For a strictly proper transfer function with no poles/zeros at infinity, I understand that it gives \$G(\infty) = 0\$. What does it do for a proper transfer function?

# Electrical – Implication of \$s \to \infty\$ in a transfer function

transfer function

#### Related Solutions

The poles in your \$H(s)\$ are \$s = -3\$ and \$s = -2\$ because they make the denominator zero. I'm not sure why you think the Bode plots suggest the poles are positive, but perhaps your confusion has to do with the fact that a Bode plot uses \$j\omega\$ as the \$x\$-axis where \$\omega\$ is the angular frequency. The poles are on the real (\$x\$) axis in the \$s\$-plane so they are symmetric about the imaginary (\$y\$) axis, meaning the Bode plot is the same whether the frequency \$\omega\$ is positive or negative.

The source of the confusion may also be due to the fact that there is an error in the second link you posted. The author uses the form

$$H(s)=A\frac{(s/z_0+1)(s/z_1+1)\cdots(s/z_n+1)}{(s/p_0+1)(s/p_1+1)⋯(s/p_n+1)}$$

for the transfer function but claims that the poles are at \$s = p_0\$, etc. This is incorrect in general because at \$s = p_0\$ the relevant term of the denominator is \$p_0/p_0 + 1 = 1 + 1 = 2 \neq 0\$. The pole is actually at \$s = -p_0\$ so that the relevant term of the denominator is \$-p_0/p_0 + 1 = -1 + 1 = 0\$. The author either meant to say that the poles were at \$s = -p_0\$, etc., or use the form \$s/p_0 - 1\$ for each term.

Once you find \$\dfrac{v_L}{v_1}\$, then just rearranging the terms will give you the standard form. Let me do this example for you.

**Step 1:** Calculate \$\dfrac{v_L}{v_1}\$

From the circuit, the following three equations can be derived directly. $$v_L = k_3v_3 \frac{R_L}{R_L + R_{s3}} \tag{1}$$ $$v_3 = -g_{m2}v_2(r_{\pi3}||\dfrac{1}{sc_{\pi3}})\tag{2}$$ $$v_2 = -(g_{m2}v_2 - g_{m1}v_1)(r_{\pi2}||\dfrac{1}{sc_{\pi2}})\tag{3.a}$$ $$ \Rightarrow v_2 = \frac{g_{m1}v_1(r_{\pi2}||\dfrac{1}{sc_{\pi2}})}{1+g_{m2}(r_{\pi2}||\dfrac{1}{sc_{\pi2}})}\tag3$$

**Step 2:** Rearrange the terms

Now from equation (1), (2) and (3), the value of \$v_L\$ can be expressed in terms of \$v_1\$. From this \$\dfrac{v_L}{v_1}\$ can be calculated. The answer will be in the following form:

$$\frac{v_L}{v_1} = K_1 \times \frac{r_{\pi3}}{1+sc_{\pi3}r_{\pi3}}\times \frac{r_{\pi2}}{1+sc_{\pi2}r_{\pi2}+g_{m2}r_{\pi2}}\tag4$$ where $$K_1= -g_{m1}g_{m2}k_3v_3 \frac{R_L}{R_L + R_{s3}}$$

Now its just a matter of rearranging the terms to convert it into standard form. Dividing numerator and denominator of equation (4) by \$(1+g_{m2}r_{\pi2})\$, $$\Rightarrow \frac{v_L}{v_1} = K_1 \times \frac{r_{\pi3}}{1+sc_{\pi3}r_{\pi3}}\times \frac{\frac{r_{\pi2}}{(1+g_{m2}r_{\pi2})}}{1+s\frac{c_{\pi2}r_{\pi2}}{(1+g_{m2}r_{\pi2})}}$$

$$\Rightarrow \frac{v_L}{v_1} = K \times \frac{1}{1+sc_{\pi3}r_{\pi3}}\times \frac{1}{1+s\frac{c_{\pi2}r_{\pi2}}{(1+g_{m2}r_{\pi2})}}\tag5$$ where $$K = K_1 \times r_{\pi3} \times \frac{r_{\pi2}}{(1+g_{m2}r_{\pi2})}$$

Now equation (5) is in standard form and value of poles can be calculated directly.

## Best Answer

For a transfer function with more finite poles than finite zeros, the gain goes to zero as the frequency increases without bound.

For a transfer function with an equal number of finite poles and finite zeros, the gain will be non-zero as the frequency increases without bound. This is given by the "D" term in the state space representation of the transfer function.