Transfer Function of The Laplace Transform

The transfer function is a key concept in signal processing because it indicates how a signal is processed as it passes through a network. It is a fitting tool for finding the network response, determining (or designing for) network stability, and network synthesis. The transfer function of a network describes how the output behaves in respect to the input. It specifies the transfer from the input to the output in the $ s $ domain, assuming no initial energy.
The transfer function H(s) is the ratio of the output response Y(s) to the input excitation X(s), assuming all initial conditions are zero.
Thus, $$H(s)=\frac{Y(s)}{X(s)} \tag{2}$$ The transfer function depends on what we define as input and output. since the input and output can be either current or voltage at any place in the circuit, there are four possible transfer functions: $$\begin{array}{c}H(s)=\text { Voltage gain }=\frac{V_{o}(s)}{V_{i}(s)} \\H(s)=\text { Current gain }=\frac{I_{o}(s)}{I_{i}(s)} \\H(s)=\text { Impedance }=\frac{V(s)}{I(s)} \\H(s)=\text { Admittance }=\frac{I(s)}{V(s)}\end{array} \tag{1}$$ Thus, a circuit can have many transfer functions. Note that $ H(s) $ is dimensionless in Eqs. (1).
Each of the transfer functions in Eq. (1) can be found in two ways. One way is to assume any convenient input $ X(s) $, use any circuit analysis technique (such as current or voltage division, nodal or mesh analysis) to find the output $ Y(s) $, and then obtain the ratio of the two. The other approach is to apply the ladder method, which involves walking our way through the circuit.
By this approach, we assume that the output is $ 1 \mathrm{~V} $ or $ 1 \mathrm{~A} $ as appropriate and use the basic laws of $ \mathrm{Ohm} $ and Kirchhoff (KCL only) to obtain the input. The transfer function becomes unity divided by the input. This approach may be more convenient to use when the circuit has many loops or nodes so that applying nodal or mesh analysis becomes cumbersome.
In the first method, we assume an input and find the output; in the second method, we assume the output and find the input. In both methods, we calculate $ H(s) $ as the ratio of output to input transforms.
The two methods rely on the linearity property, since we only deal with linear circuits in this book. Example $ 1 $ illustrates these methods. Equation (2) assumes that both $ X(s) $ and $ Y(s) $ are known. Sometimes, we know the input $ X(s) $ and the transfer function $ H(s) $. We find the output $ Y(s) $ as $$Y(s)=H(s) X(s) \tag{3}$$ and take the inverse transform to get $ y(t) $. A special case is when the input is the unit impulse function, $ x(t)=\delta(t) $, so that $ X(s)=1 $. For this case, $$Y(s)=H(s) \quad \text { or } \quad y(t)=h(t)$$ where $$h(t)=\mathcal{L}^{-1}[H(s)] \tag{4}$$ The term $ h(t) $ represents the unit impulse response - it is the time-domain response of the network to a unit impulse. Thus, Eq. (4) provides a new interpretation for the transfer function: $ H(s) $ is the Laplace transform of the unit impulse response of the network. Once we know the impulse response $ h(t) $ of a network, we can obtain the response of the network to any input signal using Eq. (3) in the $ s $ domain or using the convolution integral (see next section) in the time domain.
Example 1: The output of a linear system is $$ y(t)=10 e^{-t} \cos 4 t u(t) $$ when the input is $ x(t)=e^{-t} u(t) $. Find the transfer function of the system and its impulse response.
Solution: If $ x(t)=e^{-t} u(t) $ and $ y(t)=10 e^{-t} \cos 4 t u(t) $, then $$X(s)=\frac{1}{s+1} \quad \text { and } \quad Y(s)=\frac{10(s+1)}{(s+1)^{2}+4^{2}}$$ The unit impulse response is the output response of a circuit when the input is a unit impulse. Hence, $$H(s)=\frac{Y(s)}{X(s)}=\frac{10}{(s+1)^{2}+16}=\frac{10}{s^{2}+2 s+17}$$ To find $ h(t) $, we write $ H(s) $ as $$H(s)=\frac{10}{4} \frac{4}{(s+1)^{2}+4^{2}}$$ From Table 2, we obtain $$h(t)=2.5 e^{-t} \sin 4 t$$
Example 2: Determine the transfer function $ H(s)=V_{o}(s) / I_{o}(s) $ of the circuit in Fig. $ 1 $
Fig. 1: Example 2.
METHOD I By current division, $$I_{2}=\frac{(s+4) I_{o}}{s+4+2+1 / 2 s}$$ But $$V_{o}=2 I_{2}=\frac{2(s+4) I_{o}}{s+6+1 / 2 s}$$ Hence, $$H(s)=\frac{V_{o}(s)}{I_{o}(s)}=\frac{4 s(s+4)}{2 s^{2}+12 s+1}$$ METHOD 2 We can apply the ladder method. We let $ V_{o}=1 \mathrm{~V} $. By Ohm's law, $ I_{2}=V_{o} / 2=1 / 2 \mathrm{~A} $. The voltage across the $ (2+1 / 2 s) $ impedance is $$V_{1}=I_{2}\left(2+\frac{1}{2 s}\right)=1+\frac{1}{4 s}=\frac{4 s+1}{4 s}$$ This is the same as the voltage across the $ (s+4) $ impedance. Hence, $$I_{1}=\frac{V_{1}}{s+4}=\frac{4 s+1}{4 s(s+4)}$$ Applying $ \mathrm{KCL} $ at the top node yields $$I_{o}=I_{1}+I_{2}=\frac{4 s+1}{4 s(s+4)}+\frac{1}{2}=\frac{2 s^{2}+12 s+1}{4 s(s+4)}$$ Hence, $$H(s)=\frac{V_{o}}{I_{o}}=\frac{1}{I_{o}}=\frac{4 s(s+4)}{2 s^{2}+12 s+1}$$ as before.
Example 3: For the $ s $-domain circuit in Fig. 2, find:
(a) the transfer function $ H(s)=V_{o} / V_{i} $,
(b) the impulse response,
(c) the response when $ v_{i}(t)= $ $ u(t) \mathrm{V} $,
(d) the response when $ v_{i}(t)=8 \cos 2 t \mathrm{~V} $.
Fig. 2: Example 3.
(a) Using voltage division, $$V_{o}=\frac{1}{s+1} V_{a b} \tag{3.1}$$ But $$V_{a b}=\frac{1 \|(s+1)}{1+1 \|(s+1)} V_{i}=\frac{(s+1) /(s+2)}{1+(s+1) /(s+2)} V_{i}$$ or $$V_{a b}=\frac{s+1}{2 s+3} V_{i} \tag{3.2}$$ Substituting Eq. (3.2) into Eq. (3.1) results in $$V_{o}=\frac{V_{i}}{2 s+3}$$ Thus, the impulse response is $$H(s)=\frac{V_{o}}{V_{i}}=\frac{1}{2 s+3}$$ (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: $$h(t)=\frac{1}{2} e^{-3 t / 2} u(t)$$ (c) When $ v_{i}(t)=u(t), V_{i}(s)=1 / s $, and $$V_{o}(s)=H(s) V_{i}(s)=\frac{1}{2 s\left(s+\frac{3}{2}\right)}=\frac{A}{s}+\frac{B}{s+\frac{3}{2}}$$ where $$\begin{array}{c}A=\left.s V_{o}(s)\right|_{s=0}=\left.\frac{1}{2\left(s+\frac{3}{2}\right)}\right|_{s=0}=\frac{1}{3} \\B=\left.\left(s+\frac{3}{2}\right) V_{o}(s)\right|_{s=-3 / 2}=\left.\frac{1}{2 s}\right|_{s=-3 / 2}=-\frac{1}{3}\end{array}$$ Hence, for $ v_{i}(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 $$v_{o}(t)=\frac{1}{3}\left(1-e^{-3 t / 2}\right) u(t) \mathrm{V}$$ (d) When $ v_{i}(t)=8 \cos 2 t $, then $ V_{i}(s)=\frac{8 s}{s^{2}+4} $, and $$\begin{aligned}V_{o}(s) &=H(s) V_{i}(s)=\frac{4 s}{\left(s+\frac{3}{2}\right)\left(s^{2}+4\right)} \\&=\frac{A}{s+\frac{3}{2}}+\frac{B s+C}{s^{2}+4}\end{aligned} \tag{3.3}$$ where $$A=\left.\left(s+\frac{3}{2}\right) V_{o}(s)\right|_{s=-3 / 2}=\left.\frac{4 s}{s^{2}+4}\right|_{s=-3 / 2}=-\frac{24}{25}$$ To get $ B $ and $ C $, we multiply Eq. $ (3.3) $ by $ (s+3 / 2)\left(s^{2}+4\right) $. We get $$4 s=A\left(s^{2}+4\right)+B\left(s^{2}+\frac{3}{2} s\right)+C\left(s+\frac{3}{2}\right)$$ Equating coefficients, $$\begin{aligned}\text { Constant: } & 0 &=4 A+\frac{3}{2} C \quad \Longrightarrow \quad C=-\frac{8}{3} A \\s: & 4 &=\frac{3}{2} B+C \\s^{2}: & 0 &=A+B \quad \Longrightarrow \quad B=-A\end{aligned}$$ Solving these gives $ A=-24 / 25, B=24 / 25, C=64 / 25 $. Hence, for $ v_{i}(t)=8 \cos 2 t \mathrm{~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 $$v_{o}(t)=\frac{24}{25}\left(-e^{-3 t / 2}+\cos 2 t+\frac{4}{3} \sin 2 t\right) u(t) \mathrm{V}$$