N
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.
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.