#27 Control Theory #27 — Block Diagram Transformation (Worked Example 4)

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Read the signed diagram

   Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.

   The feedback is negative through F4. Disturbance D enters after F1, before F2 and F3.

   Seek Y=G2(G1U+D) for a well-defined scalar zero-state interconnection.

   Narration transcript

   Here is the problem. We are given a block diagram with four transfer functions F1, F2, F3 and F4. Let me describe the diagram. The input u enters a summing junction with negative feedback. From there, the signal goes through F1. After F1, there is a second summing junction where the disturbance d enters from above. After this junction, the signal goes through F2, then through F3, and the output is y. The feedback path takes y, passes it through F4, and sends it back to the first summing junction with a minus sign. Our task: transform this into the simpler form shown in Figure 2. In that form, u goes through G1, then a summing junction where d enters, then through G2 to produce y. No feedback loop in the target form. We need to find G1 and G2.

2. 2. Rules

   Series blocks multiply. Negative-feedback denominators take the form 1+GH.

   Writing every junction equation prevents branch or sign loss.

   Narration transcript

   Before we start, let us recall the block diagram rules we will use. Rule one: series connection. Two blocks in a row multiply. G1 followed by G2 gives G1 times G2. Rule two: negative feedback. A forward path G with feedback H gives G over one plus G H. And we will also use a third technique: writing the system equations directly and solving for the output. These tools are all we need.

3. 3. Combine F2 and F3

   Replace their cascade with F2F3. The disturbance junction remains before that combined block.

   Narration transcript

   Step one: combine F2 and F3 in series. Looking at the diagram, F2 and F3 are directly connected, one after the other. By the series rule, we can replace them with a single block equal to F2 times F3. Now our diagram looks simpler: u goes through a negative feedback summing junction, then through F1, then through a summing junction where d enters, then through the combined block F2 F3, and out as y. The feedback from y still goes through F4 back to the first junction.

4. 4. Correct substitution

   $\displaystyle E=U-F4Y.$

   $\displaystyle Y=F2F3\left(F1\left(U-F4Y\right)+D\right).$

   Expanding gives (1+F1F2F3F4)Y=F1F2F3U+F2F3D.

   There is no additional −U term inside the substituted expression.

   Narration transcript

   Step two: write the system equations. Let us label the signals. At the first summing junction, the error signal E equals U minus F4 times Y. After F1, the signal is F1 times E. At the second summing junction, we add the disturbance, so the signal entering F2 F3 is F1 times E plus D. The output is Y equals F2 F3 times the quantity F1 E plus D. Now substitute E equals U minus F4 Y. We get Y equals F2 F3 times the quantity F1 times U minus F1 F4 Y plus D. Expand: Y equals F1 F2 F3 U minus F1 F2 F3 F4 Y plus F2 F3 D. Move the F1 F2 F3 F4 Y term to the left side. Y times the quantity one plus F1 F2 F3 F4 equals F1 F2 F3 U plus F2 F3 D.

5. 5. Identify the blocks

   $\displaystyle G1=F1.$

   $\displaystyle G2=F2F3/\left(1+F1F2F3F4\right).$

   Factoring the right-hand side gives Y=G2(G1U+D).

   Narration transcript

   Step three: factor and identify G1 and G2. From our equation, Y times one plus F1 F2 F3 F4 equals F1 F2 F3 U plus F2 F3 D. Notice that F2 F3 is a common factor on the right side. Factor it out: Y times one plus F1 F2 F3 F4 equals F2 F3 times the quantity F1 U plus D. Divide both sides by one plus F1 F2 F3 F4. We get Y equals F2 F3 over one plus F1 F2 F3 F4, times the quantity F1 U plus D. Now compare this with the target form: Y equals G2 times the quantity G1 U plus D. By matching terms: G1 equals F1. G2 equals F2 F3 divided by one plus F1 F2 F3 F4. That is our answer.

6. 6. Verify both paths

   The U transfer is F1F2F3/(1+F1F2F3F4); the D transfer is F2F3/(1+F1F2F3F4).

   Both agree with the original junction equations; internal stability is a separate question.

   Narration transcript

   Let us recap. We started with a block diagram containing F1, F2, F3 in the forward path, F4 in the feedback path, and a disturbance d entering between F1 and F2. In three steps, we found G1 and G2. Step one: we combined F2 and F3 in series to get F2 F3. Step two: we wrote the system equations and solved for Y. Step three: we factored the result and matched it with the target form. Final answer: G1 of s equals F1. G2 of s equals F2 times F3 divided by one plus F1 F2 F3 F4. Two rules, three steps, done.

Source video: Control Theory #27 — Block Diagram Transformation (Worked Example 4) (5:54)