Block diagrams: practice problems with step-by-step solutions
Free Block diagrams practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: series, parallel and feedback, closed-loop transfer functions and errors, disturbances, stability and poles.
Free practice problemEasy
The signal R splits into two parallel blocks G₂ = 11 and G₃ = 8. Their outputs meet at a summing junction with the signs shown (+ and +), and the sum passes through G₁ = 9. Find the overall gain T = C/R.
A control system is a chain of parts: a controller, a motor, a sensor. A block diagram draws each part as a box and each signal as an arrow.
A block multiplies its input signal by its gain or transfer function G. If 3 goes into a block G = 5, then 15 comes out.
A summing junction (a circle) adds or subtracts signals. The sign next to each incoming arrow tells you which. A dot is a pickoff point: the signal is copied, not split.
This is the most important picture in control. The output C is measured through H and subtracted from the reference R. The difference E = R − H·C is the error that drives G.
This is negative feedback: when the output is too high, the error shrinks and pulls it back down.
Self-check At a summing junction R enters with + and the feedback signal B enters with −. R = 10 and B = 4. What leaves the junction?
14
40
6
−6
Answer: 6. The signs at the junction say E = R − B = 10 − 4 = 6. 14 ignores the minus sign, and −6 subtracts the wrong way round.
2Three rules to simplify any diagram
You reduce a diagram step by step until one block remains. Its gain T = C/R is the closed-loop transfer function.
Series: blocks one after the other multiply, G₁·G₂.
Parallel: blocks that feed the same summing junction add, with the signs at the junction: G₁ + G₂ or G₁ − G₂.
Feedback loop: a forward block G with negative feedback H becomes G/(1 + G·H).
The feedback rule follows from two equations. At the junction E = R − H·C, and the block gives C = G·E. So C = G·R − G·H·C, and therefore:
T = C/R = G/(1 + G·H)
With positive feedback the signal is added at the junction, and the sign in the denominator flips: T = G/(1 − G·H).
The product G·H is the loop gain. When it is large, T comes close to 1/H: the sensor, not the plant, sets the gain.
Self-check G₁ = 2 and G₂ = 5 are in series, with negative feedback H = 0.3. What is T = C/R?
0.75
10
2.5
−5
Answer: 2.5. G = 2·5 = 10 and G·H = 3, so T = 10/(1 + 3) = 2.5. The value 10 ignores the loop, −5 uses 1 − G·H, and 0.75 puts H in the numerator.
3Loops with transfer functions
Real blocks depend on s, for example a motor G(s) = b/(s + a). The same rules apply. Afterwards you clear the fractions inside fractions.
With unity feedback (H = 1), multiply the top and bottom of G/(1 + G) by (s + a):
T(s) = b/(s + a + b)
The closed-loop pole is the root of the denominator: p = −(a + b). It lies further left than the open-loop pole −a, so the loop is faster.
The time constant is τ = 1/(a + b).
The DC gain is T(0) = b/(a + b). A step of size A settles at css = A·T(0).
With a sensor gain H the denominator becomes s + a + b·H.
The steady-state error ess is what remains of R − C after settling. With unity feedback, use the error constants:
For a step of size A: ess = A/(1 + Kp), with the position constant Kp = G(0).
For a ramp r(t) = A·t: ess = A/Kv, with the velocity constant Kv = lims→0 s·G(s).
For G(s) = 8/(s(s + 4)) the factor s cancels: Kv = 8/4 = 2. A ramp r(t) = 1·t then leaves ess = 1/2 = 0.5.
For a second-order loop such as G(s) = K/(s(s + 6)), the closed loop has s² + 6s + K = 0. Two equal real poles (critical damping) need 6² = 4K, so K = 9.
Worked example
G(s) = 6/(s + 2) is placed in a unity negative feedback loop. Find the closed-loop pole, the time constant and the steady-state output for a step of size 4.
1
Close the loop
T(s) = G/(1 + G) = 6/(s + 2 + 6) = 6/(s + 8)
2
Pole and time constant
The pole is the root of s + 8 = 0.
p = −8, τ = 1/8 = 0.125 s
3
Steady-state output
Set s = 0 for the DC gain, then multiply by the step size.
T(0) = 6/8 = 0.75, css = 4·0.75 = 3
The error is 4 − 3 = 1. The constant Kp = G(0) = 3 gives the same: 4/(1 + 3) = 1.
Answer
p = −8, τ = 0.125 s and css = 3.
Self-check G(s) = 6/(s + 2) is in a unity negative feedback loop. Which number is the closed-loop pole?
8
−0.125
−2
−8
Answer: −8. T(s) = 6/(s + 8), so the pole is −8. −2 is the open-loop pole before feedback. A pole is a value of s, so it is negative here, and −0.125 mixes it up with τ = 1/8.
4The method step by step
Find the innermost loop. Reduce it first with G/(1 + G·H), or G/(1 − G·H) for positive feedback.
Combine blocks in series (multiply) and in parallel (add with the signs at the junction).
Repeat until one block remains. That block is T(s) = C/R.
Clear the fractions: multiply the top and bottom by the denominators of G(s).
Read what is asked: the poles from the denominator, T(0) for the steady state, τ = 1/(pole size) for a first-order loop.
Quick check: at s = 0 you can reduce the diagram with plain numbers. That gives T(0) directly and catches most algebra slips.
5A worked example
This is a typical medium question with a minor loop. Try each step before you open it.
Worked example
The plant G(s) = 3/(s + 1) has an inner negative feedback loop with H₁ = 1. A controller K = 4 and an outer unity negative feedback loop close the system. Find the closed-loop pole p and the DC gain T(0).
1
Look at the diagram
There are two loops. The inner loop goes around the plant only. The outer loop goes around everything, from C back to R.
2
Reduce the inner loop
Use G/(1 + G·H₁) and multiply the top and bottom by (s + 1):
Gᵢ = 3/(s + 1 + 3·1) = 3/(s + 4)
3
Series with the controller
K and Gᵢ are now in series, so multiply:
K·Gᵢ = 4·3/(s + 4) = 12/(s + 4)
4
Close the outer loop
Unity feedback around 12/(s + 4):
T(s) = 12/(s + 4 + 12) = 12/(s + 16)
The pole is p = −16, and T(0) = 12/16 = 0.75.
5
Check the answer
Redo it at s = 0 with numbers only. G(0) = 3, the inner loop gives 3/(1 + 3) = 0.75, and K makes it 3.
The outer loop gives 3/(1 + 3) = 0.75. This matches T(0).
Answer
T(s) = 12/(s + 16), so p = −16 and T(0) = 0.75.
6Common mistakes
Using 1 − G·H for negative feedback. Negative feedback gives 1 + G·H, positive feedback gives 1 − G·H.
Putting H in the numerator. T = G/(1 + G·H): H appears only in the loop gain.
Closing the outer loop before the inner loop. Always work from the inside out.
Forgetting the sensor gain H in the pole: the denominator is s + a + b·H, not s + a + b.
Giving the open-loop pole −a or the open-loop gain G(0). Feedback changes both.
Using A/Kp for a step error. The correct formula is A/(1 + Kp).
Self-check An amplifier G = 10 has positive feedback H = 0.05: the feedback signal is added at the junction. What is T = C/R?
0.5
10
20
6.67
Answer: 20. Positive feedback gives T = G/(1 − G·H) = 10/(1 − 0.5) = 20. The value 6.67 uses the negative-feedback formula 10/1.5. 10 ignores the loop, and 0.5 is only the loop gain G·H.
7How exam questions look
Easy: series, parallel and feedback. You reduce gains in a loop, parallel blocks with signs, or positive feedback. Or you close a unity loop around b/(s + a) and give the pole, τ, T(0) or css.
Medium: closed-loop transfer functions and errors. A minor loop as in the example, a sensor gain H, the gain K for critical damping, or the steady-state error for a step or a ramp.
Hard: disturbances, stability and poles. A disturbance D added at the plant input, or the largest gain Kmax for stability with the Routh–Hurwitz test.
For a disturbance, use superposition: find C from R with D = 0, then C from D with R = 0, and add the two.
The signal R splits into two parallel blocks G₂ = 7 and G₃ = 12. Their outputs meet at a summing junction with the signs shown (+ and −), and the sum passes through G₁ = 10. Find the overall gain T = C/R.
G₁ = 10G₂ = 7G₃ = 12signs: + G₂, − G₃
Find T
Show hints
Parallel blocks feeding a summing junction add with the signs at the junction.
The combination is in series with G₁, so multiply.
G₂ − G₃ = -5.
Show the answer and the step-by-step solution
Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).
Parallel blocks
G₂ − G₃ = 7 − 12 = −5
Series with G₁
T = G₁·(G₂ − G₃) = 10·(−5) = −50
Answer
T = −50
Example 2Medium
A unity negative feedback loop has G(s) = 1(s + 3)/((s + 2)(s + 4)). A step of size 2 is applied at the reference. Find the steady-state error ess.
G(s) = 1(s + 3)/((s + 2)(s + 4))R = 2
Find ess
Show hints
Position error constant Kₚ = G(0).
Steady-state error for a step of size A: A/(1 + Kₚ).
Kₚ = 1·3/(2·4) = 0.375.
Show the answer and the step-by-step solution
Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).
Position error constant
Kₚ = G(0) = 1·3/(2·4) = 0.375
Steady-state error
ess = A/(1 + Kₚ) = 2/(1 + 0.375) = 1.455
Answer
ess = 1.455
Example 3Hard
Reduce the block diagram with G₁ = 5/(s + 2), G₂ = 10/(s + 4), an inner negative feedback H₁ = 4 around G₂ and an outer negative feedback H₂ = 1. Find the DC gain T(0) of C(s)/R(s).
G₁ = 5/(s + 2)G₂ = 10/(s + 4)H₁ = 4H₂ = 1
Find T(0)
Show hints
Inner loop: G₂/(1 + G₂·H₁). Then T = G₁·Gᵢ/(1 + G₁·Gᵢ·H₂).
Altogether T = G₁G₂/(1 + G₂H₁ + G₁G₂H₂). For the DC gain set s = 0 in every block.
G₁(0) = 2.5, G₂(0) = 2.5.
Show the answer and the step-by-step solution
Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).
A control system is a chain of parts: a controller, a motor, a sensor. A block diagram draws each part as a box and each signal as an arrow.
A block multiplies its input signal by its gain or transfer function G. If 3 goes into a block G = 5, then 15 comes out.
How do I solve Block diagrams problems step by step?
Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).
What are common mistakes in Block diagrams?
Check the sign at the summing junction: the G₃ path enters with −.
Only parallel blocks add; G₁ is in series with them and multiplies.
Parallel blocks add (with their signs); they do not multiply.
What do the Block diagrams exercises cover?
Easy: series, parallel and feedback
Medium: closed-loop transfer functions and errors
Hard: disturbances, stability and poles
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.