Question

Given f (t) = [cos (3t)]^2. The value of L {f(t)} is _______________

(a) \(\frac{1}{2}(\frac{1}{s} + \frac{s}{s^2+36})\)

(b) \(\frac{1}{4}(\frac{1}{s} + \frac{s}{s^2+36})\)

(c) \(\frac{1}{5}(\frac{1}{s} + \frac{s}{s^2+36})\)

(d) \(\frac{1}{8}(\frac{1}{s} + \frac{s}{s^2+36})\)

The question was asked in class test.

Enquiry is from Common Laplace Transforms topic in portion Laplace Transform and System Design of Signals and Systems

Answer 1

Right choice is (a) \(\frac{1}{2}(\frac{1}{s} + \frac{s}{s^2+36})\)

Easiest explanation: We know that by trigonometric identity, cos^2(3t) = \(\frac{1}{2}\)(1 + cos⁡(6t))

Also, we know that, L {cos (at)} = \(\frac{s}{s^2+a^2}\)

And L {1} = \(\frac{1}{s}\)

So, L {cos^2 (3t)} = L \({\frac{1}{2}(1+cos⁡(6t))} = \frac{1}{2}(\frac{1}{s} + \frac{s}{s^2+36})\).