Skip to content

Quiz: Laplace Transform Methods

Test your understanding of the Laplace transform, transfer functions, and techniques for analyzing control systems in the s-domain.

1. What is the primary advantage of using the Laplace transform for control system analysis?

  1. It eliminates the need for any mathematical analysis
  2. It converts differential equations into algebraic equations
  3. It only works for first-order systems
  4. It provides exact solutions without approximation
Show Answer

The correct answer is B. The Laplace transform converts differential equations (which involve derivatives) into algebraic equations (which involve polynomials in $s$). Differentiation becomes multiplication by $s$, and integration becomes division by $s$. This makes the equations much easier to manipulate and solve.

Concept Tested: Laplace Transform

2. The transfer function $G(s)$ of an LTI system is defined as:

  1. The ratio of output to input in the time domain
  2. The ratio of the Laplace transform of the output to the input, assuming zero initial conditions
  3. The sum of all poles and zeros
  4. The inverse Laplace transform of the impulse response
Show Answer

The correct answer is B. The transfer function is $G(s) = Y(s)/U(s)$ where $Y(s)$ and $U(s)$ are the Laplace transforms of output and input respectively, with all initial conditions set to zero. This ratio captures the system's input-output relationship in the s-domain.

Concept Tested: Transfer Function

3. In the s-domain, the complex frequency variable $s$ is expressed as:

  1. $s = \omega + j\sigma$
  2. $s = \sigma + j\omega$
  3. $s = e^{j\omega t}$
  4. $s = \ln(\omega)$
Show Answer

The correct answer is B. The complex frequency variable is $s = \sigma + j\omega$, where $\sigma$ is the real part (related to exponential growth/decay) and $\omega$ is the imaginary part (related to oscillation frequency). This convention is standard in control systems.

Concept Tested: S-Domain

4. The Final Value Theorem states that $\lim_{t\to\infty} f(t) = $:

  1. $\lim_{s\to\infty} sF(s)$
  2. $\lim_{s\to 0} sF(s)$
  3. $\lim_{s\to 0} F(s)$
  4. $\lim_{s\to\infty} F(s)$
Show Answer

The correct answer is B. The Final Value Theorem states that $\lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s)$, provided that the limit exists (i.e., all poles of $sF(s)$ are in the left half-plane). This theorem allows us to find the steady-state value directly from the s-domain representation without taking the complete inverse transform.

Concept Tested: Final Value Theorem

5. The cover-up method is used to:

  1. Hide complex poles in a transfer function
  2. Quickly find residues for partial fraction expansion with simple poles
  3. Convert nonlinear systems to linear ones
  4. Determine system stability
Show Answer

The correct answer is B. The cover-up method (also called Heaviside's cover-up method) is a shortcut for finding residues when performing partial fraction expansion on transfer functions with simple (non-repeated) poles. You "cover up" the factor corresponding to each pole and evaluate the remaining expression at that pole location.

Concept Tested: Cover-Up Method, Residue Calculation

6. In the s-domain, multiplication of two Laplace transforms corresponds to what operation in the time domain?

  1. Addition of the time functions
  2. Multiplication of the time functions
  3. Convolution of the time functions
  4. Differentiation of one time function
Show Answer

The correct answer is C. Multiplication in the s-domain corresponds to convolution in the time domain: if $Y(s) = G(s) \cdot U(s)$, then $y(t) = g(t) * u(t) = \int_0^t g(\tau)u(t-\tau)d\tau$. This is one of the key properties that makes the Laplace transform so useful—the complex convolution integral becomes simple multiplication.

Concept Tested: Convolution in S-Domain

7. The Initial Value Theorem allows us to find:

  1. The final value of a function as $t \to \infty$
  2. The value of a function at $t = 0^+$ from its Laplace transform
  3. The poles of a transfer function
  4. The time constant of a first-order system
Show Answer

The correct answer is B. The Initial Value Theorem states that $\lim_{t\to 0^+} f(t) = \lim_{s\to\infty} sF(s)$. This allows us to determine the initial value of a function directly from its Laplace transform without computing the complete inverse transform.

Concept Tested: Initial Value Theorem

8. Partial fraction expansion is used to:

  1. Add fractions with different denominators
  2. Break a complex rational function into simpler terms for inverse Laplace transformation
  3. Find the roots of a polynomial
  4. Calculate convolution integrals
Show Answer

The correct answer is B. Partial fraction expansion decomposes a complex rational function (ratio of polynomials) into a sum of simpler fractions, each with a recognizable inverse Laplace transform. This makes it possible to transform back to the time domain using standard transform pairs.

Concept Tested: Partial Fraction Expansion

9. The Laplace transform of a unit step function $u(t)$ is:

  1. $1$
  2. $s$
  3. $1/s$
  4. $1/s^2$
Show Answer

The correct answer is C. The Laplace transform of the unit step function $u(t)$ is $\mathcal{L}{u(t)} = 1/s$. This is one of the fundamental transform pairs that should be memorized. The unit impulse has transform $1$, and the unit ramp has transform $1/s^2$.

Concept Tested: Laplace Transform

10. If differentiation in the time domain corresponds to multiplication by $s$ in the s-domain, then integration corresponds to:

  1. Multiplication by $s$
  2. Division by $s$
  3. Addition of $s$
  4. Subtraction of $s$
Show Answer

The correct answer is B. Just as differentiation in time becomes multiplication by $s$, integration in time becomes division by $s$. This is the inverse relationship: $\mathcal{L}{\int_0^t f(\tau)d\tau} = F(s)/s$. This property is central to the power of Laplace transform methods.

Concept Tested: Laplace Transform, S-Domain