2017 Exam

Question 1

Knee Frequency Definition (2)

Answer 1

cut-off freq. value above which the energy spectral density of a digital waveform falls off rapidly (-40dB/dec)

Most of energy is digital pulses and is concentrated below the knee frequency

Question 2

Freq Spectrum of unequal duty cycle

Answer 2

Question 3

Freq Spectrum of 50% duty cycle

Answer 3

Question 4

Freq. Spectrum of trapezoidal waveform

Answer 4

Question 5

WHy are large # of GND via holes used on a PCB (4)

Answer 5

• @ high freq, return current follows the path of least inductance
• lowest ind. return path lies directly underneath T.L, minimising the total loop area between the outgoing + returning current paths
• If signal jumps from 1 PCB layer to another through via hole, there is no wat for return current path to make the jump
  
  • Return current then follows a path other than directly underneath signal trace
    
    • ​more inductance = EMI = crosstalk
• ​​Plenty of via holes used to minimise inductances

Question 6

Why is there ISI in high speed tranmission system

Answer 6

Causes - reflection, crosstalk, any other source

Last transition may have not have died away by the time the next transition occurs causing ISI

Degreades performance power factor + is important when

period < 2x delay time

Question 7

Eye Diagram + Uses (3)

Answer 7

• A superposition of observations of sequential bit periods, generated as pseudo-random bit sequence
• Induction of the impact of ISI

Wide Open = good performance

Closed = BER increases

Question 8

Common Impedance Coupling Definition

Answer 8

causes crosstalk when a single conductor is used for more than 1 signal

Question 9

Common Path Noise Voltage

Answer 9

product of the returning signal current + impedance of teh common conductor

Question 10

Two ways to reduce common impedance coupling (2)

Answer 10

• // power + ground planes
  
  • provides low impedance path between pwr + GND pins
• decoupling capacitors between GND + pwr planes to provide a low impedance path

Question 11

90 angle bend effects on propagating signal

Answer 11

• angle bend causes a discontinuity between 2 T.Ls with different field patterns leadingto stored energy
  
  • can be accounted for by adding an excess capacitance to the capacitors in equivalent circuit (same for inductance) - capacitance dominates
• reduce by champfering the corner reducing line width + total capacitance

Question 12

Characteristics + differences of near + far-end crosstalk in T.L (5)

Answer 12

• Near-end (backward) - closest, Far-end (forward) - furthest
• Initial wave will generate two crosstalk pulses
  
  • near-end pulse that flows into the near-end termination
  • far-end pulse that propagates towards the far-end
• Contributions to these pulses
  
  • M.C - creates current forward + backward
  • M.I - creates current backward only
• As waves propagates along the line, continuously induces new pulses, arriving simultaneously as a single pulse (w/ initial wave) at the far end
• Arrive sequentially @ near-end, beginning w/ launch of initial wave + ending when the final contribute (created when the initial wave reached the far-end) arrives back @ near-end
  
  • i.e. 2x propagation delay of the line

Question 13

Max Jitter early or late arrival time

Answer 13

Early - when line carries opposite bit w.r.t L1 + L3

Late - carries same as L1 + L3

Question 14

USB protocol structure (3)

Answer 14

• Top of Pyramid is Host
• Connected to a node of hub
• Max of 127 devices can be connected to host
  
  • responsible for all comms on the bus incl
    
    • device addr
    • bus b/w
    • pwr requirements

Question 15

USB 2.0 Diagram

Answer 15

1. on-Twisted Power Pair
   
   1. Red - V_BUB
   2. Black - Pwr GND
2. PVC Jacket
3. Outer Shield
4. Inner Shield
5. Twisted Signalling Pair
   
   1. White - D+
   2. Green - D-
6. AWG Tined Copper Drain Wire

Question 16

USB 3.0 Diagram

Answer 16

1. UTP
2. Shield Braid
3. STP
4. STP

Question 17

2x Techniques USB 3 employs to mitigate attentuation + distortion

Answer 17

De-emphasis Amplitude of transmitted signal voltage is arranged to be higher than normal after a logic-level transition. Gives more drive when high freq. components of the signal are significant which mitigate the greater cable loss that is found @ high freq

Equalisation Perceived signal is multiplied by the inverse of teh channel transfer function so recover the transmitted s(w) using continuous time linear equalisation to flatten the freq response.

Question 18

Reduce strangt of interfering signals (4)

Answer 18

• Add ferrite ring around STP cable
• Wanted signals balanced, total current throught the ring is zero - no effect
• Interfering signal has its return path through external wiring - interfering current passing through the ring will be finite
• Impedance controlled by the ring will reduce the interfering current

Question 19

Why are screened cables used @ high freq only? (5)

Answer 19

• @ high freq, thickness of screen is greater than skin depth
• signal current flow on the inside of the screen
  
  • interference currents flow on outside
• Screen must be grounded @ both ends
• Greatest eff. if src, load + cable are screened and in their own boxes