Unit 5: The laplace transform - Subjective Questions

ECE220 — Signal And Systems • Practice Questions with Detailed Answers

20 questions

Define the Laplace Transform of a continuous-time signal $x(t)$ . Explain how it generalizes the Fourier Transform.

Distinguish between the bilateral (two-sided) and unilateral (one-sided) Laplace Transforms. When is each preferred?

Bilateral Laplace Transform:

$X(s) = \int_{-\infty}^{\infty} x(t) e^{-st}\, dt$

Unilateral Laplace Transform:

$X(s) = \int_{0^{-}}^{\infty} x(t) e^{-st}\, dt$

Comparison:

Feature	Bilateral	Unilateral
Integration limits	$-\infty$ to $\infty$	$0^-$ to $\infty$
Handles initial conditions	No inherent way	Yes, naturally
Requires ROC	Must be specified	Always right-sided
Use case	Signal/system analysis	Solving differential equations with ICs

When each is preferred:

The bilateral form is used for general signal and system analysis where signals may exist for all time.
The unilateral form is preferred for causal systems and for solving differential equations with non-zero initial conditions, since it incorporates initial values directly.

Explain the concept of the Region of Convergence (ROC) of the Laplace Transform. List its important properties.

Find the Laplace Transform of $x(t) = e^{-at}u(t)$ and specify its ROC. Also find the Laplace Transform of $x(t) = -e^{-at}u(-t)$ and its ROC. Comment on the result.

Describe the method of finding the Inverse Laplace Transform using partial fraction expansion. Illustrate with an example.

Explain the geometric evaluation of the Fourier Transform (frequency response) from the pole-zero plot of $X(s)$ .

State and prove the time-shifting property and the time-scaling property of the Laplace Transform.

State and prove the differentiation-in-time property of the Laplace Transform. How is it used to solve differential equations?

State the convolution property of the Laplace Transform and explain its significance in LTI system analysis.

Define the system function (transfer function) $H(s)$ of an LTI system. Explain how causality and stability are determined from its ROC and pole locations.

Determine whether the system with transfer function $H(s) = \dfrac{s+2}{s^2 + 3s + 2}$ is causal and stable, assuming a causal system.

State and prove the Initial Value Theorem and the Final Value Theorem of the Laplace Transform. Mention conditions for their validity.

Explain the frequency-shifting (s-domain shift) property and the multiplication by $t$ (s-domain differentiation) property of the Laplace Transform.

A causal LTI system is described by the differential equation $\dfrac{d^2y(t)}{dt^2} + 3\dfrac{dy(t)}{dt} + 2y(t) = x(t)$ . Find the transfer function $H(s)$ , the poles and zeros, and the impulse response $h(t)$ .

Compare the Laplace Transform and the Fourier Transform in terms of definition, convergence, applicability, and use in system analysis.

Comparison of Laplace and Fourier Transforms:

Aspect	Fourier Transform	Laplace Transform
Definition	$X(j\omega)=\int x(t)e^{-j\omega t}dt$	$X(s)=\int x(t)e^{-st}dt$ , $s=\sigma+j\omega$
Variable	Real frequency $\omega$	Complex frequency $s$
Convergence	Needs absolute integrability	Converges over an ROC via $e^{-\sigma t}$
Applicability	Limited (some signals have no FT)	Broader (handles growing signals)
Initial conditions	Not handled	Unilateral form handles them
Domain	Frequency spectrum	$s$ -plane (pole-zero analysis)

Key observations:

The Fourier Transform is a special case of the Laplace Transform evaluated on the $j\omega$ -axis (when the ROC includes it).
The Laplace Transform is superior for transient and stability analysis, since it captures both decay/growth and oscillation.
The Fourier Transform is superior for steady-state frequency (spectral) analysis of stable signals.
Laplace excels at solving differential equations and analyzing system behavior (poles, stability, causality).

Explain the significance of pole-zero plots in characterizing an LTI system. How do pole locations affect the system's behavior?

Describe how software simulation tools (e.g., MATLAB) are used for system representation and pole-zero analysis. Mention relevant functions and their purposes.

Find the inverse Laplace Transform of $X(s) = \dfrac{2s + 3}{s^2 + 2s + 5}$ assuming a causal signal.

Explain how the Laplace Transform is used to analyze the interconnection (cascade, parallel, and feedback) of LTI systems.

Derive the inverse Laplace Transform relation (synthesis equation) and briefly explain the Bromwich contour integral.

Question 1

Define the Laplace Transform of a continuous-time signal $x(t)$ . Explain how it generalizes the Fourier Transform.

Tap card to reveal answer

Answer 1

The Laplace Transform of a continuous-time signal $x(t)$ is defined as:

$X(s) = \int_{-\infty}^{\infty} x(t) e^{-st} \, dt$

where $s = \sigma + j\omega$ is a complex variable.

Relationship with Fourier Transform:

The Fourier Transform is given by $X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t}\, dt$ .
The Laplace Transform introduces a real exponential factor $e^{-\sigma t}$ , giving $X(s) = \int_{-\infty}^{\infty} \big[x(t)e^{-\sigma t}\big] e^{-j\omega t}\, dt$ .
Thus the Laplace Transform is the Fourier Transform of $x(t)e^{-\sigma t}$ .

Key points:

When $\sigma = 0$ (i.e., $s = j\omega$ ), the Laplace Transform reduces to the Fourier Transform (provided it exists on the $j\omega$ axis).
The factor $e^{-\sigma t}$ improves convergence, allowing analysis of signals that grow with time and lack a Fourier Transform.
This makes the Laplace Transform a more general and powerful tool for system analysis.

Tap card to flip back

Question 2

Distinguish between the bilateral (two-sided) and unilateral (one-sided) Laplace Transforms. When is each preferred?

Tap card to reveal answer

Answer 2

Bilateral Laplace Transform:

$X(s) = \int_{-\infty}^{\infty} x(t) e^{-st}\, dt$

Unilateral Laplace Transform:

$X(s) = \int_{0^{-}}^{\infty} x(t) e^{-st}\, dt$

Comparison:

Feature	Bilateral	Unilateral
Integration limits	$-\infty$ to $\infty$	$0^-$ to $\infty$
Handles initial conditions	No inherent way	Yes, naturally
Requires ROC	Must be specified	Always right-sided
Use case	Signal/system analysis	Solving differential equations with ICs

When each is preferred:

The bilateral form is used for general signal and system analysis where signals may exist for all time.
The unilateral form is preferred for causal systems and for solving differential equations with non-zero initial conditions, since it incorporates initial values directly.

Tap card to flip back

Question 3

Explain the concept of the Region of Convergence (ROC) of the Laplace Transform. List its important properties.

Tap card to reveal answer

Answer 3

The Region of Convergence (ROC) is the set of values of $s = \sigma + j\omega$ in the complex $s$ -plane for which the Laplace Transform integral converges (i.e., yields a finite value).

$\int_{-\infty}^{\infty} |x(t)| e^{-\sigma t}\, dt < \infty$

Important properties of the ROC:

The ROC consists of strips parallel to the $j\omega$ -axis, because convergence depends only on $\sigma = \text{Re}\{s\}$ .
The ROC does not contain any poles.
For a signal of finite duration, the ROC is the entire $s$ -plane (except possibly $s=0$ or $s=\infty$ ).
For a right-sided signal, the ROC lies to the right of the rightmost pole.
For a left-sided signal, the ROC lies to the left of the leftmost pole.
For a two-sided signal, the ROC is a strip bounded by poles.
A signal has a Fourier Transform if and only if its ROC includes the $j\omega$ -axis.
The ROC is a connected region.

Tap card to flip back

Question 5

Describe the method of finding the Inverse Laplace Transform using partial fraction expansion. Illustrate with an example.

Tap card to reveal answer

Answer 5

The Inverse Laplace Transform recovers $x(t)$ from $X(s)$ . For rational $X(s)$ , partial fraction expansion is the standard method.

Steps:

Ensure $X(s)$ is a proper rational function (numerator degree < denominator degree); if not, perform long division.
Factor the denominator into its poles.
Expand $X(s)$ as a sum of simple partial fractions.
Use the ROC to associate each term with a right-sided or left-sided signal.
Apply standard transform pairs.

Example: Find $x(t)$ for

$X(s) = \frac{1}{(s+1)(s+2)}, \quad \text{ROC: } \text{Re}\{s\} > -1$

Partial fractions:

$X(s) = \frac{A}{s+1} + \frac{B}{s+2}$

$A = (s+1)X(s)\big|_{s=-1} = \frac{1}{-1+2} = 1$
$B = (s+2)X(s)\big|_{s=-2} = \frac{1}{-2+1} = -1$

So $X(s) = \dfrac{1}{s+1} - \dfrac{1}{s+2}$ .

Since ROC is to the right of both poles, both terms are right-sided (causal):

$\boxed{x(t) = \left(e^{-t} - e^{-2t}\right)u(t)}$

Tap card to flip back

Question 6

Explain the geometric evaluation of the Fourier Transform (frequency response) from the pole-zero plot of $X(s)$ .

Tap card to reveal answer

Answer 6

The frequency response $X(j\omega)$ can be evaluated geometrically from the pole-zero plot when the ROC includes the $j\omega$ -axis.

Consider a rational transform:

$X(s) = M \frac{\prod_{i} (s - z_i)}{\prod_{k} (s - p_k)}$

Evaluating on the $j\omega$ -axis ( $s = j\omega$ ):

$X(j\omega) = M \frac{\prod_{i} (j\omega - z_i)}{\prod_{k} (j\omega - p_k)}$

Geometric interpretation: Each factor $(j\omega - z_i)$ or $(j\omega - p_k)$ is a vector from the pole/zero to the point $j\omega$ on the imaginary axis.

Magnitude:

$|X(j\omega)| = |M| \cdot \frac{\text{product of distances from zeros to } j\omega}{\text{product of distances from poles to } j\omega}$

Phase:

$\angle X(j\omega) = \angle M + \sum_i \angle(\text{zero vectors}) - \sum_k \angle(\text{pole vectors})$

Key insights:

When $j\omega$ is near a pole, the pole vector is short, so the magnitude peaks.
When $j\omega$ is near a zero, the zero vector is short, so the magnitude dips (approaches zero).
This method gives an intuitive picture of the shape of the frequency response without full computation.

Tap card to flip back

Question 8

State and prove the differentiation-in-time property of the Laplace Transform. How is it used to solve differential equations?

Tap card to reveal answer

Answer 8

Differentiation-in-Time Property (Bilateral):

If $x(t) \xleftrightarrow{\mathcal{L}} X(s)$ , then:

$\frac{dx(t)}{dt} \xleftrightarrow{\mathcal{L}} sX(s)$

Proof: Using the inverse transform,

$x(t) = \frac{1}{2\pi j}\int_{\sigma - j\infty}^{\sigma + j\infty} X(s) e^{st}\, ds$

Differentiating both sides w.r.t. $t$ :

$\frac{dx(t)}{dt} = \frac{1}{2\pi j}\int_{\sigma - j\infty}^{\sigma + j\infty} sX(s) e^{st}\, ds$

Thus $\dfrac{dx(t)}{dt} \xleftrightarrow{\mathcal{L}} sX(s)$ .

Unilateral form (with initial conditions):

$\mathcal{L}\left\{\frac{dx(t)}{dt}\right\} = sX(s) - x(0^-)$

Application to differential equations:

Differential equations become algebraic equations in $s$ after transformation.
Initial conditions are incorporated automatically via the unilateral property.
One solves the algebraic equation for $X(s)$ , then takes the inverse Laplace Transform to obtain $x(t)$ .
This converts calculus operations into algebra, greatly simplifying the solution process.

Tap card to flip back

Question 10

Define the system function (transfer function) $H(s)$ of an LTI system. Explain how causality and stability are determined from its ROC and pole locations.

Tap card to reveal answer

Answer 10

The system function $H(s)$ is the Laplace Transform of the impulse response $h(t)$ :

$H(s) = \int_{-\infty}^{\infty} h(t) e^{-st}\, dt = \frac{Y(s)}{X(s)}$

Causality:

An LTI system is causal if $h(t) = 0$ for $t < 0$ (right-sided).
Therefore, the ROC is a right-half plane to the right of the rightmost pole.
For a system with rational $H(s)$ , causality requires the ROC to lie to the right of the rightmost pole and the degree of the numerator $\le$ degree of the denominator.

Stability (BIBO):

A system is BIBO stable if $\int_{-\infty}^{\infty}|h(t)|\,dt < \infty$ .
This occurs if and only if the ROC of $H(s)$ includes the $j\omega$ -axis ( $\text{Re}\{s\}=0$ ).

Combined condition (Causal + Stable):

For a causal AND stable system, all poles must lie in the left-half of the $s$ -plane ( $\text{Re}\{p_k\} < 0$ ).
This ensures the ROC (right of rightmost pole) includes the $j\omega$ -axis.

Tap card to flip back

Question 12

State and prove the Initial Value Theorem and the Final Value Theorem of the Laplace Transform. Mention conditions for their validity.

Tap card to reveal answer

Answer 12

Initial Value Theorem (IVT):

For a causal signal $x(t)$ with no impulses at $t=0$ :

$x(0^+) = \lim_{s \to \infty} sX(s)$

Final Value Theorem (FVT):

$\lim_{t \to \infty} x(t) = \lim_{s \to 0} sX(s)$

Proof sketch of IVT: Using the differentiation property (unilateral):

$\int_{0^-}^{\infty}\frac{dx}{dt}e^{-st}\,dt = sX(s) - x(0^-)$

As $s \to \infty$ , the integral $\to 0$ (the exponential suppresses contributions), giving:

$\lim_{s\to\infty} sX(s) = x(0^+)$

Proof sketch of FVT: As $s \to 0$ :

$\lim_{s\to 0}\int_{0^-}^{\infty}\frac{dx}{dt}e^{-st}\,dt = \int_{0^-}^{\infty}\frac{dx}{dt}\,dt = x(\infty) - x(0^-)$

Since the LHS equals $\lim_{s\to 0}[sX(s) - x(0^-)]$ , we obtain $x(\infty) = \lim_{s\to 0} sX(s)$ .

Conditions for validity:

IVT: valid if $x(t)$ and its derivative are Laplace-transformable and $X(s)$ is a proper rational function.
FVT: valid only if all poles of $sX(s)$ lie in the left-half plane (i.e., $x(t)$ actually converges to a finite limit). Otherwise it gives a misleading result.

Tap card to flip back

Question 15

Compare the Laplace Transform and the Fourier Transform in terms of definition, convergence, applicability, and use in system analysis.

Tap card to reveal answer

Answer 15

Comparison of Laplace and Fourier Transforms:

Aspect	Fourier Transform	Laplace Transform
Definition	$X(j\omega)=\int x(t)e^{-j\omega t}dt$	$X(s)=\int x(t)e^{-st}dt$ , $s=\sigma+j\omega$
Variable	Real frequency $\omega$	Complex frequency $s$
Convergence	Needs absolute integrability	Converges over an ROC via $e^{-\sigma t}$
Applicability	Limited (some signals have no FT)	Broader (handles growing signals)
Initial conditions	Not handled	Unilateral form handles them
Domain	Frequency spectrum	$s$ -plane (pole-zero analysis)

Key observations:

The Fourier Transform is a special case of the Laplace Transform evaluated on the $j\omega$ -axis (when the ROC includes it).
The Laplace Transform is superior for transient and stability analysis, since it captures both decay/growth and oscillation.
The Fourier Transform is superior for steady-state frequency (spectral) analysis of stable signals.
Laplace excels at solving differential equations and analyzing system behavior (poles, stability, causality).

Tap card to flip back

Question 16

Explain the significance of pole-zero plots in characterizing an LTI system. How do pole locations affect the system's behavior?

Tap card to reveal answer

Answer 16

A pole-zero plot displays the locations of poles ( $\times$ ) and zeros ( $\circ$ ) of $H(s)$ in the complex $s$ -plane. It provides a compact, visual characterization of an LTI system.

Significance:

The plot (along with a gain factor and the ROC) completely specifies a rational system function.
It reveals stability, causality, and frequency response at a glance.

Effect of pole locations on behavior:

Poles in the left-half plane ( $\text{Re}\{p\}<0$ ): produce decaying natural responses $\Rightarrow$ stable behavior.
Poles in the right-half plane ( $\text{Re}\{p\}>0$ ): produce growing responses $\Rightarrow$ unstable behavior.
Poles on the $j\omega$ -axis: produce sustained oscillations or constant terms $\Rightarrow$ marginally stable.
Complex-conjugate poles: produce oscillatory (sinusoidal) responses; the imaginary part sets the oscillation frequency and the real part sets the decay rate.
Poles closer to the $j\omega$ -axis: produce resonant peaks in magnitude response and slower decay.

Effect of zeros:

Zeros near the $j\omega$ -axis cause dips (nulls) in the frequency response.
Zeros shape but do not affect stability.

Tap card to flip back

Question 17

Describe how software simulation tools (e.g., MATLAB) are used for system representation and pole-zero analysis. Mention relevant functions and their purposes.

Tap card to reveal answer

Answer 17

Software simulation tools such as MATLAB (or Python with SciPy) allow engineers to represent LTI systems and analyze them without tedious manual computation.

System representation:

A transfer function is defined by numerator and denominator coefficient vectors.
tf(num, den) creates a transfer function object.
zpk(z, p, k) creates the system from zeros, poles, and gain.

Pole-zero analysis functions (MATLAB):

roots(poly) — computes the roots (poles/zeros) of a polynomial.
pzmap(sys) — plots the pole-zero map in the $s$ -plane.
pole(sys) / zero(sys) — return system poles and zeros.
residue(num, den) — performs partial fraction expansion for inverse Laplace.

Response and frequency analysis:

impulse(sys) — plots the impulse response $h(t)$ .
step(sys) — plots the step response.
bode(sys) — plots magnitude and phase (frequency response).
freqs(num, den, w) — evaluates the analog frequency response.

Benefits:

Rapidly checks stability (pole positions) and transient behavior.
Enables visualization of pole-zero plots and responses.
Reduces algebraic errors and supports design iteration.

Tap card to flip back

Question 19

Explain how the Laplace Transform is used to analyze the interconnection (cascade, parallel, and feedback) of LTI systems.

Tap card to reveal answer

Answer 19

The Laplace Transform simplifies analysis of interconnected LTI systems because time-domain convolution becomes $s$ -domain multiplication.

1. Cascade (Series) Connection:

Output of one system feeds the input of the next.
Overall system function:

$H(s) = H_1(s)\, H_2(s)$

Transfer functions multiply.

2. Parallel Connection:

Same input applied to both; outputs are summed.
Overall system function:

$H(s) = H_1(s) + H_2(s)$

Transfer functions add.

3. Feedback Connection:

Forward path $H_1(s)$ with feedback path $H_2(s)$ .
For negative feedback:

$H(s) = \frac{H_1(s)}{1 + H_1(s)H_2(s)}$

For positive feedback:

$H(s) = \frac{H_1(s)}{1 - H_1(s)H_2(s)}$

Significance:

Complex block diagrams reduce to algebraic manipulation of transfer functions.
The poles of the resulting $H(s)$ determine the stability of the overall interconnected system, which is central to control-system design.

Tap card to flip back

Question 20

Derive the inverse Laplace Transform relation (synthesis equation) and briefly explain the Bromwich contour integral.

Tap card to reveal answer

Answer 20

Derivation of the Inverse Laplace Transform:

Recall that the Laplace Transform is the Fourier Transform of $x(t)e^{-\sigma t}$ :

$X(\sigma + j\omega) = \int_{-\infty}^{\infty}\big[x(t)e^{-\sigma t}\big]e^{-j\omega t}\,dt$

This is the Fourier Transform of $g(t) = x(t)e^{-\sigma t}$ . Applying the inverse Fourier Transform:

$x(t)e^{-\sigma t} = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\sigma + j\omega)\,e^{j\omega t}\,d\omega$

Multiplying both sides by $e^{\sigma t}$ :

$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\sigma + j\omega)\,e^{(\sigma + j\omega)t}\,d\omega$

Substituting $s = \sigma + j\omega$ , so $ds = j\,d\omega$ (with $\sigma$ fixed in the ROC):

$\boxed{x(t) = \frac{1}{2\pi j}\int_{\sigma - j\infty}^{\sigma + j\infty} X(s)\,e^{st}\,ds}$

Bromwich Contour Integral:

The integral is evaluated along a vertical line $\text{Re}\{s\} = \sigma$ in the complex plane, called the Bromwich contour.
The value of $\sigma$ must lie within the ROC so that convergence is guaranteed.
In practice, the integral is evaluated using the residue theorem:

$x(t) = \sum (\text{residues of } X(s)e^{st} \text{ at the poles})$

This is why partial fraction expansion (which computes residues) is the standard practical method for finding inverse transforms.

Tap card to flip back

1 / 20

Unit4 Unit6

Unit 5: The laplace transform - Subjective Questions

Did this save you a night before the exam?