Question

The Laplace transform of the function 6e^5tcos(2t) – e^7t is ______________

(a) $\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}$

(b) $\frac{6(s-5)}{(s-5)^2+4} + \frac{1}{s-7}$

(c) $\frac{6(s+5)}{(s+5)^2+4} – \frac{1}{s-7}$

(d) $\frac{6(s+5)}{(s+5)^2+4} + \frac{1}{s-7}$

I got this question in quiz.

This is a very interesting question from Exponential Fourier Series and Fourier Transforms in portion Fourier Series of Signals and Systems

Answer 1

Right option is (a) $\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}$

The explanation is: We know that, Laplace transform of e^at = $\frac{1}{s-a}$

Here, a=7, so L {e^7t} = $\frac{1}{s-7}$

And the Laplace transform of e^at cos (b t) = $\frac{(s-a)}{(s-a)^2+b^2}$

Here, a=5 and b=2, so L {6e^5t cos (2t)} = $\frac{6(s-5)}{(s-5)^2+4}$

∴ L {6e^5tcos(2t) – e^7t} = $\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}$.