Cascade Form Realization of Continuous-Time Systems

Signals and Systems Electronics & Electrical Digital Electronics

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.

Cascade Form Realisation of CT Systems

In the cascade form realisation of continuous-time systems, the transfer function of the system is expressed as the product of several transfer functions of first order and second order, and each of these transfer functions are then realised using integrators and summers and finally, the all the realised structures are cascaded, i.e., connected in series.

The output of the first structure is connected to the input of the second one, and the output of the second to the input of the third and so on. Therefore, in the cascade form realisation of continuous-time systems, the input is connected to the first structure and the output is taken from the last one.

Following example explains the realization of continuous-time systems in the cascade form.

Numerical Example

Using the cascade form realisation, realise the continuous time system described by the transfer function.

$$\mathrm{\mathit{H\left ( s \right )\mathrm{\,=\,}\frac{\mathrm{3}\left ( s^{\mathrm{2}}\mathrm{\,+\,}\mathrm{3}s\mathrm{\,+\,}\mathrm{2} \right )}{s^{\mathrm{3}}\mathrm{\,+\,}\mathrm{12}s^{\mathrm{2}}\mathrm{\,+\,}\mathrm{47}s\mathrm{\,+\,}\mathrm{60}}}}$$

Solution

The given transfer function is,

Expressing the given transfer function as the product of several transfer functions as,

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

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

Let,

$$\mathrm{\mathit{H_{\mathrm{1}}\left ( s \right )\mathrm{\,=\,}\left ( \frac{\mathrm{3}}{s\mathrm{\,+\,}\mathrm{3}} \right )\mathrm{\,=\,}\frac{\mathrm{3}s^{-\mathrm{1}}}{\mathrm{1\mathrm{\,+\,}3}s^{\mathrm{-1}}}}} $$

$$\mathrm{\mathit{H_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}\left ( \frac{s\mathrm{\,+\,}\mathrm{2}}{s\mathrm{\,+\,}\mathrm{4}} \right )\mathrm{\,=\,}\frac{\mathrm{1\mathrm{\,+\,}2}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}4}s^{-\mathrm{1}}}}}$$

$$\mathrm{\mathit{H_{\mathrm{3}}\left ( s \right )\mathrm{\,=\,}\left ( \frac{s\mathrm{\,+\,}\mathrm{1}}{s\mathrm{\,+\,}\mathrm{5}} \right )\mathrm{\,=\,}\frac{\mathrm{1\mathrm{\,+\,}}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}5}s^{-\mathrm{1}}}}}$$

Each of these transfer functions can be realised as follows −

Step 1

Realisation of $\mathrm{\mathit{H_{\mathrm{1}}\left ( s \right )}}$ −

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

$$\mathrm{\mathit{\frac{X_{\mathrm{1}}\left ( s \right )}{A\left ( s \right )}\mathrm{\,=\,}\mathrm{3}s^{-\mathrm{1}}}}$$

$$\mathrm{\mathit{\Rightarrow X_{\mathrm{1}}\left ( s \right )\mathrm{\,=\,}\mathrm{3}s^{-\mathrm{1}}A\left ( s \right )}}$$

And

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

$$\mathrm{\mathit{\Rightarrow A\left ( s \right )\frac{}{}\mathrm{\,=\,}X\left ( s \right )-\mathrm{3}s^{-\mathrm{1}}A\left ( s \right )}}$$

Step 2

Realisation of $\mathrm{\mathit{H_{\mathrm{2}}\left ( s \right )}}$ −

$$\mathrm{\mathit{H_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}\frac{X_{\mathrm{2}}\left ( s \right )}{X_{\mathrm{1}}\left ( s \right )}\mathrm{\,=\,}\frac{X_{\mathrm{2}}\left ( s \right )}{A_{\mathrm{1}}\left ( s \right )}\frac{A_{\mathrm{1}}\left ( s \right )}{X_{\mathrm{1}}\left ( s \right )}\mathrm{\,=\,}\frac{\mathrm{1\mathrm{\,+\,}2}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}4}s^{-\mathrm{1}}}}}$$

$$\mathrm{\mathit{\Rightarrow \frac{X_{\mathrm{2}}\left ( s \right )}{A_{\mathrm{1}}\left ( s \right )}\mathrm{\,=\,}\mathrm{1}\mathrm{\,+\,}\mathrm{2}s^{\mathrm{-1}}}}$$

$$\mathrm{\mathit{\Rightarrow X_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}A_{\mathrm{1}}\left ( s \right )\mathrm{\,+\,}\mathrm{2}s^{\mathrm{-1}}A\left ( s \right )}}$$

And,

$$\mathrm{\mathit{\frac{A_{\mathrm{1}}\left ( s \right )}{X_{\mathrm{1}}\left ( s \right )}\mathrm{\,=\,}\frac{\mathrm{1}}{\mathrm{1\mathrm{\,+\,}4}s^{\mathrm{-1}}}}}$$

$$\mathrm{\mathit{\Rightarrow A_{\mathrm{1}}\left ( s \right )\mathrm{\,=\,}X_{\mathrm{1}}\left ( s \right )-\mathrm{4}s^{\mathrm{-1}}A_{\mathrm{1}}\left ( s \right )}}$$

Step 3

Realisation of $\mathrm{\mathit{H_{\mathrm{3}}\left ( s \right )}}$ −

$$\mathrm{\mathit{H_{\mathrm{3}}\left ( s \right )\mathrm{\,=\,}\frac{Y\left ( s \right )}{X_{\mathrm{2}}\left ( s \right )}\mathrm{\,=\,}\frac{Y\left ( s \right )}{A_{\mathrm{2}}\left ( s \right )}\frac{A_{\mathrm{2}}\left ( s \right )}{X_{\mathrm{2}}\left ( s \right )}\mathrm{\,=\,}\frac{\mathrm{1\mathrm{\,+\,}}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}5}s^{-\mathrm{1}}}}}$$

$$\mathrm{\mathit{\Rightarrow \frac{Y\left ( s \right )}{A_{\mathrm{2}}\left ( s \right )}\mathrm{\,=\,}\mathrm{1\mathrm{\,+\,}}s^{\mathrm{-1}}}}$$

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

And,

$$\mathrm{\mathit{\frac{A_{\mathrm{2}}\left ( s \right )}{X_{\mathrm{2}}\left ( s \right )}\mathrm{\,=\,}\frac{\mathrm{1}}{\mathrm{1\mathrm{\,+\,}5}s^{\mathrm{-1}}}}}$$

$$\mathrm{\mathit{\Rightarrow A_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}X_{\mathrm{2}}\left ( s \right )-\mathrm{5}s^{\mathrm{-1}}A_{\mathrm{2}\left ( s \right )}}}$$

Step 4

Now, the above three structures are to be connected in cascade to get the cascaded realisation of the system transfer function H(s) as shown in the figure below.

Manish Kumar Saini

Updated on: 21-Jan-2022

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