Find the Laplace transform of:
f(t) = cos^(2)(2t)
cos^2(2t)
L{cos(t)} = s/(s^2+1)
L{f(at)} = 1/a f(s/a), so
L{cos(4t)} = 1/4 (s/4)/(s^2/16+1) = s/(s^2+6)
L{cos^2(2t)} = L{(1+cos(4t))/2}
= 1/2 (L{1} + L{cos(4t)})
= 1/2 (1/s + s/(s^2+16))
= (s^2+8) / (s^3+16s)
Thank you!!
L(cos^2 2t)=s^2+8/s(s^2+16)
Why did the cosine go to school? Because it wanted to learn how to transform itself into a Laplace!
The Laplace transform of f(t) = cos^2(2t) can be found using the trigonometric identity cos^2(x) = (1 + cos(2x))/2.
So, let's break it down:
f(t) = (1 + cos(4t))/2
Now, we can take the Laplace transform:
L{f(t)} = L{(1 + cos(4t))/2}
Applying the linearity property of the Laplace transform:
L{f(t)} = (1/2) * [L{1} + L{cos(4t)}]
The Laplace transform of the constant 1 is 1/s, where s is the complex variable.
The Laplace transform of cos(4t) can be found using the formula:
L{cos(at)} = s/(s^2 + a^2)
So, substituting in our values:
L{f(t)} = (1/2) * [1/s + s/(s^2 + 4^2)]
Simplify:
L{f(t)} = (1/2) * [1/s + s/(s^2 + 16)]
And there you have it - the Laplace transform of f(t) = cos^2(2t) is (1/2) * [1/s + s/(s^2 + 16)].
Now that I've transformed this problem into a laughing matter, do you need any additional assistance?
To find the Laplace transform of f(t) = cos^2(2t), we can apply the basic Laplace transform properties and the Laplace transform table.
Step 1: Express the given function in terms of the exponential function.
We know that cos^2(2t) can be written as (1/2)(1 + cos(4t)). This is achieved by using the identity cos^2(x) = (1/2)(1 + cos(2x)).
Step 2: Apply the basic Laplace transform property.
The Laplace transform of f(t) = (1/2)(1 + cos(4t)) is given by:
L{(1/2)(1 + cos(4t))} = (1/2)L{1} + (1/2)L{cos(4t)}
Step 3: Use the Laplace transform table.
The Laplace transform of 1 is given by L{1} = 1/s and the Laplace transform of cos(4t) is given by L{cos(4t)} = s/(s^2 + 16).
Substituting these values into the equation from step 2, we get:
L{(1/2)(1 + cos(4t))} = (1/2)(1/s) + (1/2)(s)/(s^2 + 16)
Simplifying further, we have:
L{(1/2)(1 + cos(4t))} = 1/(2s) + s/(2s^2 + 32)
Therefore, the Laplace transform of f(t) = cos^2(2t) is 1/(2s) + s/(2s^2 + 32).