# Direct Form-I Realization of Continuous-Time Systems

## Realization of Continuous-Time System

Realisation of a continuous-time LTI system means obtaining a network corresponding to the differential equation or transfer function of the system.

The transfer function of the system can be realised either by using integrators or differentiators. Due to certain drawbacks, the differentiators are not used to realise the practical systems. Therefore, only integrators are used for the realization of continuous-time systems. The adder and multipliers are other two elements which are used realise the continuous-time systems.

## Direct Form-I Realization of CT Systems

The direct form-I realization is the simplest and most straight forward structure for the realization of a continuous-time system. In the direct form-I realization of a continuous time system, the differential equation or transfer function describing the system is directly implemented using the separate integrators for the input and output variables. Hence, for realising the system using direct form-I, more number of integrators are required. Therefore, it is more complex.

The direct form-I realization provides a direct relation between time domain and s-domain equations. For direct form realization of continuous-time system, the output variables are expressed in terms of all other terms in the equations.

## Numerical Example

Using the direct form-I realization, realize the system described by the following transfer function −

$$\mathrm{\mathit{H\left ( s \right )\mathrm{\,=\,}\frac{Y\left ( s \right )}{X\left ( s \right )}\mathrm{\,=\,}\frac{s^{\mathrm{2}}\mathrm{\,+\,}\mathrm{4}s\mathrm{\,+\,}\mathrm{3}}{s^{\mathrm{2}}\mathrm{\,+\,}\mathrm{2}s\mathrm{\,+\,}\mathrm{5}}}}$$

### Solution

In order to realise the system described by the transfer function H(s), first express the numerator and denominator of H(s) in powers of 𝑠−1 as −

$$\mathrm{\mathit{H\left ( s \right )\mathrm{\,=\,}\frac{Y\left ( s \right )}{X\left ( s \right )}\mathrm{\,=\,}\frac{s^{\mathrm{2}}\mathrm{\,+\,}\mathrm{4}s\mathrm{\,+\,}\mathrm{3}}{s^{\mathrm{2}}\mathrm{\,+\,}\mathrm{2}s\mathrm{\,+\,}\mathrm{5}}\mathrm{\,=\,}\frac{\mathrm{1}\mathrm{\,+\,}\mathrm{4}s^{\mathrm{-1}}\mathrm{\,+\,}\mathrm{3}s^{\mathrm{-2}}}{\mathrm{1}\mathrm{\,+\,}\mathrm{2}s^{\mathrm{-1}}\mathrm{\,+\,}\mathrm{5}s^{\mathrm{-2}}} }}$$

By cross multiplying, we have,

$$\mathrm{\mathit{Y\left ( s \right )\left [ \mathrm{1}\mathrm{\,+\,}\mathrm{2}s^{\mathrm{-1}}\mathrm{\,+\,}\mathrm{5}s^{\mathrm{-2}} \right ]\mathrm{\,=\,}X\left ( s \right )\left [ \mathrm{1}\mathrm{\,+\,}\mathrm{4}s^{\mathrm{-1}}\mathrm{\,+\,}\mathrm{3}s^{\mathrm{-2}} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow Y\left ( s \right )\mathrm{\,+\,}\mathrm{2}s^{\mathrm{-1}}Y\left ( s \right )\mathrm{\,+\,}\mathrm{5}s^{\mathrm{-2}}Y\left ( s \right )\mathrm{\,=\,}X\left ( s \right )\mathrm{\,+\,}\mathrm{4}s^{\mathrm{-1}}X\left ( s \right )\mathrm{\,+\,}\mathrm{3}s^{\mathrm{-2}}X\left ( s \right )}}$$

$$\mathrm{\mathit{\Rightarrow Y\left ( s \right )\mathrm{\,=\,}X\left ( s \right )\mathrm{\,+\,}\mathrm{4}s^{\mathrm{-1}}X\left ( s \right )\mathrm{\,+\,}\mathrm{3}s^{\mathrm{-2}}X\left ( s \right )-\mathrm{2}s^{\mathrm{-1}}Y\left ( s \right )-\mathrm{5}s^{\mathrm{-2}}Y\left ( s \right )}}$$

Let,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{\,+\,}\mathrm{4}s^{\mathrm{-1}}X\left ( s \right )\mathrm{\,+\,}\mathrm{3}s^{\mathrm{-2}}X\left ( s \right )\mathrm{\,=\,}A\left ( s \right)}}$$

$$\mathrm{\mathit{\therefore Y\left ( s \right )\mathrm{\,=\,}A\left ( s \right )-\mathrm{2}s^{\mathrm{-1}}Y\left ( s \right )-\mathrm{5}s^{\mathrm{-2}}Y\left ( s \right )}}$$

This equation can be realised as follows −

### Step 1

Realise A(s) as −

### Step 2

Realise Y(s) in terms of A(s) as −

### Step 3

Combine the above two results to get the direct form-I realisation as −