Lab · Fixture Course

01Thermal resistance

Sim

The plate cooling from both faces

Move $t$ to watch the plate cool, or $\alpha$ to make it another material. The table reads it where W01.2 does; the map shows the whole plate over time.

Assumed: supplied here, not in the Materials: $t$ tuned from $0$ to $120\ \text{s}$ in steps of $1$ · $\alpha$ tuned from $4$ to $40\ \text{mm}^2/\text{s}$ in steps of $1$

T

(°C)

x

(mm)

Fig.The temperature across the wall at the chosen time, between where it starts and where it settles, both dashed.

t

(s)30

\alpha

(mm²/s)12

$x$ (mm)	$T$ (°C)
0	20.0
10	37.6
20	44.9
30	37.6
40	20.0

$\mathrm{Fo} = \frac{\alpha t}{L^2}$ = 0.225: the mid-plane is at 44.9 °C, the quarter points at 37.6 °C

Drag the start $\theta^{(0)}$ on the contours of $J$ , change $\alpha$ or the number of steps. The table fills step by step; past the largest $\alpha$ that converges, the steps overshoot and $J$ grows.

Assumed: supplied here, not in the Materials: $\theta_0$ and $\theta_1$ tuned from $-1$ to $3$ in steps of $0.1$ · $\alpha$ tuned from $0.01$ to $1$ in steps of $0.01$ · up to $30$ steps

\theta_1

\theta_0

Fig.Contours of

J(\theta)

. Drag

\theta^{(0)}

to start somewhere else.

y

x

Fig.The data, and the line after 2 steps

\alpha

0.10

\theta_0

at the start0.0

\theta_1

at the start0.0

Steps2

$k$	$\theta_0$	$\theta_1$	$J(\theta)$
0	0.0000	0.0000	4.3333
1	0.2667	0.3667	2.5231
2	0.4700	0.6456	1.4741

After 2 steps: $\theta_0$ = 0.4700, $\theta_1$ = 0.6456, $J(\theta)$ = 1.4741

Step-through

The descent path, step by step

Where the first two steps of W03.1 take $\theta$ on the plane of $\theta_0$ and $\theta_1$ .

\theta_1

\theta_0

\theta^{(0)}

minimum

Fig.Descent starts at

\theta^{(0)} = (0, 0)

, far from the minimum

Step 1 / 3

A step-through, not a live sim: no independent recompute could check a model of this figure, so it steps through the Materials' own values.

Derived: worked out here, no official key: the minimum, $(1.1667, 1.5)$

Python

The same line by the normal equation

Real Python with numpy: the line gradient descent steps towards, solved in one step by $\theta = (X^{\mathsf{T}} X)^{-1} X^{\mathsf{T}} y$ . Edit the code and run it live.

y

x

h_\theta(x)

Fig.The three points and the least-squares line

h_\theta(x) = \theta_0 + \theta_1 x

The plot and output the code gave at build. Run live downloads Python (16.5 MB) and runs the code as you edit it.

Output

theta0 = 1.1667
theta1 = 1.5000
J(theta) = 0.0278

Code

# The line through the three points by the normal equation, solved in one step with numpy:
# theta = (X^T X)^-1 X^T y. Change the points or the model and run it again.
import numpy as np

x = np.array([0.0, 1.0, 2.0])
y = np.array([1.0, 3.0, 4.0])

X = np.column_stack([np.ones_like(x), x])  # a column of ones for theta0
theta = np.linalg.solve(X.T @ X, X.T @ y)
cost = np.sum((X @ theta - y) ** 2) / (2 * len(x))

print(f"theta0 = {theta[0]:.4f}")
print(f"theta1 = {theta[1]:.4f}")
print(f"J(theta) = {cost:.4f}")

ends = np.array([0.0, 3.0])
plot = [{"id": f"point-{i + 1}", "kind": "point", "at": [xi, yi]} for i, (xi, yi) in enumerate(zip(x, y))]
plot.append({"id": "fit", "kind": "line", "through": [[e, theta[0] + theta[1] * e] for e in ends]})

04Full adder

Sim

The full adder in the figure

Tap a row of the truth table, or an input, to set $A$ , $B$ and $C_{in}$ . Every wire carrying a $1$ is inked; a $0$ stays in pencil.

Fig.The figure's circuit. A wire carrying a 1 is inked; a 0 stays in pencil.

Inputs

S = 0, Cout = 1

ABCin	S	Cout
	0	0
	1	0
	1	0
	0	1
	1	0
	0	1
	0	1
	1	1

05Derivatives

Sim

Secants closing on the tangent

Slide $P$ along the curve to move the tangent, or $Q$ to change the run $h$ . The table's secants, over $h$ , a tenth of it and a hundredth, close on the slope the red pen rings.

Assumed: supplied here, not in the Materials: $a$ tuned from $0$ to $2.5$ in steps of $0.05$ · $h$ tuned from $0.1$ to $1$ in steps of $0.05$

y

x

Fig.

f(x)

, its tangent at

P

and the secants towards

Q

. Slide

P

or

Q

along the curve.

a

2.00

h

1.00

$h$	$f(a + h)$	$\frac{f(a + h) - f(a)}{h}$
1	7.0000	6.0000
0.1	1.2310	2.3100
0.01	1.0203	2.0301
$h \to 0$	1.0000	2.0000

The secants close on $f'(a)$ = 2.0000: the tangent at $P$ is y = 2.0000x − 3.0000

06Tank level in STL

Sim

The tank listing, running

Set the level transmitter's raw value and the switches, then step a statement or run a whole scan. The trace shows the accumulators, AR 1, the status word and every byte each statement writes.

ORGANIZATION_BLOCK OB 1
TITLE = Horizontal cylindrical tank: level to volume
// Tank radius r = 1.0 m, length 4.0 m.
// PIW 256: level transmitter, 0..27648 = empty..full (-27648..27648 if I 0.1 = bipolar).
// MD 60..72: four volume set-points in litres, written by the HMI.
BEGIN
NETWORK
TITLE = 1 Zero-point correction
L PIW 256 // raw level from the analog module
L 200 // zero-point offset found at commissioning
+I // INT add: wraps past 32767 and sets OV
T MW 20 // corrected raw level
A OV // did the addition overflow?
S M 1.2 // latch: offset overflow alarm
NETWORK
TITLE = 2 Scale to metres with FC105
CALL FC 105
IN :=MW20
HI_LIM :=2.000000e+000 // full = tank diameter, 2.0 m
LO_LIM :=0.000000e+000 // empty = 0.0 m
BIPOLAR:=I0.1
RET_VAL:=MW22 // W#16#0008 = input out of range
OUT :=MD24 // level h in metres
NETWORK
TITLE = 3 Wetted cross-section: A = r2 acos(d/r) - d sqrt(2rh - h2)
L 1.000000e+000 // r = 1.0 m
T MD 12
L MD 12
L MD 24
-R // d = r - h
T MD 44
L MD 12
/R // d / r
ACOS // half the wetted angle, rad
L MD 12
SQR // r squared
*R
T MD 48 // r2 acos(d/r)
L MD 12
L 2.000000e+000
*R // 2r
L MD 24
*R // 2rh
L MD 24
SQR // h squared
-R
ABS // clears -0.0 when h = 0 or h = 2r
SQRT // half the chord
L MD 44
*R // d sqrt(2rh - h2)
T MD 52
L MD 48
L MD 52
-R // wetted area A, m2
T MD 28
NETWORK
TITLE = 4 Volume in litres
L MD 28
L 4.000000e+000 // tank length 4.0 m
*R
L 1.000000e+003 // m3 to litres
*R
T MD 32 // volume V, litres
NETWORK
TITLE = 5 Level band: how many set-points has V passed?
L 0
T MW 36 // band := 0
LAR1 P#60.0 // AR1 -> first set-point, MD 60
L 4 // four set-points
NEXT: T MW 38 // loop counter
L MD 32 // volume
L MD [AR1,P#0.0] // set-point under AR1
>R
JCN SKIP
L MW 36
L 1
+I
T MW 36 // band := band + 1
SKIP: +AR1 P#4.0 // next set-point
L MW 38
LOOP NEXT
L MW 20
L 10
*I // tenths of a count, 32 bits, for the trend log
T MD 40
NETWORK
TITLE = 6 Alarms, latched until I 0.0 acknowledges
A I 0.0 // acknowledge button
R M 1.0
R M 1.1
R M 1.2
L MW 22 // RET_VAL from FC105
L 0
<>I
S M 1.0 // latch: level signal out of range
L MW 36 // level band
L 4
>=I
S M 1.1 // latch: above the high-high set-point
O M 1.0
O M 1.1
O M 1.2
= Q 4.1 // common alarm lamp
NETWORK
TITLE = 7 High-volume lamp (exam trap)
A I 0.2 // pump running: opens a logic string ...
L MD 32
L 1.000000e+004
>R // ... but a compare writes RLO outright
= Q 4.0 // so Q 4.0 = (V > 10 000 l), whatever I 0.2 is
END_ORGANIZATION_BLOCK

PIW 256 (INT)

Scan 1 · statement 113 of 113

Statement111 BE

ACCU 110000.016#461C_4000

ACCU 26283.185516#45C4_597C

AR 1P#76.0

/FC0
RLO0
STA1
OR0
OS0
OV0
CC01
CC10
BR?

A ? is left by FC105: Siemens doesn't publish what the block leaves there, so the trace doesn't guess until a statement writes it.

Writtennothing

Operand	Value
MW 20	13824
MW 22	W#16#0000
MD 24	1.0
MD 28	1.5707964
MD 32	6283.1855
MW 36	2
MD 40	L#138240
Q 4.0	0
Q 4.1	0
M 1.0	0
M 1.1	0
M 1.2	0

Line	Statement	ACCU 1	RLO
9	L PIW 256	13624	0
10	L 200	200	0
11	+I	13824	0
12	T MW 20	13824	0
13	A OV	13824	0
14	S M 1.2	13824	0
17	CALL FC 105	?	?
26	L 1.000000e+000	1.0	?
27	T MD 12	1.0	?
28	L MD 12	1.0	?
29	L MD 24	1.0	?
30	-R	0.0	?
31	T MD 44	0.0	?
32	L MD 12	1.0	?
33	/R	0.0	?
34	ACOS	1.5707964	?
35	L MD 12	1.0	?
36	SQR	1.0	?
37	*R	1.5707964	?
38	T MD 48	1.5707964	?
39	L MD 12	1.0	?
40	L 2.000000e+000	2.0	?
41	*R	2.0	?
42	L MD 24	1.0	?
43	*R	2.0	?
44	L MD 24	1.0	?
45	SQR	1.0	?
46	-R	1.0	?
47	ABS	1.0	?
48	SQRT	1.0	?
49	L MD 44	0.0	?
50	*R	0.0	?
51	T MD 52	0.0	?
52	L MD 48	1.5707964	?
53	L MD 52	0.0	?
54	-R	1.5707964	?
55	T MD 28	1.5707964	?
58	L MD 28	1.5707964	?
59	L 4.000000e+000	4.0	?
60	*R	6.2831855	?
61	L 1.000000e+003	1000.0	?
62	*R	6283.1855	?
63	T MD 32	6283.1855	?
66	L 0	0	?
67	T MW 36	0	?
68	LAR1 P#60.0	0	?
69	L 4	4	?
70	T MW 38	4	?
71	L MD 32	6283.1855	?
72	L MD [AR1,P#0.0]	1000.0	?
73	>R	1000.0	1
74	JCN SKIP	1000.0	1
75	L MW 36	0	1
76	L 1	1	1
77	+I	1	1
78	T MW 36	1	1
79	+AR1 P#4.0	1	1
80	L MW 38	4	1
81	LOOP NEXT	3	1
70	T MW 38	3	1
71	L MD 32	6283.1855	1
72	L MD [AR1,P#0.0]	3000.0	1
73	>R	3000.0	1
74	JCN SKIP	3000.0	1
75	L MW 36	1	1
76	L 1	1	1
77	+I	2	1
78	T MW 36	2	1
79	+AR1 P#4.0	2	1
80	L MW 38	3	1
81	LOOP NEXT	2	1
70	T MW 38	2	1
71	L MD 32	6283.1855	1
72	L MD [AR1,P#0.0]	10000.0	1
73	>R	10000.0	0
74	JCN SKIP	10000.0	1
79	+AR1 P#4.0	10000.0	1
80	L MW 38	2	1
81	LOOP NEXT	1	1
70	T MW 38	1	1
71	L MD 32	6283.1855	1
72	L MD [AR1,P#0.0]	12000.0	1
73	>R	12000.0	0
74	JCN SKIP	12000.0	1
79	+AR1 P#4.0	12000.0	1
80	L MW 38	1	1
81	LOOP NEXT	0	1
82	L MW 20	13824	1
83	L 10	10	1
84	*I	L#138240	1
85	T MD 40	L#138240	1
88	A I 0.0	L#138240	0
89	R M 1.0	L#138240	0
90	R M 1.1	L#138240	0
91	R M 1.2	L#138240	0
92	L MW 22	0	0
93	L 0	0	0
94	<>I	0	0
95	S M 1.0	0	0
96	L MW 36	2	0
97	L 4	4	0
98	>=I	4	0
99	S M 1.1	4	0
100	O M 1.0	4	0
101	O M 1.1	4	0
102	O M 1.2	4	0
103	= Q 4.1	4	0
106	A I 0.2	4	1
107	L MD 32	6283.1855	1
108	L 1.000000e+004	10000.0	1
109	>R	10000.0	0
110	= Q 4.0	10000.0	0
111	BE	10000.0	0

07Machine parts

3D part

The flanged hub

The hub of W07.1 as an exact solid: a flange with four bolt holes, a hub, and a bore through both. Turn it to see the holes from below.

Drawing the part…

The part's dimensions, in millimetres
Dimension	mm	Read as
	Ø80	stated
	10	stated
	Ø40	stated
	30	scaled off the drawing
	Ø20	stated
	Ø8	assumed
	60	stated

08Root locus and the step response

Sim

The loop on its root locus

Drag a closed-loop pole along the locus, or slide $K$ , and watch the step response, the Bode plot and the Nyquist plot follow. The red pen marks the overshoot and the settling time.

Assumed: supplied here, not in the Materials: $K$ tuned from $0.5$ to $20$ in steps of $0.5$

Fig.The loop as the figure draws it: the gain and the plant under unity negative feedback.

j\omega

\sigma

Fig.The root locus: drag a closed-loop pole along it to set the gain. Crosses are the open-loop poles.

K

4.0

$\omega_n$ (rad/s)	2.000
$\zeta$	0.500
$\sigma$	−1.000
$\omega_d$ (rad/s)	1.732
overshoot (%)	16.303
$T_p$ (s)	1.814
$T_r$ (s)	0.819
$T_s$ (s)	4.038
$\omega_c$ (rad/s)	1.572
PM (°)	51.827

At $K$ = 4.0 the response overshoots by 16.303 % and settles in 4.038 s.

09A blending station in SCL

Sim

The blending station, running

Set the load cells and the buttons, then step a statement or run a whole scan. Each scan calls the Blender block once; the trace shows what every statement decided and wrote.

// A blending station: three silos dose into one hopper, then a mixer runs the batch.
TYPE "Silo"
STRUCT
target_kg : REAL; // the recipe's weight from this silo
raw : INT; // load cell: 0..27648 = 0..500 kg
kg : REAL; // the weight it reads
short : BOOL; // still under its target
END_STRUCT
END_TYPE
DATA_BLOCK "Recipe"
VAR
silo : ARRAY[1..3] OF "Silo";
batches : INT; // batches mixed so far
END_VAR
BEGIN
silo[1].target_kg := 120.0;
silo[2].target_kg := 80.0;
silo[3].target_kg := 120.0;
END_DATA_BLOCK
FUNCTION ToKg : REAL
VAR_INPUT
raw : INT;
END_VAR
BEGIN
ToKg := INT_TO_REAL(raw) / 27648.0 * 500.0;
END_FUNCTION
FUNCTION_BLOCK "Blender"
VAR_INPUT
start : BOOL; // the operator's start button
raw1 : INT; // the three load cells
raw2 : INT;
raw3 : INT;
mixed : BOOL; // the mixer's timer has run out
END_VAR
VAR_OUTPUT
valve : INT; // the silo dosing now, 0 = none
total_kg : REAL; // in the hopper
fill_pct : REAL; // of the batch's weight
bags : INT; // 25 kg bags the hopper fills
mixer : BOOL;
END_VAR
VAR
step : INT; // 0 IDLE, 1 DOSE, 2 MIX
last_start : BOOL; // start, one scan ago
END_VAR
VAR_TEMP
i : INT;
target : REAL;
END_VAR
BEGIN
"Recipe".silo[1].raw := raw1;
"Recipe".silo[2].raw := raw2;
"Recipe".silo[3].raw := raw3;
total_kg := 0.0;
target := 0.0;
FOR i := 1 TO 3 DO
"Recipe".silo[i].kg := ToKg(raw := "Recipe".silo[i].raw);
"Recipe".silo[i].short := "Recipe".silo[i].kg < "Recipe".silo[i].target_kg;
total_kg := total_kg + "Recipe".silo[i].kg;
target := target + "Recipe".silo[i].target_kg;
END_FOR;
fill_pct := total_kg / target * 100.0;
bags := REAL_TO_INT(total_kg / 25.0); // 312.5 kg is 12.5 bags (a Divergence: the exam's answer differs from a real S7)
valve := 0;
CASE step OF
0: // IDLE: wait for start to rise
IF start AND NOT last_start THEN
step := 1;
END_IF;
1: // DOSE: open the first silo still short
FOR i := 1 TO 3 DO
IF "Recipe".silo[i].short THEN
valve := i;
EXIT;
END_IF;
END_FOR;
IF valve = 0 THEN
step := 2;
END_IF;
2: // MIX: run the mixer until its timer runs out
mixer := NOT mixed;
IF mixed THEN
"Recipe".batches := "Recipe".batches + 1;
step := 0;
END_IF;
ELSE
step := 0;
END_CASE;
last_start := start;
END_FUNCTION_BLOCK

raw1 (INT)

raw2 (INT)

raw3 (INT)

Scan 1 · statement 27 of 27

Statement92 last_start := start;

Writtenlast_start := FALSE

Variable	Value
step	0
valve	0
total_kg	312.5
fill_pct	97.65625
bags	12
mixer	FALSE
Recipe.silo[1].kg	117.1875
Recipe.silo[2].kg	78.125
Recipe.silo[3].kg	117.1875
Recipe.batches	0

Line	Statement	Result
54	"Recipe".silo[1].raw := raw1;	Recipe.silo[1].raw := 6480
55	"Recipe".silo[2].raw := raw2;	Recipe.silo[2].raw := 4320
56	"Recipe".silo[3].raw := raw3;	Recipe.silo[3].raw := 6480
57	total_kg := 0.0;	total_kg := 0.0
58	target := 0.0;	target := 0.0
59	FOR i := 1 TO 3 DO	into the loop
60	"Recipe".silo[i].kg := ToKg(raw := "Recipe".silo[i].raw);	Recipe.silo[1].kg := 117.1875
61	"Recipe".silo[i].short := "Recipe".silo[i].kg < "Recipe".silo[i].target_kg;	Recipe.silo[1].short := TRUE
62	total_kg := total_kg + "Recipe".silo[i].kg;	total_kg := 117.1875
63	target := target + "Recipe".silo[i].target_kg;	target := 120.0
64	END_FOR;	again
60	"Recipe".silo[i].kg := ToKg(raw := "Recipe".silo[i].raw);	Recipe.silo[2].kg := 78.125
61	"Recipe".silo[i].short := "Recipe".silo[i].kg < "Recipe".silo[i].target_kg;	Recipe.silo[2].short := TRUE
62	total_kg := total_kg + "Recipe".silo[i].kg;	total_kg := 195.3125
63	target := target + "Recipe".silo[i].target_kg;	target := 200.0
64	END_FOR;	again
60	"Recipe".silo[i].kg := ToKg(raw := "Recipe".silo[i].raw);	Recipe.silo[3].kg := 117.1875
61	"Recipe".silo[i].short := "Recipe".silo[i].kg < "Recipe".silo[i].target_kg;	Recipe.silo[3].short := TRUE
62	total_kg := total_kg + "Recipe".silo[i].kg;	total_kg := 312.5
63	target := target + "Recipe".silo[i].target_kg;	target := 320.0
64	END_FOR;	done
65	fill_pct := total_kg / target * 100.0;	fill_pct := 97.65625
66	bags := REAL_TO_INT(total_kg / 25.0);	bags := 12
67	valve := 0;	valve := 0
68	CASE step OF	0: line 69
70	IF start AND NOT last_start THEN	FALSE
92	last_start := start;	last_start := FALSE

Derived: worked out here, no official key: $12.5$ bags, held as $12$
Divergence: The exam answer is $13$ bags, the Professor's. REAL_TO_INT rounds $12.5$ to the even number on a real S7, so the CPU holds $12$ .

10Two conveyors in ladder

Sim

The two conveyors, running

Press Start and Stop, then run time on. A rung carrying power is inked, the TON's gauge fills as ET counts up to PT, and the timing chart measures the delay from Q 4.0 to Q 4.1.

Fig.The figure's networks. A rung carrying power is inked; a timer's bar fills as it runs.