# Transfer function algebra

Depending on the complexity of the control system, the block diagram can be quite complicated and difficult to work with. A block diagram with several transfer function can be simplified using easily derivable transformations. After several transformation we’ll end up with an equivalent transfer function, input and output.

Transfer functions can be arranged in series or parallel. The series arrangement of transfer functions is also called cascade, because the blocks are coming one after the other. The arrangement in parallel can be of two types, depending on the loop: with forward loop or feedback loop.

If you want to recall what a transfer function is and how to calculated it, read the article How to find the transfer function of a system.

### Transfer functions in series (cascade)

Two or more transfer functions are arranged in series (cascade) when the output of the first transfer function is the input of the following transfer function.

In the example above we have the H1(s) transfer function which has the input u1(s) and the output y1(s). The second transfer function H2(s) has the input u2(s) and the output y2(s). Notice that the input u2(s) is equal with the output y1(s).

Generally speaking, any finite number of transfer functions blocks connected in series (cascade) can be algebraically combined by multiplication of the transfer functions. For several transfer functions H1(s), H2(s), H3(s), … Hn(s), the resulting transfer functions H(s) is:

$H(s) = H_{1}(s) \cdot H_{2}(s) \cdot H_{3}(s) \cdot \text{ … } H_{n}(s)=\prod_{i=1}^{n}H_{i}(s) \tag{1}$

For our particular case, with two transfer functions, the resulting transfer function H(s) will be:

$H(s) = H_{1}(s) \cdot H_{2}(s) \tag{2}$

We can easily demonstrate this result using the definition of a transfer function and the relationship between inputs and outputs.

Both transfer functions are defined as:

$H_{1}(s) = \frac{y_{1}(s)}{u_{1}(s)} \tag{3}$ $H_{2}(s) = \frac{y_{2}(s)}{u_{2}(s)} \tag{4}$

We also know that the input u2(s) is equal with the output y1(s):

$u_{2}(s) = y_{1}(s) \tag{5}$

Also, the overall transfer function of the system H(s), can be written as:

$H(s) = \frac{y_{2}(s)}{u_{1}(s)} \tag{6}$

since the input in the connected transfer functions is u1(s) and the final output is y2(s).

From (3) and (4) if we extract y2(s) and u1(s) and replace in (6), we get:

$H(s) = \frac{y_{2}(s)}{u_{1}(s)} = \frac{H_{2}(s) \cdot u_{2}(s)}{\frac{y_{1}(s)}{H_{1}(s)}}=H_{1}(s) \cdot H_{2}(s) \cdot \frac{u_{2}(s)}{y_{1}(s)} \tag{7}$

Replacing (5) in (7) gives (2).

$\bbox[#FFFF9D]{H(s) = H_{1}(s) \cdot H_{2}(s)}$

Example. Transfer functions connected in series

To prove the equivalent transfer function calculation is correct we are going to use two transfer functions and an Xcos block diagram simulation. First we are going to calculate based on (2) the equivalent transfer function expression. Second, in Xcos, we are going to simulate the series connection and the equivalent transfer function and plot their outputs on the same scope.

Let’s define H1(s) and H2(s) as:

$\begin{split} H_{1}(s) = \frac{1}{s+1}\\ H_{2}(s) = \frac{1}{s+2} \end{split}$

Applying (2) gives the equivalent transfer function:

$H(s) = \frac{1}{s^2 + 3 \cdot s + 2}$

The Xcos block diagram model for the series transfer functions and equivalent one is:

The Step input block outputs 1 at 1 s. The simulation is run for 10 s and the Clock block samples the data at 0.01 s. Running the simulation will output the following graphical window:

As expected the two outputs are identical which proves that the equivalent transfer function H(s) is the correct representation of the series transfer functions connection.

### Transfer function in parallel (forward loop)

In a parallel connection of transfer functions the same input is fed to all the transfer functions and all the outputs are summed together. The equivalent transfer function is this case is the algebraic sum of all the transfer functions.

In a parallel connection block diagram all the inputs u1(s), u2(s), u3(s), … un(s) are equal to each other and the final output y(s) is the algebraic sum of all the outputs y1(s), y2(s), y3(s), … yn(s). In this case we can write the general expression of the equivalent transfer function for a forward loop parallel connection as:

$H(s) = H_{1}(s) + H_{2}(s) + H_{3}(s) + \text{ … } H_{n}(s)=\sum_{i=1}^{n}H_{i}(s) \tag{8}$

For out particular case, with two transfer functions connected in parallel (forward loop), the equivalent transfer function is:

$H(s) = H_{1}(s) \pm H_{2}(s) \tag{9}$

Notice that, depending on the sign of the outputs, the transfer functions can be added or subtracted.

Again, we can easily demonstrate this result using the definition of a transfer function and the relationship between inputs and outputs. Both transfer functions are defined as:

$H_{1}(s) = \frac{y_{1}(s)}{u_{1}(s)} \tag{10}$ $H_{2}(s) = \frac{y_{2}(s)}{u_{2}(s)} \tag{11}$

We also know that the input u2(s) is equal with the input u1(s) and equal with the equivalent input u(s):

$u(s) = u_{1}(s) = u_{2}(s) \tag{12}$

The equivalent output y(s) is equal with the sum between y1(s) and y2(s):

$y(s) = y_{1}(s) + y_{2}(s) \tag{13}$

Also, the overall transfer function of the system H(s), can be written as:

$H(s) = \frac{y(s)}{u(s)} \tag{14}$

From (10) and (11) we extract y1(s) and y2(s) and we replace them in (13), which gives:

$y(s) = H_{1}(s) \cdot u_{1}(s) + H_{2}(s) \cdot u_{2}(s) = u(s) \cdot \left ( H_{1}(s) + H_{2}(s) \right ) \tag{15}$

Replacing (15) in (14) gives the equivalent transfer function for a parallel connection (forward loop):

$\bbox[#FFFF9D]{H(s) = H_{1}(s) + H_{2}(s)}$

Example. Transfer functions connected in parallel (forward loop)

Let’s define H1(s) and H2(s) as:

$\begin{split} H_{1}(s) = \frac{1}{s+1}\\ H_{2}(s) = \frac{1}{s+2} \end{split}$

Applying (9) gives the equivalent transfer function:

$H(s) = \frac{2 \cdot s + 3}{s^2 + 3 \cdot s + 2}$

The Xcos block diagram model for the parallel (forward loop) transfer functions and equivalent one is:

The Step input block outputs 1 at 1 s. The simulation is run for 10 s and the Clock block samples the data at 0.01 s. Running the simulation will output the following graphical window:

As expected the two outputs are identical which proves that the equivalent transfer function H(s) is the correct representation of the parallel (forward loop) transfer functions connection.

### Transfer function in parallel (feedback loop)

Another common arrangement of transfer functions is in parallel with feedback loop. This means that the output of first transfer function y1(s) is fed back as input to the second transfer function. Further, the output of the second transfer function y2(s) is added/subtracted from the input u(s).

This configuration is also called the canonical form of a feedback control system. For two given transfer functions H1(s) and H2(s) connected in parallel with a feedback loop, the equivalent transfer function H(s) is:

$H(s) = \frac{H_{1}(s)}{1 \mp H_{1}(s) \cdot H_{2}(s)} \tag{16}$

We can demonstrate this result using the definition of a transfer function and the relationship between inputs and outputs. Both transfer functions are defined as:

$H_{1}(s) = \frac{y_{1}(s)}{u_{1}(s)} \tag{17}$ $H_{2}(s) = \frac{y_{2}(s)}{u_{2}(s)} \tag{18}$

We know that the input u2(s) is equal with the output y1(s) and equal with the equivalent output y(s):

$y(s) = y_{1}(s) = u_{2}(s) \tag{19}$

The input u1(s) is equal with the sum between the output y2(s) and the equivalent input u(s):

$u_{1}(s) = u(s) + y_{2}(s) \tag{20}$

Also, the overall transfer function of the system H(s), can be written as:

$H(s) = \frac{y(s)}{u(s)} \tag{21}$

Using (19) and (20), we can rewrite H(s) as:

$H(s) = \frac{y_{1}(s)}{u_{1}(s)-y_{2}(s)} \tag{22}$

From (17) and (18) we extract y1(s) and y2(s) which are replaced in (22) giving:

$H(s) = \frac{u_{1}(s) \cdot H_{1}(s)}{u_{1}(s)-u_{2}(s) \cdot H_{2}(s)}= \frac{H_{1}(s)}{ 1 – \frac{u_{2}(s)}{u_{1}(s)} \cdot H_{2}(s) } \tag{23}$

From (19) and (23) results the expression of the equivalent transfer function for a parallel connection with feedback loop:

$\bbox[#FFFF9D]{H(s) = \frac{H_{1}(s)}{1 \mp H_{1}(s) \cdot H_{2}(s)}}$

Example. Transfer functions connected in parallel (feedback loop)

Let’s define H1(s) and H2(s) as:

$\begin{split} H_{1}(s) = \frac{1}{s+1}\\ H_{2}(s) = \frac{1}{s+2} \end{split}$

Applying (16) gives the equivalent transfer function:

$H(s) = \frac{s^2 + 3 \cdot s + 2}{s^3 + 4 \cdot s^2 + 4 \cdot s + 1}$

The Xcos block diagram model for the parallel (feedback loop) transfer functions and equivalent one is:

The Step input block outputs 1 at 1 s. The simulation is run for 10 s and the Clock block samples the data at 0.01 s. Running the simulation will output the following graphical window:

As expected the two outputs are identical which proves that the equivalent transfer function H(s) is the correct representation of the parallel (feedback loop) transfer functions connection.

### Unity feedback system

A particular case of the parallel connection with feedback loop is when the second transfer function H2(s) is equal to 1. In this case we have a unity feedback system and the block diagram is:

The equivalent transfer function for the unity feedback system is:

$\bbox[#FFFF9D]{H(s) = \frac{H_{1}(s)}{1 \mp H_{1}(s)}}$

Using these transformations, a complicated block diagram of transfer functions can be reduced to a single equivalent transfer function. The advantage of having a single transfer function is that several analysis methods can be applied in order to understand the controllability and observability of the system.

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