## Question: (s)/T(s). Find the transfer function , TU) θ1(1) θ2(t)

## Answer:

From the given diagram, we can write the following equations:

```
T = K1 * θ1 + K2 * θ2
θ2 = s * θ1 / D + T
```

Substituting the second equation into the first equation, we get:

```
T = K1 * θ1 + K2 * (s * θ1 / D + T)
```

Simplifying, we get:

```
T = (K1 + K2 * s / D) * θ1
```

Therefore, the transfer function is:

```
G(s) = θ2(s) / T(s) = (s / D) / (K1 + K2 * s / D)
```

This can be simplified to:

```
G(s) = s / (KD + K2 * s)
```

where K = K1 + K2.tunesharemore_vertadd_photo_alternate

The question is asking for the transfer function, which is the ratio of the output signal to the input signal, in this case, the angular velocity *θ*2(*t*) to the torque *T*(*s*).

The data from the image shows that the system is a simple feedback system with a proportional controller. The proportional controller has a gain of *K* and the feedback gain is 1/*D*.

The transfer function of the system is as follows:

```
G(s) = \frac{\theta_2(s)}{T(s)} = \frac{K}{1 + DKs}
```

This can be simplified to the following form:

```
G(s) = \frac{\omega_2(s)}{T(s)} = \frac{1}{\tau s + 1}
```

where *τ*=*D*/*K*.

## Example

Let’s assume that the proportional controller has a gain of *K*=10 and the feedback gain is 1/*D*=1. Then, the transfer function of the system would be as follows:

```
G(s) = \frac{\omega_2(s)}{T(s)} = \frac{10}{10s + 1}
```

This means that the system has a time constant of *τ*=0.1 seconds. This means that the output signal will reach 63% of its final value within 0.1 seconds of a step input.

## Conclusion

The transfer function of the system is as follows:

```
G(s) = \frac{\omega_2(s)}{T(s)} = \frac{1}{\tau s + 1}
```

where *τ*=*D*/*K*.