Question

Consider a causal LTI system with frequency response \( H(j\omega) = \frac{1{j\omega + 3} \). This system produces the output to an input \( x(t) \) as \( y(t) = e^{-3t}u(t) - e^{-4t}u(t) \). The input \( x(t) \) is

• A. \((2e^{-4t} - 3e^{-3t})u(t)\)
• B. \(e^{-4t}u(t)\)
• C. \((e^{-3t} - e^{-4t})u(t)\)
• D. \((e^{-4t} - e^{-3t})u(t)\)

Hint:

Working with Laplace variables (\(S = j\omega\)) often makes tracking the algebra cleaner and faster during competitive tests: \[ H(S) = \frac{1}{S+3}, \quad Y(S) = \frac{1}{S+3} - \frac{1}{S+4} = \frac{1}{(S+3)(S+4)} \] \[ X(S) = \frac{Y(S)}{H(S)} = \frac{1}{S+4} \implies x(t) = e^{-4t}u(t) \]

Answer 1

The Correct Option is B

Understanding the Concept: For a Linear Time-Invariant (LTI) system, the relationship between input, system characteristics, and output is given by continuous convolution in the time domain: \(y(t) = x(t) * h(t)\). Applying the Continuous-Time Fourier Transform (or Laplace Transform) converts this convolution operation into a simple algebraic multiplication in the frequency domain: \[ Y(j\omega) = X(j\omega) \cdot H(j\omega) \implies X(j\omega) = \frac{Y(j\omega)}{H(j\omega)} \] By finding the mathematical representations of \(Y(j\omega)\) and \(H(j\omega)\), we can systematically solve for the unknown input spectrum \(X(j\omega)\) and then transform back into the time domain.

Step 1: Identify the frequency response of the system.
The problem directly states: \[ H(j\omega) = \frac{1}{j\omega + 3} \]

Step 2: Find the Fourier Transform of the given output \(y(t)\).
The given output signal equation is: \[ y(t) = e^{-3t}u(t) - e^{-4t}u(t) \] Using the standard Fourier Transform pair property \(e^{-at}u(t) \iff \frac{1}{j\omega + a}\) (for \(\text{Re}\{a\} > 0\)): \[ \mathcal{F}\{e^{-3t}u(t)\} = \frac{1}{j\omega + 3} \] \[ \mathcal{F}\{e^{-4t}u(t)\} = \frac{1}{j\omega + 4} \] Combining these linear terms via superposition: \[ Y(j\omega) = \frac{1}{j\omega + 3} - \frac{1}{j\omega + 4} \] Let us combine these terms over a single common denominator to simplify our algebra: \[ Y(j\omega) = \frac{(j\omega + 4) - (j\omega + 3)}{(j\omega + 3)(j\omega + 4)} = \frac{1}{(j\omega + 3)(j\omega + 4)} \]

Step 3: Solve for the input spectrum \(X(j\omega)\).
Using our frequency domain system equation: \[ X(j\omega) = \frac{Y(j\omega)}{H(j\omega)} \] Substitute our derived expression for \(Y(j\omega)\) and the given \(H(j\omega)\): \[ X(j\omega) = \frac{\frac{1}{(j\omega + 3)(j\omega + 4)}}{\frac{1}{j\omega + 3}} \] Canceling out the common factor \(\frac{1}{j\omega + 3}\) from both the numerator and denominator simplifies the expression to: \[ X(j\omega) = \frac{1}{j\omega + 4} \]

Step 4: Take the Inverse Fourier Transform to recover \(x(t)\).
Applying the standard transform pair in reverse: \[ \mathcal{F}^{-1}\left\{ \frac{1}{j\omega + 4} \right\} = e^{-4t}u(t) \] Therefore, the input signal is: \[ x(t) = e^{-4t}u(t) \] This perfectly matches option (B).