I Ω=0 and its magnitude for a unit pulse

[PainterGuy]

Hi,

I was trying to find the magnitude of DC signal in a shifted unit pulse signal. The unshifted pulse lasts from -π to π and then it is shifted by π duration.

In Method #1 I have used FT formula and ended up with magnitude of 6.2832 for DC signal, i.e. ω=0.

In Method #2 I have used Laplace transform to derive the FT by setting σ=0 and ended up with magnitude of 6.5938 for the DC signal, i.e. ω=0.

What is contributing toward the error of 6.5938-6.2832=0.3106?

Could you please help me with it? Thank you!

Method #1:

Finding magnitude of DC signal:

Method #2:

Finding magnitude of DC signal:

 

[PainterGuy]

I think I was able to trace the problem. Method #1 produces the right answer.

There is a mistake Method #2. I have the step where the mistake occurred but even after fixing it, I couldn't simplify it any further. Could you please help me with it?

Please note that I didn't write lim_ω→0 with every step. Thank you

 

[Fred Wright]

Yes, you got the right answer in method #1. For method #2 you have,$$
\frac{1}{s}-\frac{e^{2\pi s}}{s}\rightarrow \frac{1}{j\omega}-\frac{e^{2\pi j \omega}}{j\omega} =e^{j\pi \omega}(\frac{e^{-j\pi \omega}-e^{j\pi \omega}}{j\omega})=-e^{j\pi \omega}\frac{2\sin(\pi \omega)}{\omega}$$Taking the absolute value, you recover the answer from method #1

 

[PainterGuy]

Yes, you got the right answer in method #1. For method #2 you have,$$
\frac{1}{s}-\frac{e^{2\pi s}}{s}\rightarrow \frac{1}{j\omega}-\frac{e^{2\pi j \omega}}{j\omega} =e^{j\pi \omega}(\frac{e^{-j\pi \omega}-e^{j\pi \omega}}{j\omega})=-e^{j\pi \omega}\frac{2\sin(\pi \omega)}{\omega}$$Taking the absolute value, you recover the answer from method #1

Thank you!

But I was purposely trying to avoid exponentials. Is there any way to simplify the trigonometric expression in Method #2 further?

 

[PainterGuy]

Hi,

I was able to fix Method #2 and got the correct answer.