How to find Laplace transform of cos(at+b) where a, b are constants

Introduction : In this we derive the Laplace transform of cos(at+b) where a, b are constants. We will also obtain Laplace transform of cos(at+b) in MATLAB.

Question: find Laplace transform of

\[f(t) = cos(at + b)\]

Solution :

We have function

\[f(t) = cos(at + b)\]

First expand this function using formula of cos(C+D)=cos(C)cos(D)-sin(C)sin(D)

\[cos(at + b)= \cos \left( {at} \right)\cos \left( b \right) – \sin \left( {at} \right)\sin \left( b \right)\]

now apply Laplace transform operator L both sides

\[L\left[ {cos(at + b)} \right] = \cos \left( b \right)L\left[ {\cos \left( {at} \right)} \right] – \sin \left( b \right)L\left[ {\sin \left( {at} \right)} \right]\]

Substituting formulas of Laplace transform of cos(at) and sin(at)

\[ = \cos \left( b \right)\frac{s}{{{s^2} + {a^2}}} – \sin \left( b \right)\frac{a}{{{s^2} + {a^2}}}\]

Simplifying the expression we get the result

\[ L\left[ {cos(at + b)} \right]= \frac{{\left( {s\cos \left( b \right) – a\sin \left( b \right)} \right)}}{{{s^2} + {a^2}}}\]

% Laplace transform of cos(at+b) in MATLAB where a, b are constants
syms a b t s
laplace(cos(a*t+b),t,s)

% output results

ans = (sym)

  -a⋅sin(b) + s⋅cos(b)
  ────────────────────
         2    2       
        a  + s