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I am trying to understand the connection between Laplace transform ($s$-plane), and frequency domain calculation.

Let's take the Fourier transform of $\cos(\omega_0t)$, which equals to $\pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]$. So clearly the frequency domain has only two non-zero values at two particular frequencies, and others are zero. Fine!

Now lets do the Laplace transform of the same function: $\cos(\omega_0t)$, which gives me $F(s) = \frac{s}{s^2 + \omega_0^2}=\frac{s}{(s+j\omega_0)(s-j\omega_0)}$.

enter image description here

So there are two poles on the $y$-axis ($j\omega$ axis) as in above plot; $\beta$ is $\omega_0$ here. Now if I move upwards from the origin, I am staying on the frequency axis, and for every point except the pole locations, I am getting a non-zero value [from $F(s)]$, whereas only the pole locations should give me non-zero values (as can be seen from the fourier transform), and zero for all others points.

But this is not happening here. Where exactly I am going wrong here.

Your guidance will be greatly appreciated.

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You're comparing the transforms of two different functions. You consider the Fourier transform of the function $x_1(t)=\cos(\omega_0 t)$, but you took the Laplace transform of the function $x_2(t)=\cos(\omega_0t)u(t)$, where $u(t)$ is the unit step function:

$$X_1(j\omega)=\int_{-\infty}^{\infty}x_1(t)e^{-j\omega t}dt\\ X_2(s)=\int_{0}^{\infty}x_2(t)e^{-st}dt=\int_{0}^{\infty}x_1(t)e^{-st}dt$$

Note the difference in the lower integration limits.

The (bilateral) Laplace transform of the function $x_1(t)=\cos(\omega_0t)$, $-\infty<t<\infty$, does not exist, whereas the Fourier transform of the function $x_2(t)=\cos(\omega_0t)u(t)$ does exist:

$$\mathcal{F}\{\cos(\omega_0t)u(t)\}=\frac{\pi}{2}[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]+\frac{j\omega}{\omega_0^2-\omega^2}\tag{1}$$

However, it cannot be obtained by simply replacing $s$ by $j\omega$ in the expression for $X_2(s)$ because $X_2(s)$ has poles on the imaginary axis, and replacing $s$ by $j\omega$ only gives the correct expression for the Fourier transform if the imaginary axis is inside the region of convergence.

Note that in general for causal functions there are three cases concerning the relationship between the Laplace transform and the Fourier transform:

  1. If the region of convergence (ROC) contains the $j\omega$-axis (i.e., all poles are in the left half-plane), the Fourier transform is simply obtained by the substitution $s=j\omega$.

  2. If there are poles on the $j\omega$-axis, but no poles in the right half-plane, then the Fourier transform contains Dirac delta impulses (as in the example above) plus a term that can be obtained from the Laplace transform by setting $s=j\omega$. Note that in $(1)$ the right-most term is simply $X_2(j\omega)$.

  3. If there are poles in the right half-plane, the Fourier transform of the corresponding causal signal does not exist.

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enter image description here

Bilateral laplace transform of cosine doesn't exist. and hence we use unilateral LT of cos(wt) which is by the way also same as Bilateral laplace transform of (cosine multiplied by unit step signal).Hence finding LT of Cosine is equivalent to finding it by multiplieng with unit step signal.

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  • $\begingroup$ Please use Latex to format equations. $\endgroup$ – Matt L. Jul 18 '18 at 19:48
  • $\begingroup$ Could you try to point out what your answer adds to the accepted answer? $\endgroup$ – Matt L. Jul 18 '18 at 19:51

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