Hi

For an RC circuit (Vout across capacitor), the impulse response is:

h(t) = 1/RC * exp[-t/RC] * u(t), where u(t) is the unit step.

The transfer function of the circuit is:

H(s) = 1/RC * 1/(s + 1/RC)

Here, the the pole is at Re(s) = -1/RC, and Im(s) = 0. The location of the pole is dictated by the physical parameters of the circuit, namely the size of R and C.

What I don't follow is the validity of obtaining the frequency response:

We set Re(s) = 0, and Im(s) = jw

That way, we get:

H(jw) = 1/RC * 1/(jw + 1/RC)

Two questions:
1. It was Re(s) of the pole which told us about the physical parameters of the system (R and C) - how can we just set it to zero?

2. When we set s = jw, what is it that makes the sinusoid be applied to the input of the RC circuit, rather than be applied to some other node like in between R and C?

Thanks.

 

"Fourier transforms and the closely related Laplace transforms are widely used in solving differential equations."
See:
http://en.wikipedia.org/wiki/Fourier_transform

Any signal in time can be expressed as series of sinussoidal functions. We get a Fourier series for time functions that are periodic. The Fourier series goes to the Fourier transformation when the time period of the time function goes to infinite.

With the Fourier transformation we get functions in the fequency domain like e^-i*omega*t

With the Laplace transformation we get functions in the frequency domain like e^-s*t

where s = sigma + i*omega

Both Fourier and Laplace transformations share some interesting properties with differentiation and integration.

If you have a linear system and apply a sinusoidal function to it you will get a sinusoidal function at the output.
There will only be a different amplitude and phase at the output. Thus differentiation in the time domain is equivalent to a multiplication in the frequency domain.

For example you could wobble a linear system and measure it's transfer function. (Change frequency at the input and measure amplitude and phase at the output) The result would be a bode diagram, or you could say the Fourier transformation.

What is true for a single sinus at the input is also true for a sum of sinus (N*omega0*t)

The Laplace transformation now has the same mathematical advantage to convert a differentiation in time to a multiplication in the frequency domain.

But the real part of the complex frequency (sigma)
s = sigma + i*omega

helps to solve the integral of the transformation because the integral does converge better than if you only had a imaginary frequency.

 

You can use Laplace and inverse Laplace transforms directly. Equation for such simply function are ready and maybe found thro Internet.