Frontmatter

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287 ms

In-class Activity
md"""
# In-class Activity
"""
4.9 ms

Required Reading Material

Sometimes to simplify our system or block diagrams; we have to move things around while preserving semantic equivalency so that we can apply the simplications we know to be true. What this means will become clear as we go on forward. This material is based on: https://electronicscoach.com/block-diagram-reduction-rules.html.

Shifting of take-off point ahead of the block

A take-off or pick-off point is a place in the block diagram from which we branch out a variable to reuse it. Suppose we have a combination of take-off point and block as shown below:
md"""
## Required Reading Material

Sometimes to simplify our system or block diagrams; we have to move things around while preserving semantic equivalency so that we can apply the simplications we know to be true. What this means will become clear as we go on forward. This material is based on: [https://electronicscoach.com/block-diagram-reduction-rules.html](https://electronicscoach.com/block-diagram-reduction-rules.html).

#### Shifting of take-off point ahead of the block

A _take-off_ or _pick-off_ point is a place in the block diagram from which we branch out a variable to reuse it. Suppose we have a combination of take-off point and block as shown below:
"""
389 μs
LocalResource("rule1.jpeg", :width=>350)
146 μs

If we need to shift the take-off point ahead of the block, then we must keep '$p$' as it is

Here $p = X(s)$

So even after shifting p must be $X(s)$ and for this, we have to add a block with which is reciprocal of the originally present block.

As the actual block is $G(s)$ so the new/additional block will have a $1/G(s)$.
md"""
If we need to shift the take-off point ahead of the block, then we must keep '$p$' as it is

Here $p = X(s)$

So even after shifting p must be $X(s)$ and for this, we have to add a block with which is reciprocal of the originally present block.

As the actual block is $G(s)$ so the new/additional block will have a $1/G(s)$.
"""
416 μs
LocalResource("rule2.jpeg", :width=>400)
131 μs

Shifting of take-off point behind the block

Suppose there is a take-off point ahead of block as given below:
md"""
#### Shifting of take-off point behind the block

Suppose there is a take-off point ahead of block as given below:
"""
323 μs
LocalResource("rule3.jpeg", :width=>380)
97.2 μs

In order to move the take-off point behind the block, we need to keep the value of $p$ same. Here $p = X(s)G(s)$.

But with backward movement $p$ will become $X(s)$. So, we have to add another block with the same value to balance things out. This will make the value of $p = X(s)G(s)$
md"""
In order to move the take-off point behind the block, we need to keep the value of $p$ same. Here $p = X(s)G(s)$.

But with backward movement $p$ will become $X(s)$. So, we have to add another block with the same value to balance things out. This will make the value of $p = X(s)G(s)$

"""
239 μs
LocalResource("rule4.jpeg", :width=>400)
123 μs

Shifting of summing point ahead of the block

Suppose we have a configuration of summing point and block as given below:
md"""
#### Shifting of summing point ahead of the block

Suppose we have a configuration of summing point and block as given below:
"""
463 μs
LocalResource("rule5.jpeg", :width=>400)
103 μs

If the summing point is to be moved from backward to forward of the block, then $Y(s)$ will become $X(s)G(s)+p$. However earlier $Y(s)$ was $[X(s)+p]G(s)$.

So, to have unaltered output, we need to add a block having exact value as the original one.
md"""
If the summing point is to be moved from backward to forward of the block, then $Y(s)$ will become $X(s)G(s)+p$. However earlier $Y(s)$ was $[X(s)+p]G(s)$.

So, to have unaltered output, we need to add a block having exact value as the original one.
"""
238 μs
LocalResource("rule6.jpeg", :width=>450)
140 μs

Shifting of summing point behind the block

Suppose we have a combination where we have a summing point present after the block as shown below:
md"""
#### Shifting of summing point behind the block

Suppose we have a combination where we have a summing point present after the block as shown below:

"""
243 μs
LocalResource("rule7.jpeg",:width=>350)
110 μs

We need to move this summing point behind the position of the block without changing the response. So, for this, a block with value which is reciprocal of the original block is to be inserted in the configuration in series.
md"""
We need to move this summing point behind the position of the block without changing the response. So, for this, a block with value which is reciprocal of the original block is to be inserted in the configuration in series.
"""
333 μs
LocalResource("rule8.jpeg", :width=>400)
203 μs

Problems

For the two diagrams shown below. Use properties we have discussed in class and/or the above rules to simplify the diagrams.

Problem 1
md"""
## Problems

For the two diagrams shown below. Use properties we have discussed in class and/or the above rules to simplify the diagrams.

#### Problem 1
"""
225 μs

Simplify the block diagram into one equivalent transfer function.
md"""
Simplify the block diagram into **one** equivalent transfer function.

"""
208 μs
LocalResource("problem1.png")
58.5 ms

Problem 2
md"""
#### Problem 2
"""
145 μs

Simply block diagram into one equivalent transfer function.
md"""
Simply block diagram into **one** equivalent transfer function.

"""
195 μs
LocalResource("problem2.png")
261 μs