Question

$\text{Consider a causal LTI system with frequency response for a particular}\text{input x(t), this system is observed to produce the output as}y\left(t\right)=e^{-3t}u\left(t\right)-e^{-4t}u\left(t\right)\text{then the}H\left(j\omega \right)=\frac{1}{3+j\omega }\text{input x(t) is}$

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

Answer 1

The correct option is A $e^{-4t}u\left(t\right)$



$\text{Given the Causal LTI system,}H\left(j\omega \right)=\frac{1}{3+j\omega }\text{and output},y\left(t\right)=e^{-3t}u\left(t\right)-e^{-4t}u\left(t\right)\text{We know that}H\left(j\omega \right)=\frac{Y\left(j\omega \right)}{X\left(j\omega \right)}Y\left(j\omega \right)=\frac{1}{3+j\omega }-\frac{1}{4+j\omega }=\frac{1}{(3+j\omega )(4+j\omega )}\therefore X\left(j\omega \right)=\frac{Y\left(j\omega \right)}{H\left(j\omega \right)}=\frac{1}{4+j\omega }\text{By inverse Fourier transform of X(j}\omega \text{), we have,}x\left(t\right)=e^{-4t}u\left(t\right)$