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$$The transfer function of a linear system is Practice Problem 16.7$$

H(s) = \frac{2s}{s+6}

Find the output *y*(*t*) due to the input 45*e* −3*t u*(*t*) and its impulse response. **Answer:** −90*e*⁻3*^t* + 180*e*⁻6*^t* , *t* ≥ 0, 2*δ*(*t*) − 12*e*⁻6*^t u*(*t*). The unit impulse response is the output response of a circuit when the input is a unit impulse. I 1 Example 16.8 Determine the transfer function *H*(*s*) = *Vo*(*s*)∕*Io*(*s*) of the circuit in Fig. 16.18. # **Solution:** ■ **METHOD 1** By current division,

I_2 = \frac{(s+4)I_o}{s+4+2+1/2s}

$$But + ‒ Vo 2 Ω 4 Ω s$$

V_o = 2I_2 = \frac{2(s+4)I_o}{s+6+1/2s}

$$Hence,$$

H(s) = \frac{V_o(s)}{I_o(s)} = \frac{4s(s+4)}{2s^2 + 12s + 1}

$$■ **METHOD 2** We can apply the ladder method. We let *Vo* = 1 V. By Ohm's law, *I*2 = *Vo*∕2 = 1∕2 A. The voltage across the (2 + 1∕2*s*) impedance is$$

V_1 = I_2 \left( 2 + \frac{1}{2s} \right) = 1 + \frac{1}{4s} = \frac{4s + 1}{4s}

$$This is the same as the voltage across the (*s* + 4) impedance. Hence,$$

I_1 = \frac{V_1}{s+4} = \frac{4s+1}{4s(s+4)}

$$Applying KCL at the top node yields$$

I_o = I_1 + I_2 = \frac{4s + 1}{4s(s + 4)} + \frac{1}{2} = \frac{2s^2 + 12s + 1}{4s(s + 4)}

$$Hence,$$

H(s) = \frac{V_o}{I_o} = \frac{1}{I_o} = \frac{4s(s+4)}{2s^2 + 12s + 1}

$$as before. Practice Problem 16.8 Find the transfer function *H*(*s*) = *I*1(*s*)∕*Io*(*s*) in the circuit of Fig. 16.18. Answer:$$

\frac{4s+1}{2s^2+12s+1}

. V(s) + ‒ **Figure 16.18** For Example 16.8. # **Solution:** (a) Using voltage division,

V_o = \frac{1}{s+1} V_{ab}

$$(16.9.1) But$$

V_{ab} = \frac{1|| (s+1)}{1+1|| (s+1)} V_i = \frac{(s+1)/(s+2)}{1+(s+1)/(s+2)} V_i

$$or$$

V_{ab} = \frac{s+1}{2s+3} V_i

$$(16.9.2) Substituting Eq. (16.9.2) into Eq. (16.9.1) results in$$

V_o = \frac{V_i}{2s + 3}

$$Thus, the transfer function is$$

H(s) = \frac{V_o}{V_i} = \frac{1}{2s + 3}

$$(b) We may write *H*(*s*) as$$

H(s) = \frac{1}{2} \frac{1}{s + \frac{3}{2}}

$$Its inverse Laplace transform is the required impulse response:$$

h(t) = \frac{1}{2}e^{-3t/2}u(t)

$$(c) When *vi*(*t*) = *u*(*t*), *Vi*(*s*) = 1∕*s*, and$$

V_o(s) = H(s)V_i(s) = \frac{1}{2s(s + \frac{3}{2})} = \frac{A}{s} + \frac{B}{s + \frac{3}{2}}

$$where$$

A = sV_o(s)|{s=0} = \frac{1}{2(s + \frac{3}{2})}|{s=0} = \frac{1}{3}

$$$$

B = \left(s + \frac{3}{2}\right) V_o(s)|{s=-3/2} = \frac{1}{2s}|{s=-3/2} = -\frac{1}{3}

$$Hence, for *vi*(*t*) = *u*(*t*),$$

V_o(s) = \frac{1}{3} \left( \frac{1}{s} - \frac{1}{s + \frac{3}{2}} \right)

$$and its inverse Laplace transform is$$

v_o(t) = \frac{1}{3}(1 - e^{-3t/2})u(t) \text{ V}

$$For Example 16.9. <span id="page-750-0"></span>(d) When$$

v_i(t) = 8 \cos 2t

, then $V_i(s) = \frac{8s}{s^2 + 4}$ , and \n

V_o(s) = H(s)V_i(s) = \frac{4s}{(s + \frac{3}{2})(s^2 + 4)}

$$\n$$

= \frac{A}{s + \frac{3}{2}} + \frac{Bs + C}{s^2 + 4}

$$\n(16.9.3) where$$

A = \left(s + \frac{3}{2}\right) V_o(s) \Big|{s = -3/2} = \frac{4s}{s^2 + 4} \Big|{s = -3/2} = -\frac{24}{25}

$$To get *B* and *C*, we multiply Eq. (16.9.3) by (*s* + 3∕2)(*s* 2 + 4). We get$$

4s = A(s^{2} + 4) + B(s^{2} + \frac{3}{2}s) + C(s + \frac{3}{2})

$$Equating coefficients, Constant:$$

0 = 4A + \frac{3}{2}C

$\Rightarrow$ $C = -\frac{8}{3}A$ \ns: $4 = \frac{3}{2}B + C$ \ns²: $0 = A + B$ $\Rightarrow$ $B = -A$ Solving these gives *A* = −24∕25, *B* = 24∕25, *C* = 64∕25. Hence, for *vi*(*t*) = 8 cos 2*t* V,

V_o(s) = \frac{-\frac{24}{25}}{s + \frac{3}{2}} + \frac{24}{25} \frac{s}{s^2 + 4} + \frac{32}{25} \frac{2}{s^2 + 4}

$$and its inverse is$$

v_o(t) = \frac{24}{25} \left( -e^{-3t/2} + \cos 2t + \frac{4}{3} \sin 2t \right) u(t) \text{ V}

$$**Figure 16.20** **Figure 16.21** A linear system with *m* inputs and *p* outputs. Practice Problem 16.9 Rework Example 16.9 for the circuit shown in Fig. 16.20. **Answer:** (a)$$

2/(s + 4)

, (b) $2e^{-4t}u(t)$ , (c) $\frac{1}{2}(1 - e^{-4t})u(t)$ V, (d) $3.2(-e^{-4t} + \cos 2t + \frac{1}{2} \sin 2t)u(t)$ V.