Embedded Systems

Assignment 7

Jan Hendrik Dithmar

2031259

Pascal Gwosdek

2505221

Assignment 7  Dithmar (2031259), Gwosdek (2505221)

7.1 Markov processes

a. The probability s_{2}(A) is calculated like this:

s_{2}(A) = 0.2 · 0.2 + 0.8 · 0.6 = 0.04 + 0.48 = 0.52

b.

s_{lim}(A) = s_{lim}(A) · 0.2 + s_{lim}(B) · 0.6
⇔ 0.8s_{lim}(A) = 0.6s_{lim}(B)
⇒ s_{lim}(A) = 0.75s_{lim}(B)
s_{lim}(A) + s_{lim}(B) = 1
⇒ s_{lim}(A) + \frac{1}{0.75}s_{lim}(A) = 1
⇔ \frac{1.75}{0.75}s_{lim}(A) = 1
s_{lim}(A) = \frac{0.75}{1.75}
s_{lim}(A) = \frac{3}{7}

⇒ s_{lim}(B) = \frac{4}{7}

c. s_{lim}(A) depends on s_{lim}(B). We saw that in the previous part. Furthermore we also know from the previous part that s_{lim}(A) + s_{lim}(B) = 1. These two facts suffice to calculate the probabilities without knowing the initial probabilities.

7.2 A/D conversion

• x = 2.25V → 0101_{2}

Comparison c x < / ⩾ c? Output
3.00V  <  0
2.00V  ⩾  1
2.50V  <  0
2.25V  ⩾  1

• x = 3.75V → 1011_{2}

Comparison c x < / ⩾ c? Output
3.00V  ⩾  1
4.00V  <  0
3.50V  ⩾  1
3.75V  ⩾  1

• x = 1.80V → 0011_{2}

Comparison c x < / ⩾ c? Output
3.00V  <  0
2.00V  <  0
1.50V  ⩾  1
1.75V  ⩾  1

7.3 Digital circuits

a. One possible instantiation is the following: • a = 0 • b = 0 • c = 1 • d = 0 • e = 1 • f = 1

b. • ¬x:

• x ∨ y:

7.4 D/A conversion

a. • Assume x⃗ = 00...0. ⇒ U = 0 = U_{min} (∀R′) ✓

• Assume x⃗ = 11...1.

−V = R′ · I
= \frac{V_{ref}}{128R} · R′ · (128 + 64 + 32 + 16 + 8 + 4 + 2 + 1)
= 5V · R′ · \frac{255}{128R} =^{!} 10V
⇔ R′ · \frac{255}{128R} = 2
⇔ R′ = \frac{256}{255}R
= (1 + \frac{1}{255}) kΩ
≈ 1.003922kΩ

b. • Case x⃗ = 011...1:

I_{1} = \frac{V_{ref}}{128R} · 127
= \frac{127}{128} · \frac{5V}{kΩ}
= 4.9609375mA

• Case x⃗ = 100...0:

I_{2} = V_{ref} · \frac{128}{128} · \frac{1}{R}
= 5mA

c.

I_{01...1} < I_{10...0}
⇔ \frac{V_{ref}}{R̂_{01...1}} < \frac{V_{ref}}{R̂_{10...0}}
⟹^{R̂_{i}>0 ∀i} R̂_{10...0} < R̂_{01...1}
Let now be R(x7)′ = R(x7) · d

⇔ R · d < (\frac{1}{128R} + \frac{1}{64R} + \frac{1}{32R} + \frac{1}{16R} + \frac{1}{8R} + \frac{1}{4R} + \frac{1}{2R})^{−1}
⇔ R · d < \frac{128R}{127}
⇔ d < \frac{128}{127} ≈ 1.007874

⇒ R(x7) may increasingly deviate by at most 7.7874‰, i.e. must necessarily be smaller than 1.007874kΩ.

7.5 SDL/Flexray

Please have a look at the handwritten solution which is attached.