Parallel 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.

Parallel Form Realisation of Continuous-Time Systems

In the parallel form realisation of continuous-time systems, the transfer function of the system is expressed into its partial fractions and each factor is then realised using integrators and adders. Finally, all the realised structures are connected in parallel, i.e., the input signal is applied to each one of those structures and all the output signals are added together.

Following example explains the realisation of continuous-time systems in parallel form.

Numerical Example

Realise the continuous-time system described by the following transfer function in parallel form.

Solution

The given transfer function of the system is

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

The partial fraction of H(s) are,

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

Now, the coefficients A, B and C are determined as follows −

$$\mathrm{\mathit{A\mathrm{\,=\,}\left [ \left ( s\mathrm{\,+\,}\mathrm{2} \right )H\left ( s \right ) \right ]_{s\mathrm{\,=\,}-\mathrm{2}}\mathrm{\,=\,}\left [ \frac{s\left ( s\mathrm{\,+\,}\mathrm{1} \right )}{\left ( s\mathrm{\,+\,}\mathrm{3} \right )\left ( s\mathrm{\,+\,}\mathrm{4} \right )} \right ]_{s\mathrm{\,=\,}-\mathrm{2}}}}$$

$$\mathrm{\mathit{\therefore A\mathrm{\,=\,}\frac{\left ( -\mathrm{2} \right )\left ( -\mathrm{2\mathrm{\,+\,}1} \right )}{\mathrm{\left ( -2\mathrm{\,+\,}3 \right )\left ( -2\mathrm{\,+\,}4 \right )}}\mathrm{\,=\,}\mathrm{\frac{2}{2}\mathrm{\,=\,}1}}}$$

$$\mathrm{\mathit{B\mathrm{\,=\,}\left [ \left ( s\mathrm{\,+\,}\mathrm{3} \right )H\left ( s \right ) \right ]_{s\mathrm{\,=\,}-\mathrm{3}}\mathrm{\,=\,}\left [ \frac{s\left ( s\mathrm{\,+\,}\mathrm{1} \right )}{\left ( s\mathrm{\,+\,}\mathrm{2} \right )\left ( s\mathrm{\,+\,}\mathrm{4} \right )} \right ]_{s\mathrm{\,=\,}-\mathrm{3}}}}$$

$$\mathrm{\mathit{\therefore B\mathrm{\,=\,}\frac{\left ( -\mathrm{3} \right )\left ( -\mathrm{3\mathrm{\,+\,}1} \right )}{\mathrm{\left ( -3\mathrm{\,+\,}2 \right )\left ( -3\mathrm{\,+\,}4 \right )}}\mathrm{\,=\,}\mathrm{-6}}}$$

$$\mathrm{\mathit{C\mathrm{\,=\,}\left [ \left ( s\mathrm{\,+\,}\mathrm{4} \right )H\left ( s \right ) \right ]_{s\mathrm{\,=\,}-\mathrm{4}}\mathrm{\,=\,}\left [ \frac{s\left ( s\mathrm{\,+\,}\mathrm{1} \right )}{\left ( s\mathrm{\,+\,}\mathrm{2} \right )\left ( s\mathrm{\,+\,}\mathrm{3} \right )} \right ]_{s\mathrm{\,=\,}-\mathrm{4}}}}$$

$$\mathrm{\mathit{\therefore C\mathrm{\,=\,}\frac{\left ( -\mathrm{4} \right )\left ( -\mathrm{4\mathrm{\,+\,}1} \right )}{\mathrm{\left ( -4\mathrm{\,+\,}2 \right )\left ( -4\mathrm{\,+\,}3 \right )}}\mathrm{\,=\,}\mathrm{6}}}$$

Therefore, the transfer function is,

$$\mathrm{\mathit{H\left ( s \right )\mathrm{\,=\,}\frac{\mathrm{1}}{\left ( s\mathrm{\,+\,}\mathrm{2} \right )}-\frac{\mathrm{6}}{\left ( s\mathrm{\,+\,}\mathrm{3} \right )}\mathrm{\,+\,}\frac{\mathrm{6}}{\left ( s\mathrm{\,+\,}\mathrm{4} \right )} }}$$

Let,

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

$$\mathrm{\mathit{H_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}-\frac{\mathrm{6}}{\left ( s\mathrm{\,+\,}\mathrm{3} \right )}\mathrm{\,=\,}\frac{-\mathrm{6}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}3}s^{-\mathrm{1}}}}}$$

$$\mathrm{\mathit{H_{\mathrm{3}}\left ( s \right )\mathrm{\,=\,}\frac{\mathrm{6}}{\left ( s\mathrm{\,+\,}\mathrm{4} \right )}\mathrm{\,=\,}\frac{\mathrm{6}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}4}s^{-\mathrm{1}}}}}$$

These transfer functions can be realised as follows −

Step 1

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

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

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

$$\mathrm{\mathit{\therefore Y_{\mathrm{1}}\left ( s \right )\mathrm{\,=\,}s^{-\mathrm{1}}A_{\mathrm{1}}\left ( s \right )}}$$

And,

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

$$\mathrm{\mathit{\therefore A_{\mathrm{1}}\left ( s \right )\frac{}{}\mathrm{\,=\,}X\left ( s \right )-\mathrm{2}s^{-\mathrm{1}}A_{\mathrm{1}}\left ( s \right )}}$$

Step 2

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

$$\mathrm{\mathit{H_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}\frac{Y_{\mathrm{2}}\left ( s \right )}{X\left ( s \right )}\mathrm{\,=\,}\frac{Y_{\mathrm{2}}\left ( s \right )}{A_{\mathrm{2}}\left ( s \right )}\frac{A_{\mathrm{2}}\left ( s \right )}{X\left ( s \right )}\mathrm{\,=\,}\frac{\mathrm{-6}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}3}s^{-\mathrm{1}}}}}$$

$$\mathrm{\mathit{\Rightarrow \frac{Y_{\mathrm{2}}\left ( s \right )}{A_{\mathrm{2}}\left ( s \right )}\mathrm{\,=\,}\mathrm{-6}s^{-\mathrm{1}}}}$$

$$\mathrm{\mathit{\therefore Y_{\mathrm{2}}\left ( s \right )\mathrm{\,=\,}\mathrm{-6}s^{-\mathrm{1}}A_{\mathrm{2}}\left ( s \right )}}$$

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

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

Step 3

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

$$\mathrm{\mathit{H_{\mathrm{3}}\left ( s \right )\mathrm{\,=\,}\frac{Y_{\mathrm{3}}\left ( s \right )}{X\left ( s \right )}\mathrm{\,=\,}\frac{Y_{\mathrm{3}}\left ( s \right )}{A_{\mathrm{3}}\left ( s \right )}\frac{A_{\mathrm{3}}\left ( s \right )}{X\left ( s \right )}\mathrm{\,=\,}\frac{\mathrm{6}s^{\mathrm{-1}}}{\mathrm{1\mathrm{\,+\,}4}s^{-\mathrm{1}}}}}$$

$$\mathrm{\mathit{\Rightarrow \frac{Y_{\mathrm{3}}\left ( s \right )}{A_{\mathrm{3}}\left ( s \right )}\mathrm{\,=\,}\mathrm{6}s^{-\mathrm{1}}}}$$

$$\mathrm{\mathit{\therefore Y_{\mathrm{3}}\left ( s \right )\mathrm{\,=\,}\mathrm{6}s^{-\mathrm{1}}A_{\mathrm{3}}\left ( s \right )}}$$

And,

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

$$\mathrm{\mathit{A_{\mathrm{3}}\left ( s \right )\frac{}{}\mathrm{\,=\,}X\left ( s \right )-\mathrm{4}s^{-\mathrm{1}}A_{\mathrm{3}}\left ( s \right )}}$$