transform can analyze a broader class of signals and systems, including unstable ones, which the Fourier transform cannot.

The Fourier transform of a signal is found by restricting the complex variable 's' in its Laplace transform to the imaginary axis, i.e., by setting (assuming the ROC includes the axis).

Software tools commonly represent a system's transfer function as a ratio of two polynomials in 's'. Users typically input the coefficients of these numerator and denominator polynomials to define the system.

This is a standard Laplace transform pair. The function has a pole at . For a causal system, the ROC is , and the corresponding time-domain signal is the decaying exponential .

A key condition for the stability of an LTI system is that the ROC of its transfer function must include the entire imaginary axis ( axis). This ensures that the Fourier transform of the impulse response exists and is finite.

The time-shifting property states that a delay of in the time domain corresponds to multiplication by in the s-domain.

The complex variable 's' has a real part, , and an imaginary part, . The real part, , represents the rate of exponential decay (if ) or growth (if ), while the imaginary part, , represents the frequency of oscillation.

A two-sided signal can be seen as the sum of a right-sided part and a left-sided part. The ROC for its Laplace transform is the intersection of the ROCs of these two parts, which results in a vertical strip between two poles.

Zeros of a transfer function are the roots of the numerator polynomial. At these specific values of 's', the transfer function becomes zero.

The Initial Value Theorem states that . It provides a way to determine the value of a signal at without needing to compute the inverse Laplace transform.

This requires the frequency differentiation property of the Laplace transform, which states that . First, find the Laplace transform of , which is . Now, differentiate with respect to and multiply by -1.
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The poles of the transform are at and . For a right-sided (causal) signal, the Region of Convergence (ROC) is a right-half plane located to the right of the rightmost pole. The rightmost pole is at . Therefore, the ROC is .

First, factor the denominator: . Then, use partial fraction expansion: .
To find A, cover the term and substitute : .
To find B, cover the term and substitute : .
So, . Since the signal is causal, the inverse Laplace transform is .

The differentiation property of the Laplace transform is . Since the signal is causal, .
Therefore, .
Alternatively, one could find , differentiate it to get , and then find its transform, which is .

For a system to be BIBO stable, all its poles must lie in the left-half of the s-plane. The poles are the roots of the characteristic equation . For a second-order polynomial , all roots are in the left-half plane if and only if all coefficients are positive. Therefore, we must have and . The second condition is always true. The first condition gives .

The transfer function is . The frequency response is .
Geometrically, the magnitude is the ratio of the vector distance from the zero at the origin to the point on the imaginary axis, to the vector distance from the pole at to .
At DC (), the distance to the zero is 0, so .
As , the distances from both the pole and zero to become very large and almost equal, so approaches a constant value .
Since the filter attenuates low frequencies and passes high frequencies, it is a high-pass filter.

The ROC is a vertical strip between two poles, located at and . A strip ROC always corresponds to a two-sided signal. The term with ROC corresponds to a right-sided part . The term with ROC corresponds to a left-sided part . The combination is a two-sided signal.

The Final Value Theorem states that , provided that all poles of are in the left-half plane.
Here, . The poles of are the roots of , which are . Since these poles are in the left-half plane, the theorem is applicable.
Now, we evaluate the limit: .

To find the transfer function, we take the Laplace transform of the differential equation, assuming zero initial conditions.



The transfer function is .

This is an application of the time-scaling property of the Laplace transform. The property states that if , then for a constant , the transform of is . In this case, , so the transform of is .

Poles on the -axis (the imaginary axis) that are simple (not repeated) indicate marginal stability. The real part of the poles is zero, so there is no exponential decay or growth. The imaginary part corresponds to an oscillation. Therefore, the impulse response will be a sustained sinusoid of the form , which neither decays nor grows, characterizing a marginally stable system.

The Laplace transform for a general left-sided exponential is with ROC . Our signal is , which is equivalent to . Here, . Therefore, the Laplace transform is , and the ROC is .

The input is , so its Laplace transform is . The output transform is . The inverse Laplace transform of for a causal system is .

The condition for means the signal is anti-causal or left-sided. For a left-sided signal, the ROC is a left-half plane to the left of the leftmost pole. The poles are at and . The leftmost pole is at . Therefore, the ROC must be .

The Initial Value Theorem states that . The theorem is applicable if the degree of the numerator of is less than the degree of the denominator. Here, degree(num)=2 and degree(den)=3, so it applies.
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Now, we evaluate the limit as : .

The zero at the origin () ensures that the frequency response at DC () is zero, since the distance from the zero to the point is zero. The pair of complex poles at will cause a peak or resonance in the frequency response at a frequency near their imaginary part, which is . A system that has zero response at DC and a peak at a non-zero frequency is a band-pass filter.

This is a direct application of the time-shifting property of the Laplace transform. The property states that if a signal is shifted by in time to produce , its Laplace transform is multiplied by . Thus, .

First, find the Laplace transforms. . The input is the same, so . The output transform is . This is a standard Laplace transform pair. The inverse Laplace transform of is . Therefore, .

The denominator has complex roots. We complete the square: . So, . We use the transform pair . Here, and . We need a '2' in the numerator, so we write . Taking the inverse transform, we get .

The poles of the transfer function are at . For an LTI system to be BIBO stable, its Region of Convergence (ROC) must include the -axis (i.e., the line ). The three possible ROCs for these poles are (causal, unstable), (anti-causal, unstable), and the strip (non-causal). Only the ROC includes the -axis, which corresponds to a stable (but non-causal) system.

First, for the original signal to be two-sided, its ROC must be the strip between its poles at and . Thus, the ROC of is . The frequency shifting property of the Laplace transform states that if , then . Here, . Therefore, . The poles of are at and . The ROC of is the ROC of shifted by . So, the new ROC is , which simplifies to . This vertical strip between the poles at and corresponds to a two-sided signal.

For a system to be stable, all its poles must lie in the Left Half-Plane (LHP) of the s-plane. The poles are the roots of the characteristic equation . The roots are given by the quadratic formula: .
Case 1: Underdamped/critically damped (). The poles are . The real part is , which is in the LHP. So the system is stable for .
Case 2: Overdamped (). The poles are real: and . The pole is always negative. For to be in the LHP, we need . So for this case, stability requires .
Combining both cases, the system is stable if .

Let . Using partial fraction expansion, . Solving for coefficients gives . Thus, . Since the signal is causal, the inverse Laplace transform is . The given transform is . Using the time-shifting property, . For , the unit step function is equal to 1. Therefore, for .

The magnitude of the frequency response is proportional to the product of the lengths of vectors from zeros to the point on the imaginary axis, divided by the product of the lengths of vectors from poles to . As , the point moves very far up the imaginary axis.
The length of the vector from the zero at the origin to is .
The lengths of the vectors from the poles at to are and . For very large , both lengths are approximately and .
Therefore, . So, the magnitude approaches 0 proportionally to .

The properties of the signal (real and even) and the symmetric pole locations and ROC are consistent. For the bilateral Laplace transform, the differentiation property is . The term is part of the unilateral transform property. Multiplication by in the s-domain does not change the location of the poles of . It only adds a zero at (unless already has a pole there to be cancelled). The Region of Convergence for contains the ROC of . Since no poles are added or removed on the boundaries of the ROC, the ROC for remains . Thus, has the same poles and ROC as .

The output is . A simpler method is to use superposition. The input is a positive step at and a negative step at . The step response of the system is . By linearity and time-invariance, the response to is . For , both step functions are active. So, . The coefficient is therefore .

The impulse response has two parts. The term is non-zero for , making the system non-causal. To determine stability, we find the system's ROC and check if it includes the -axis (). The transform of the anti-causal part is with ROC . The transform of the causal part is with ROC . The ROC of the entire system is the intersection of these two ROCs, which is . Since this ROC includes the imaginary axis (), the system is stable. Therefore, the system is non-causal and stable.

First, we use the Final Value Theorem (FVT) to find K. The poles are at and . Since all non-origin poles are in the LHP, the FVT is applicable. FVT: . We have . This gives , so . The full transform is . Now we use the Initial Value Theorem (IVT). The number of poles (3) is greater than the number of zeros (1), so IVT is applicable. IVT: . Since the degree of the denominator (3) is greater than the degree of the numerator (2), the limit is 0. Therefore, .

The phase of the frequency response is given by the angle of the numerator minus the angle of the denominator: . This can be written as . We want to find such that . Using the identity , we have . Let and . Then and . So, . This leads to the equation , or . This is a perfect square: . Thus, the solution is rad/s.

The characteristic equation of the system is . For stability of a second-order system, all coefficients must be positive, which means and . The combined condition for stability is . For the system to be critically damped, the roots of the characteristic equation must be real and equal, which means the discriminant must be zero. The discriminant is . Setting , we solve for . Using the quadratic formula, . This gives two possible values for : and . We must satisfy the stability condition . Only is greater than 1. Therefore, this is the only value for which the system is both stable and critically damped.

The behavior of a system's response is determined by its pole locations. An oscillatory response corresponds to complex conjugate poles. A response that grows without bound indicates instability. A system is unstable if it has poles in the Right Half-Plane (RHP). Combining these, an unbounded oscillatory response is caused by a pair of complex conjugate poles located in the RHP (i.e., with a positive real part). A single pole on the positive real axis would cause unbounded growth but no oscillation. Poles on the -axis would cause sustained (not growing) oscillations. Poles in the LHP would result in a stable, bounded response.

First, we find the Laplace transform and ROC of . We can write . The transform of the causal part is with ROC . The transform of the anti-causal part is with ROC . The transform is the sum of these, and its ROC is the intersection of the individual ROCs, which is the strip . The convolution property states . The ROC of the result of a convolution is at least the intersection of the individual ROCs. Since we are convolving with itself, the ROC of is the same as the ROC of , which is . The poles of are double poles at , and since is also a two-sided signal, its ROC must be the strip between these poles.

The inverse system has the transfer function . This system has a pole at and a zero at . For an LTI system to be causal, its ROC must be a right-half plane to the right of its rightmost pole. Here, the rightmost pole is at , so for causality, the ROC must be . For an LTI system to be stable, its ROC must include the imaginary axis (). The ROC required for causality, , does not include the imaginary axis. Therefore, this inverse system cannot be both stable and causal simultaneously. The original system is non-minimum phase (it has a zero in the RHP), and the inverse of a non-minimum phase system cannot be both stable and causal.

The given integral is in the form of a convolution integral, . By comparing the two expressions, we can identify for (due to the upper limit of integration), and 0 otherwise. This is equivalent to saying . The Laplace transform of this impulse response is . According to the convolution property of the Laplace transform, if , then . Therefore, .

The transfer function is . The phase of the frequency response is . We evaluate this at . The first term is . This vector points from the origin to , which is in the second quadrant. Its angle is . The second term is . This vector points from the origin to , which is in the first quadrant. Its angle is . The total phase is the difference: .

The output signal is . As , both exponential terms decay to zero. The rate of decay is determined by the magnitude of the negative exponent. The term decays faster than . Therefore, the term is the dominant term, meaning it is the last to become negligible. This dominant behavior corresponds to the pole of the overall transfer function that is closest to the -axis. The system pole is at and the input pole is at . The pole at is closer to the imaginary axis, and thus it dictates the slowest-decaying component of the response.

The Laplace transform exists if the defining integral converges. Let . For , the integral is . For this integral to converge, the magnitude of the integrand, , must be integrable. As , the term grows faster than the linear term for any finite value of . This means the exponent will eventually become positive and grow, causing the integral to diverge. Because there is no value of for which the integral converges, the ROC is empty and the Laplace transform does not exist.

The Region of Convergence (ROC) determines the nature of the time-domain signal. An ROC that is a vertical strip in the s-plane, bounded by poles, corresponds to a two-sided signal. The part of the signal associated with poles to the left of the ROC is right-sided (causal), and the part associated with poles to the right of the ROC is left-sided (anti-causal). In this case, the ROC is to the right of the pole at , so the component related to this pole ( form) is right-sided. The ROC is to the left of the pole at , so the component related to this pole ( form) is left-sided. Therefore, the signal is a sum of a right-sided and a left-sided component, making it a two-sided signal.

The impulse response is the derivative of the step response . We can find by differentiating with respect to time. Using the product rule . Here, . First, , so the term is zero. Next, we find the derivative . Therefore, . Alternatively, taking the Laplace transform of the step response gives . Since , we get . The inverse Laplace transform of is .

The form of the denominator, , suggests damped sinusoidal functions. Here, and . We use the transform pairs: and . We must manipulate the numerator, , to fit these forms. We write as .
So, . The first term corresponds to . For the second term, we need a 3 in the numerator to match the sine transform. We write it as . This corresponds to . Combining them, we get .