Let $f(t)$ be a function of $t$ defined for all positive values of $t$. The Laplace transform of $f(t)$ denoted by $L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt$, provided that the integral exists where $s$ is a parameter which may be a real or complex number. The Laplace transform of $f(t) = \sin 2t \sin 4t$ is
Laplace Transform of $\sin 2t \sin 4t$
To find the Laplace transform of the function $f(t) = \sin 2t \sin 4t$, we first simplify the function using trigonometric identities.
Trigonometric Simplification
- Use the product-to-sum identity: $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$.
- Let $A = 4t$ and $B = 2t$.
- Then, $2 \sin 4t \sin 2t = \cos(4t - 2t) - \cos(4t + 2t)$.
- $2 \sin 4t \sin 2t = \cos(2t) - \cos(6t)$.
- Therefore, $f(t) = \sin 2t \sin 4t = \frac{1}{2}(\cos 2t - \cos 6t)$.
Calculating Laplace Transform
Now, we apply the Laplace transform operator $L\{\cdot\}$ to the simplified function:
$L\{\sin 2t \sin 4t\} = L\left\{\frac{1}{2}(\cos 2t - \cos 6t)\right\}$
Using the linearity property of Laplace transforms:
$L\{\sin 2t \sin 4t\} = \frac{1}{2} [L\{\cos 2t\} - L\{\cos 6t\}]$
Recall the standard Laplace transform formula for cosine: $L\{\cos(at)\} = \frac{s}{s^2 + a^2}$.
- For $\cos 2t$, $a=2$, so $L\{\cos 2t\} = \frac{s}{s^2 + 2^2} = \frac{s}{s^2 + 4}$.
- For $\cos 6t$, $a=6$, so $L\{\cos 6t\} = \frac{s}{s^2 + 6^2} = \frac{s}{s^2 + 36}$.
Substitute these back into the equation:
$L\{\sin 2t \sin 4t\} = \frac{1}{2} \left[\frac{s}{s^2 + 4} - \frac{s}{s^2 + 36}\right]$
Combine the fractions:
$L\{\sin 2t \sin 4t\} = \frac{s}{2} \left[\frac{(s^2 + 36) - (s^2 + 4)}{(s^2 + 4)(s^2 + 36)}\right]$
$L\{\sin 2t \sin 4t\} = \frac{s}{2} \left[\frac{s^2 + 36 - s^2 - 4}{(s^2 + 4)(s^2 + 36)}\right]$
$L\{\sin 2t \sin 4t\} = \frac{s}{2} \left[\frac{32}{(s^2 + 4)(s^2 + 36)}\right]$
$L\{\sin 2t \sin 4t\} = \frac{16s}{(s^2 + 4)(s^2 + 36)}$
Final Result
The Laplace transform of $f(t) = \sin 2t \sin 4t$ is $\frac{16s}{(s^2 + 4)(s^2 + 36)}$. This corresponds to Option 1 (A).
Important Questions from Laplace Transform
Which of the following is the final value of the impulse response of the system whose transfer function is
(2s + 1)/(s 4 + 8s 3 + 16s 2 + s)
Find the Laplace transform for the following time domain.
y(t) = -2te -t + 4e -t - 4e -2t
Match List I with List II
List – I
List – II
f(t)
F(S)
A.
e -at
I.
\(\rm \frac{s}{s^2+ \omega^2}\)
B.
te at
II.
\(\rm \frac{\omega}{s^2+ \omega^2}\)
C.
sinωt
III.
\(\rm \frac{1}{(s- a)^2}\)
D.
cosωt
IV.
\(\rm \frac{1}{(s+ a)}\)
Choose the correct answer from the options given below:
The Laplace transform of sin h (at) is
The unilateral Laplace transform of f(t) is \(\frac{1}{s^2+s+1}\). The unilateral Laplace transform of t f(t) is
