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- Control Systems - Discussion

Block diagram algebra is nothing but the algebra involved with the basic elements of the block diagram. This algebra deals with the pictorial representation of algebraic equations.

There are three basic types of connections between two blocks.

Series connection is also called **cascade connection**. In the following figure, two blocks having transfer functions $G_1(s)$ and $G_2(s)$ are connected in series.

For this combination, we will get the output $Y(s)$ as

$$Y(s)=G_2(s)Z(s)$$

Where, $Z(s)=G_1(s)X(s)$

$$\Rightarrow Y(s)=G_2(s)[G_1(s)X(s)]=G_1(s)G_2(s)X(s)$$

$$\Rightarrow Y(s)=\lbrace G_1(s)G_2(s)\rbrace X(s)$$

Compare this equation with the standard form of the output equation, $Y(s)=G(s)X(s)$. Where, $G(s) = G_1(s)G_2(s)$.

That means we can represent the **series connection** of two blocks with a single block. The transfer function of this single block is the **product of the transfer functions** of those two blocks. The equivalent block diagram is shown below.

Similarly, you can represent series connection of ‘n’ blocks with a single block. The transfer function of this single block is the product of the transfer functions of all those ‘n’ blocks.

The blocks which are connected in **parallel** will have the **same input**. In the following figure, two blocks having transfer functions $G_1(s)$ and $G_2(s)$ are connected in parallel. The outputs of these two blocks are connected to the summing point.

For this combination, we will get the output $Y(s)$ as

$$Y(s)=Y_1(s)+Y_2(s)$$

Where, $Y_1(s)=G_1(s)X(s)$ and $Y_2(s)=G_2(s)X(s)$

$$\Rightarrow Y(s)=G_1(s)X(s)+G_2(s)X(s)=\lbrace G_1(s)+G_2(s)\rbrace X(s)$$

Compare this equation with the standard form of the output equation, $Y(s)=G(s)X(s)$.

Where, $G(s)=G_1(s)+G_2(s)$.

That means we can represent the **parallel connection** of two blocks with a single block. The transfer function of this single block is the **sum of the transfer functions** of those two blocks. The equivalent block diagram is shown below.

Similarly, you can represent parallel connection of ‘n’ blocks with a single block. The transfer function of this single block is the algebraic sum of the transfer functions of all those ‘n’ blocks.

As we discussed in previous chapters, there are two types of **feedback** — positive feedback and negative feedback. The following figure shows negative feedback control system. Here, two blocks having transfer functions $G(s)$ and $H(s)$ form a closed loop.

The output of the summing point is -

$$E(s)=X(s)-H(s)Y(s)$$

The output $Y(s)$ is -

$$Y(s)=E(s)G(s)$$

Substitute $E(s)$ value in the above equation.

$$Y(s)=\left \{ X(s)-H(s)Y(s)\rbrace G(s) \right\}$$

$$Y(s)\left \{ 1+G(s)H(s)\rbrace = X(s)G(s) \right\}$$

$$\Rightarrow \frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}$$

Therefore, the negative feedback closed loop transfer function is $\frac{G(s)}{1+G(s)H(s)}$

This means we can represent the negative feedback connection of two blocks with a single block. The transfer function of this single block is the closed loop transfer function of the negative feedback. The equivalent block diagram is shown below.

Similarly, you can represent the positive feedback connection of two blocks with a single block. The transfer function of this single block is the closed loop transfer function of the positive feedback, i.e., $\frac{G(s)}{1-G(s)H(s)}$

There are two possibilities of shifting summing points with respect to blocks −

- Shifting summing point after the block
- Shifting summing point before the block

Let us now see what kind of arrangements need to be done in the above two cases one by one.

Consider the block diagram shown in the following figure. Here, the summing point is present before the block.

Summing point has two inputs $R(s)$ and $X(s)$. The output of it is $\left \{R(s)+X(s)\right\}$.

So, the input to the block $G(s)$ is $\left \{R(s)+X(s)\right \}$ and the output of it is –

$$Y(s)=G(s)\left \{R(s)+X(s)\right \}$$

$\Rightarrow Y(s)=G(s)R(s)+G(s)X(s)$ **(Equation 1)**

Now, shift the summing point after the block. This block diagram is shown in the following figure.

Output of the block $G(s)$ is $G(s)R(s)$.

The output of the summing point is

$Y(s)=G(s)R(s)+X(s)$ **(Equation 2)**

Compare Equation 1 and Equation 2.

The first term $‘G(s) R(s)’$ is same in both the equations. But, there is difference in the second term. In order to get the second term also same, we require one more block $G(s)$. It is having the input $X(s)$ and the output of this block is given as input to summing point instead of $X(s)$. This block diagram is shown in the following figure.

Consider the block diagram shown in the following figure. Here, the summing point is present after the block.

Output of this block diagram is -

$Y(s)=G(s)R(s)+X(s)$ **(Equation 3)**

Now, shift the summing point before the block. This block diagram is shown in the following figure.

Output of this block diagram is -

$Y(S)=G(s)R(s)+G(s)X(s)$ **(Equation 4)**

Compare Equation 3 and Equation 4,

The first term $‘G(s) R(s)’$ is same in both equations. But, there is difference in the second term. In order to get the second term also same, we require one more block $\frac{1}{G(s)}$. It is having the input $X(s)$ and the output of this block is given as input to summing point instead of $X(s)$. This block diagram is shown in the following figure.

There are two possibilities of shifting the take-off points with respect to blocks −

- Shifting take-off point after the block
- Shifting take-off point before the block

Let us now see what kind of arrangements are to be done in the above two cases, one by one.

Consider the block diagram shown in the following figure. In this case, the take-off point is present before the block.

Here, $X(s)=R(s)$ and $Y(s)=G(s)R(s)$

When you shift the take-off point after the block, the output $Y(s)$ will be same. But, there is difference in $X(s)$ value. So, in order to get the same $X(s)$ value, we require one more block $\frac{1}{G(s)}$. It is having the input $Y(s)$ and the output is $X(s)$. This block diagram is shown in the following figure.

Consider the block diagram shown in the following figure. Here, the take-off point is present after the block.

Here, $X(s)=Y(s)=G(s)R(s)$

When you shift the take-off point before the block, the output $Y(s)$ will be same. But, there is difference in $X(s)$ value. So, in order to get same $X(s)$ value, we require one more block $G(s)$. It is having the input $R(s)$ and the output is $X(s)$. This block diagram is shown in the following figure.

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