In Reply to: POLE posted by PL John on 08/19/03 at 12:13 PM:
Hi, PL John.
I think I need to suggest a different look at poles and zeros because pole-zero analysis applies to impedances and admittances as well as to transfer functions.
For example, let's take the simple transfer function case of a single resistor and capacitor arranged as a low-pass filter. A signal source we shall call E1 is referenced to ground and is connected to a resistor R that goes over to an output node we shall call E2. The capacitor C is then connected from E2 to ground.
We use the notation S = sigma + j * W in which we mean that a "frequency" is not just a scalar value, but a vector value. The "sigma" part is the real part of that vector value while the "W" part (We call this omega, but I don't have the Greek letter available on the keyboard.) is equal to 2 * pi * F. The term "pi" is 3.14159265..., the term "F" is the thing we commonly understand as frequency in cycles per second, or in Hertz (Hz), and the term "j" is the square-root of minus one.
Okay, then. If we have a sinusoidal source from E1, the sigma part of the E1's output is zero. Thus, we can write the transfer function of this lowpass filter in terms of the "S" parameter as follows:
The impedance value of the resistor is simply R. We write the impedance value of the capacitor as 1/( j * W * C), or more conveniently, 1 / ( S * C ).
Just like with a voltage divider, the transfer function is given as:
H(S) = ( 1 / ( S * C ) ) / ( R + ( 1 / ( S * C ) ) = 1 / ( 1 + S * R * C )
Then, in terms of frequency, this comes to H( j W ) = 1 / ( 1 + j * W * R * C ).
In terms of magnitude:
H ( W ) = 1 / Sqrt ( 1 + ( W * R * C ) ^2 ) = 1 / Sqrt ( 1 + ( 2 * pi * F * R * C ) ^2 ) .
Note here that there is no frequency, F, in Hz at which the denominator of this transfer function becomes zero. Therefore, the transfer function itself can never become infinitity for any F. Instead, the denominator is a vector sum whose numerical value is always well behaved.
Enough for the moment?