#26 Control Theory #26 — Block Diagram Simplification (Worked Example 3)

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

1. 1. Read both inputs

   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 first integrator has negative feedback gain K1. The disturbance enters through its own integrator before parallel gains K2 and K3/s.

   Find the form Y=G2(G1U+D), under zero initial conditions.

   Narration transcript

   Here is the problem. We are given a block diagram with three gain blocks K1, K2, and K3, plus a disturbance input d. The signal u is the input, and y is the output. Let me describe the diagram. The input u enters a summing junction with negative feedback. From there, the signal goes through an integrator, one over s. The output of this integrator feeds back through the gain K1 to the first summing junction. That is a negative feedback loop. After the integrator, the signal reaches a second summing junction. The disturbance d enters this junction through its own separate integrator, one over s. From the second summing junction, the signal splits into two parallel paths. The upper path goes through K2 directly. The lower path goes through K3, and then through another integrator one over s. The two paths recombine at a final summing junction to produce the output y. Our task: simplify this into the form u, G1, summing junction with d, G2, y. We need to find G1 of s and G2 of s explicitly.

2. 2. Interconnection rules

   Series blocks multiply; additive parallel blocks add.

   A well-defined negative-feedback loop has transfer G/(1+GH).

   Narration transcript

   Before we start simplifying, let us quickly recall the three 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: parallel connection. Two blocks sharing the same input and their outputs are added give G1 plus G2. Rule three: negative feedback loop. A forward path G with feedback H gives G over one plus G times H. These three rules are all we need. Let us apply them step by step.

3. 3. Reduce the first loop

   $\displaystyle \frac{\frac{1}{s}}{1+\frac{K1}{s}}=1/\left(s+K1\right).$

   Narration transcript

   Step one: resolve the feedback loop. Look at the left side of the diagram. The forward path is an integrator, one over s. The feedback path is the gain K1. Using the feedback rule, the closed loop transfer function is the forward path divided by one plus forward times feedback. That gives us one over s, divided by one plus one over s times K1. Simplify the denominator: one plus K1 over s equals s plus K1, all over s. So the fraction becomes one over s, divided by s plus K1 over s. The s in the numerator and denominator cancel, and we get one over s plus K1. After this step, the feedback loop is replaced by a single block: one over the quantity s plus K1.

4. 4. Combine parallel paths

   $\displaystyle K2+\frac{K3}{s }= \left(K2s+K3\right)/s.$

   Narration transcript

   Step two: combine the parallel paths. After the second summing junction, we have two paths in parallel. The upper path is K2 by itself. The lower path is K3 followed by an integrator one over s. Using the series rule, K3 times one over s gives K3 over s. Now using the parallel rule, the combined block is K2 plus K3 over s. To write this as a single fraction: K2 times s over s plus K3 over s equals K2 s plus K3, all over s. So the two parallel paths become one block: K2 s plus K3, divided by s.

5. 5. Move the disturbance integrator

   Factor 1/s out of the entire junction output, multiplying the U branch by s.

   G1=s/(s+K1) and G2=(K2s+K3)/s².

   Narration transcript

   Step three: rearrange to match the target form. Right now our diagram looks like this: u goes through one over s plus K1, then reaches a summing junction where d enters through an integrator one over s, and then the combined signal goes through K2 s plus K3 over s to produce y. But the target form requires d to enter the summing junction directly, without any integrator. So we need to move that one over s block. Here is the trick: if we remove one over s from d's path, we must compensate. We multiply the path before the junction by s, which is the inverse of one over s. And we multiply the path after the junction by one over s. Before the junction: one over s plus K1, times s, equals s over s plus K1. After the junction: K2 s plus K3 over s, times one over s, equals K2 s plus K3, over s squared. Now d enters the junction directly, and we have our target form.

6. 6. Correct both transfer checks

   $\displaystyle G1G2 =\frac{K2s+K3}{s\left(s+K1\right)}$

   The numerator s cancels one of the two denominator factors s. Retaining s² here would introduce an extra origin pole.

   The disturbance transfer remains G2=(K2s+K3)/s².

   Narration transcript

   Here is the final answer. G1 of s equals s divided by s plus K1. G2 of s equals K2 s plus K3, divided by s squared. Let us verify this makes sense. For the input u: the signal goes through G1 then G2. G1 times G2 equals s over s plus K1, times K2 s plus K3 over s squared. That gives K2 s plus K3, over s squared times s plus K1. This should match the original path from u to y through all the blocks, and it does. For the disturbance d: it enters the junction directly, then goes through G2. So the disturbance transfer function is G2 equals K2 s plus K3 over s squared. Both check out.

7. 7. Review

   The two target blocks reproduce both original input/output maps.

   External zero-state equivalence alone does not establish internal stability or equivalence for arbitrary initial states.

   Narration transcript

   Let us recap. We started with a block diagram containing three gains K1, K2, K3, three integrators, and a disturbance input d. In three steps, we simplified it. Step one: we closed the K1 feedback loop to get one over s plus K1. Step two: we combined the K2 and K3 parallel paths to get K2 s plus K3 over s. Step three: we moved the disturbance integrator by multiplying the forward path by s and the plant path by one over s. Final result: G1 of s equals s over s plus K1. G2 of s equals K2 s plus K3 over s squared. Three rules, three steps, done.

Source video: Control Theory #26 — Block Diagram Simplification (Worked Example 3) (6:48)