Microelectronics Open Source Book

### [Bode Plots of Arbitrary System Functions with Real Poles and Zeros]{#bode-plots-of-arbitary-system-functions-with-real-poles-and-zeros}

For the case of real negative poles and zeros, and letting $s = j\omega$ , @eq-3.4 becomes

$$
H(j\omega) = K \frac{\left( 1 + j\frac{\omega}{\omega_{z1}} \right) \left( 1 + j\frac{\omega}{\omega_{z2}} \right) \dots \left( 1 + j\frac{\omega}{\omega_{zm}} \right)}{\left( 1 + j\frac{\omega}{\omega_{p1}} \right) \left( 1 + j\frac{\omega}{\omega_{p2}} \right) \dots \left( 1 + j\frac{\omega}{\omega_{pn}} \right)}
$${#eq-3.7}

where $\omega_{p1}, \omega_{p2},... \omega_{pn}$ are the pole frequencies and
$\omega_{z1}, \omega_{z2},..., \omega_{zm}$ are the zero frequencies. For
instance, in [Example 3-2](#example-3-2), we have

$$
H(j\omega) = K \frac{\left( 1 + j\frac{\omega}{\omega_{z1}} \right) }{\left( 1 + j\frac{\omega}{\omega_{p1}} \right) }
$${#eq-3.8}

where

$$
\omega_{z1} = \frac{1}{RC} \quad and \quad \omega_{p1} = \frac{1}{2RC}
$${#eq-3.9}

To determine the Bode plot from @eq-3.7, we
must assess the effect of each binomial term on the
magnitude and phase of the system function. If the
frequency is such that $\omega \ll \omega_{zi}$ or $\omega_{pi}$, then the
respective binomial term will have little effect on the
magnitude and phase of the system function, as it
will simply multiply it by unity. On the other hand, if
the frequency is such that $\omega \ll \omega_{zi}$ or $\omega_{pi}$, the system
function, magnitude, and phase will be altered. To
see this, we evaluate the magnitude and phase of a
general binomial term for a left half plane pole or
zero and $\omega \gg \omega_i$

$$
\left| 1 + j\frac{\omega}{\omega_i}\right| = \sqrt{1 + \left(\frac{\omega}{\omega_i}\right)^2} \simeq \frac{\omega}{\omega_i} 
$${#eq-3.10}

$$
\angle \left(1+ j \frac{\omega}{\omega_i}\right) = tan^{-1} \left(\frac{\omega}{\omega_i}\right) \simeq 90^\circ
$${#eq-3.11}

Therefore, if the binomial term is in the numerator
of the generalized system function (corresponding to
a LHP zero), the magnitude will be multiplied by
${\omega}/{\omega_i}$, and a phase angle of $90^\circ$ will be added to the
total phase. If the binomial term is located in the
denominator (LHP pole), the magnitude will be
multiplied by $1/({\omega}/{\omega_i})$ and a phase angle of $90^\circ$  will
be subtracted from the total phase. For a RHP zero,
it follows that the magnitude will be multiplied by
${\omega}/{\omega_i}$, and a phase angle of $90^\circ$  will be subtracted from
the total phase.