Transfer Functions and Block Diagrams

Transfer Function

For an LTI system with zero initial conditions,

$$G(s) = \frac{Y(s)}{U(s)}$$

$G(s)$ is the Laplace transform of the impulse response.

DC gain The steady-state gain for a constant input. With $u(t) = u_\infty$, $U(s) = u_\infty / s$, and if $G$ is stable the final value theorem gives

$$\text{DC gain} = \frac{y_\infty}{u_\infty} = \lim_{s \to 0} G(s) = G(0)$$

Second-Order Systems

$$\ddot{y}(t) + 2\zeta\omega_n \dot{y}(t) + \omega_n^2 y(t) = G_0 \omega_n^2 u(t) \quad \Longrightarrow \quad G(s) = \frac{G_0\, \omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$

with damping ratio $\zeta$, natural frequency $\omega_n$ and DC gain $G_0$. For an underdamped system ($0 < \zeta < 1$) the step response overshoot is

$$M_p = \exp\left(-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}\right)$$

Block Diagram Algebra

• Series $G_1$ then $G_2$ gives $G_1 G_2$.
• Parallel $X_2 = X_1 G_1 + X_1 G_2$, so $X_2 / X_1 = G_1 + G_2$.
• Feedback Forward path $G$, feedback path $H$, summing junction $X_1 \pm H X_2$.

$$X_2 = G(X_1 \pm H X_2) \quad \Longrightarrow \quad \frac{X_2}{X_1} = \frac{G}{1 \mp GH}$$

so negative feedback gives $G / (1 + GH)$.

Circuits

$$i(t) = C \frac{dv(t)}{dt}, \qquad v(t) = L \frac{di(t)}{dt}$$

so in the Laplace domain a capacitor has impedance $1/(Cs)$, an inductor $Ls$ and a resistor $R$.

An ideal op-amp has infinite gain and draws no input current, so with negative feedback $V_+ \approx V_-$. For the inverting amplifier (input through $R_1$, feedback through $R_2$, $V_+$ grounded), Kirchhoff's current law at the inverting node with $V_- = 0$ gives

$$\frac{0 - V_{in}}{R_1} + \frac{0 - V_{out}}{R_2} = 0 \quad \Longrightarrow \quad \frac{V_{out}}{V_{in}} = -\frac{R_2}{R_1}$$

For stages separated by op-amps, the overall transfer function is the product of the individual transfer functions. E.g. two RC stages separated by a unity-gain inverting amplifier give

$$\frac{V_2(s)}{V_{in}(s)} = \frac{1}{R_2 C_2 s + 1} \cdot \frac{-1}{R_1 C_1 s + 1}$$