Training Systems Engineering Feedback Systems & Block Diagrams
2 / 5

Feedback Systems & Block Diagrams

24 min Systems Engineering

Feedback Systems & Block Diagrams

Feedback is the core concept in control and systems engineering. A block diagram represents the system architecture showing how signals flow.

Closed-Loop Transfer Function

For a system with forward gain $G(s)$ and feedback gain $H(s)$:

$$T(s) = \frac{G(s)}{1 + G(s)H(s)} \quad \text{(negative feedback)}$$

Block Diagram Reduction Rules
  • Series: $G_{\text{total}} = G_1 \cdot G_2$
  • Parallel: $G_{\text{total}} = G_1 + G_2$
  • Feedback loop: $T = G/(1 + GH)$
Steady-State Error

For a unity feedback system with input $R(s)$:

$$e_{ss} = \lim_{s \to 0} \frac{sR(s)}{1 + G(s)}$$

Example 1

$G(s) = \dfrac{10}{s + 2}$, unity feedback ($H = 1$). Find the closed-loop transfer function.

$$T(s) = \frac{10/(s+2)}{1 + 10/(s+2)} = \frac{10}{s + 2 + 10} = \frac{10}{s + 12}$$

The closed-loop pole moved from $s = -2$ to $s = -12$ — feedback makes the system faster.

Example 2

Two blocks in series: $G_1(s) = \frac{5}{s+1}$, $G_2(s) = \frac{2}{s+3}$. Unity feedback. Find $T(s)$.

$G = G_1 G_2 = \frac{10}{(s+1)(s+3)}$

$$T(s) = \frac{10/(s+1)(s+3)}{1 + 10/(s+1)(s+3)} = \frac{10}{s^2 + 4s + 3 + 10} = \frac{10}{s^2 + 4s + 13}$$

Example 3

Find the steady-state error for a step input $R(s) = 1/s$ with $G(s) = \frac{10}{s+2}$.

$$e_{ss} = \lim_{s \to 0} \frac{s \cdot 1/s}{1 + 10/(s+2)} = \frac{1}{1 + 10/2} = \frac{1}{6} = 0.167$$

16.7% steady-state error. Adding an integrator ($1/s$) in $G$ eliminates this.

Practice Problems

1. $G(s) = 20/(s+5)$, $H = 1$. Find $T(s)$.
2. $G = 5/(s+1)$, $H = 2$. Find $T(s)$.
3. Find the DC gain of $T(s)$ in #1 (set $s = 0$).
4. Three blocks in series: $G_1 = 2, G_2 = 3/(s+1), G_3 = 1/(s+4)$. Find $G_{\text{total}}$.
5. For #4 with unity feedback, find $T(s)$.
6. Steady-state error for step input with $G(s) = 50/s(s+10)$.
7. What type of system (Type 0, 1, 2) is $G(s) = 50/s(s+10)$?
8. Two blocks in parallel: $G_1 = 3/(s+2)$ and $G_2 = 1/(s+5)$. Find $G_{\text{total}}$.
9. A negative feedback system has open-loop gain $K = 100$. Find the closed-loop gain with unity feedback.
10. Moving a block past a summing junction: if $G$ is before the junction, how to move it after?
11. Characteristic equation of #2: find the closed-loop poles.
12. If the open-loop system $G(s) = K/s^2$ has unity feedback, find the closed-loop $T(s)$ and comment on stability.
Show Answer Key

1. $T(s) = 20/(s+25)$

2. $T(s) = 5/(s+1+10) = 5/(s+11)$ — note $GH = 10/(s+1)$, so $T = 5/(s+11)$

3. $T(0) = 20/25 = 0.8$

4. $G = 6/((s+1)(s+4))$

5. $T = 6/(s^2+5s+4+6) = 6/(s^2+5s+10)$

6. Type 1 system (one integrator): $e_{ss} = 0$ for step input

7. Type 1 (one $s$ factor in denominator of $G$)

8. $G = \frac{3(s+5) + (s+2)}{(s+2)(s+5)} = \frac{4s+17}{s^2+7s+10}$

9. $T = 100/101 \approx 0.99$

10. Add a $1/G$ block on the other input to the summing junction

11. $s + 1 + 10 = 0$; pole at $s = -11$

12. $T = K/(s^2+K)$. Poles at $s = \pm j\sqrt{K}$ — on the imaginary axis → marginally stable (undamped oscillation).

System Reliability — Series & Parallel Decision Analysis & Trade Studies