Issue 284: EQ Answers

PROBLEM 1
Can you name all of the signals in the original 25-pin RS-232 connector?

ANSWER 1
Pins 9, 10, 11, 18, and 25 are unassigned/reserved. The rest are:

Pin Abbreviation Source Description
1 PG - Protective ground
2 TD DTE Transmitted data
3 RD DCE Received data
4 RTS DTE Request to send
5 CTS DCE Clear to send
6 DSR DCE Data Set Ready
7 SG - Signal ground
8 CD DCE Carrier detect
12 SCD DCE Secondary carrier detect
13 SCTS DCE Secondary clear to send
14 STD DTE Secondary transmitted data
15 TC DCE Transmitter clock
16 SRD DCE Secondary received data
17 RC DCE Receiver clock
19 SRTS DTE Secondary request to send
20 DTR DTE Data terminal ready
21 SQ DCE Signal quality
22 RI DCE Ring indicator
23 - DTE Data rate selector
24 ETC DTE External transmitter clock

 

PROBLEM 2
What is the key difference between a Moore state machine and a Mealy state machine?

ANSWER 2
The key difference between Moore and Mealy is that in a Moore state machine, the outputs depend only on the current state, while in a Mealy state machine, the outputs can also be affected directly by the inputs.

 

PROBLEM 3
What are some practical reasons you might choose one state machine over the other?

ANSWER 3
In practice, the difference between Moore and Mealy in most situations is not very important. However, when you’re trying to optimize the design in certain ways, it sometimes is.

Generally speaking, a Mealy machine can have fewer state variables than the corresponding Moore machine, which will save physical resources on a chip. This can be important in low-power designs.

On the other hand, a Moore machine will typically have shorter logic paths between flip-flops (total combinatorial gate delays), which will enable it to run at a higher clock speed than the corresponding Mealy machine.

 

PROBLEM 4
What is the key feature that distinguishes a DSP from any other general-purpose CPU?

ANSWER 4
Usually, the key distinguishing feature of a DSP when compared with a general-purpose CPU is that the DSP can execute certain signal-processing operations with few, if any, CPU cycles wasted on instructions that do not compute results.

One of the most basic operations in many key DSP algorithms is the MAC (multiply-accumulate) operation, which is the fundamental step used in matrix dot and cross products, FIR and IIR filters, and fast Fourier transforms (FFTs). A DSP will typically have a register and/or memory organization and a data path that enables it to do at least 64 MAC operations (and often many more) on unique data pairs in a row without any clocks wasted on loop overhead or data movement. General-purpose CPUs do not generally have enough registers to accomplish this without using additional instructions to move data between registers and memory.

Issue 282: EQ Answers

PROBLEM 1
Construct an electrical circuit to find the values of Xa, Xb, and Xc in this system of equations:

21Xa – 10Xb – 10Xc = 1
–10Xa + 22Xb – 10Xc = –2
–10Xa – 10Xb + 20Xc = 10

Your circuit should include only the following elements:

one 1-Ω resistor
one 2-Ω resistor
three 10-Ω resistors
three ideal constant voltage sources
three ideal ammeters

The circuit should be designed so that each ammeter displays one of the values Xa, Xb, or Xc. Given that the Xa, Xb, and Xc values represent currents, what kind of circuit analysis yields equations in this form?

ANSWER 1
You get equations in this form when you do mesh analysis of a circuit. Each equation represents the sum of the voltages around one loop in the mesh.

PROBLEM 2
What do the coefficients on the left side of the equations represent? What about the constants on the right side?

ANSWER 2
The coefficients on the left side of each equation represent resistances. Resistance multiplied by current (the unknown Xa, Xb, and Xc values) yields voltage.
The “bare” numbers on the right side of each equation represent voltages directly (i.e., independent voltage sources).

PROBLEM 3
What is the numerical solution for the equations?

ANSWER 3
To solve the equations directly, start by solving the third equation for Xc and substituting it into the other two equations:

Xc = 1/2 Xa + 1/2 Xb + 1/2

21Xa – 10Xb – 5Xa – 5Xb – 5 = 1
–10Xa + 22Xb – 5Xa – 5Xb – 5 = –2

16Xa – 15Xb = 6
–15Xa + 17Xb = 3

Solve for Xa by multiplying the first equation by 17 and the second equation by 15 and then adding them:

272Xa – 255Xb = 102
–225Xa + 255Xb = 45

47Xa = 147 → Xa = 147/47

Solve for Xb by multiplying the first equation by 15 and the second equation by 16 and then adding them:

240Xa – 225Xb = 90
–240Xa + 272Xb = 48

47Xb = 138 → Xb = 138/47

Finally, substitute those two results into the equation for Xc:

Xc = 147/94 + 138/94 + 47/94 = 332/94 = 166/47

PROBLEM 4
Finally, what is the actual circuit? Draw a diagram of the circuit and indicate the required value of each voltage source.

ANSWER 4
The circuit is a mesh comprising three loops, each with a voltage source. The common elements of the three loops are the three 10-Ω resistors, connected in a Y configuration (see the figure below).

cc281_eq_fig1The values of the voltage sources in each loop are given directly by the equations, as shown. To verify the numeric solution calculated previously, you can calculate all of the node voltages around the outer loop, plus the voltage at the center of the Y, and ensure they’re self-consistent.

We’ll start by naming Va as ground, or 0 V:

Vb = Va + 2 V = 2 V

Vc = Vb + 2 Ω × Xb = 2V + 2 Ω × 138/47 A = 370/47 V = 7.87234 V

Vd = Vc + 1 Ω × Xa = 370/47 V + 1 Ω × 147/47A = 517/47 V = 11.000 V

Ve = Vd – 1 V = 11.000 V – 1.000 V = 10.000 V

Va = Ve – 10 V = 0 V

which is where we started.

The center node, Vf, should be at the average of the three voltages Va, Vc, and Ve:

0 V + 370/47 V + 10 V/3 = 840/141 V = 5.95745 V

We should also be able to get this value by calculating the voltage drops across each of the three 10-Ω resistors:

Va + (Xc – Xb) × 10 Ω = 0 V + (166 – 138)/47A × 10 Ω = 280/47 V = 5.95745 V

Vc + (Xb – Xa) × 10 Ω = 370/47V + (138-147)/47A × 10 Ω = 280/47 V = 5.95745 V

Ve + (Xa – Xc) × 10 Ω = 10 V + (147-166)/47 A × 10 Ω = 280/47 V = 5.95745 V

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