## Find the transfer function G(s)=VL(s)/V(s) for each of the networks shown below

## Answer:

The transfer function for each of the networks shown in the image is as follows:

**Network A:**

```
G(s) = VL(s)/V(s) = R/(R+Ls)
```

**Network B:**

```
G(s) = VL(s)/V(s) = 1/(1+RCs)
```

**Network C:**

```
G(s) = VL(s)/V(s) = R/(R+Ls) * 1/(1+RCs)
```

**Solution:**

To obtain the transfer function for Network A, we can use the following steps:

- Apply Kirchhoff’s voltage law to the loop containing the voltage source, resistor, and inductor:

```
V = R * i(t) + L * di(t)/dt
```

- Take the Laplace transform of both sides of the equation:

```
V(s) = R * I(s) + L * sI(s)
```

- Solve for I(s):

```
I(s) = V(s) / (R + Ls)
```

- The transfer function is defined as the ratio of the output voltage to the input voltage, so:

```
G(s) = VL(s)/V(s) = I(s) = R / (R + Ls)
```

The transfer function for Network B can be obtained in a similar manner. However, instead of an inductor, we have a capacitor in the circuit. The Laplace transform of the capacitor current is:

```
I_C(s) = C * s * V(s)
```

Therefore, the transfer function for Network B is:

```
G(s) = VL(s)/V(s) = I_C(s) / V(s) = 1 / (1 + RCs)
```

The transfer function for Network C can be obtained by combining the transfer functions of Networks A and B:

```
G(s) = G_A(s) * G_B(s) = R / (R + Ls) * 1 / (1 + RCs)
```

Therefore, the transfer function for Network C is:

`G(s) = VL(s)/V(s) = R / (R + Ls) * 1 / (1 + RCs)`