1) Fixed point [15 points] a) What are the primary reasons we might use fixed point rather than floating point? [2]

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1 473 Fall 2018 Homework 2 Answers Due on Gradescope by 5pm on December 11 th. 165 points. Notice that the last problem is a group assignment (groups of 2 or 3). Digital Signal Processing and other specialized processors [25 points] 1) Fixed point [15 points] a) What are the primary reasons we might use fixed point rather than floating point? [2] Chips with hardware floating point support typically cost more and draw more power. As such, we sometimes don t have that hardware support. Using floating point instructions without hardware support (so software emulation of floating point) generally involves a huge number of integer instructions per floating point instruction. (Do note that a C compiler compiling for a machine without floating point support will generate the integer instructions needed to do floating point, so it isn t that doing this is difficult.) Fixed point instructions have a lot less overhead to do as integer instructions. b) Say you multiply two signed 16-bit Q15 numbers as standard integers. What size/form will the result be assuming that the size of an int on your machine is 32 bits and no casting was used? [2] 32-bit Q30 c) When could the above multiply overflow? [2] This question has two possible (correct) answers as it depends on the definition of overflow in this context. As at 32-bit Q30, there is no overflow. But once you convert the result back to 16-bit Q15, the result of -1 times -1 (which is 1) is out of range. d) Write a C function Q15mult which takes two values, A and B, each declared as an int16_t number and returns an int16_t number as the product of A and B. All three numbers are treated as Q15. There will be a fair bit of casting. Round to the nearest value (you can break a tie as you wish). Assume that the default int size on your computer is 16 bits (it does matter). [9] int16_t Q15mult (int16_t A, int16_t B) { return ((int32_t)a*b + 0x4000) >> 15; } 2) On page 11 of the lecture 12 slides there is an example problem. Work out the answer. You can assume there are only three b terms. [5 points] (0.5* * *1) = -0.5 (0.5* * *-1) = 0.25 (0.5* * *0) = 0.25 repeats

2 (0.5* * *1) = -0.5 (The question is a bit unclear so just picking one of those lines is also being taken as a correct answer). 3) An embedded processor is arguably a form of a specialized processor. What are the primary differences between an embedded processor and a typical desktop processor? [5 points] The primary differences are in the I/O. In particular an embedded processor typically has a wide variety of I/O options including general purpose I/O pins, I2C, SPI, UART, CAN, Analog inputs and outputs etc. A typical desktop processor has fewer but much wider interfaces (to DRAM or some other breakout chip). Internally a desktop processor tends to have a lot of support for OSes and CPUintensive applications. Things like memory management units, large caches, etc. And just a lot more CPU power. (We ll take any answer which gets at the I/O issue). Wireless communications [50 points] 4) The vast majority of the wireless spectrum in the US (and in the rest of the world) isn t available to us as designers of most consumer devices. [5 points] a) Why is that? [2] In order to make sure things work, the Federal government has selected which frequencies are available to which applications. Most of that frequency spectrum is reserved for uses other than consumer devices. b) What frequencies can we generally use for most consumer devices in the US? In Europe? [3] In the US there are quite a few. If you get at least 2, we ll give you full credit. What we are looking at here are the unlicensed bands. You can also reasonably argue that cell phones are consumer devices and discuss those frequencies. 2.4 to about 2.5GHz. (2401MHz to 2473 seems to be what WIFI uses in the US MHz GHz GHz GHz Europe is a bit trickier MHz is used by WiFi. Use in the 5GHz range is harder to understand and varies about inside of Europe (Switzerland has its own rules for example as does Russia). We ll take anything for Europe that seems reasonable.

3 5) We generally think of sourcing encoding and channel encoding as very different things. [5 points] a) Define those terms in your own words. [2] Source codes compress naturally redundant messages for efficient storage/transmission. Channel codes systematically add redundancy to enable high rate transmissions with as few errors as possible 1. [3] b) One of them generally adds bits to the data and another one removes them. Explain the how doing each of those things is useful/needed. [3] Source encoding removes bits and this is helpful as it makes the data that needs to be sent smaller while keeping the overall data intact. Channel encoding adds bits to allow the receiver to error correct in the event data is lost during the transfer. 6) Using the example on slide 28 of lecture 13, do the following [10 points] a) Find what parity bits would be sent with the data [2] P1 = (0, 1, 1) = 0 P2 = (0, 1, 1) = 0 P3 = (1, 1, 1) = 1 b) Say our receiver received the following 7-bit packets (d 1, d 2, d 3, d 4, p 1, p 2,p 3 in that order) and performed error correction. For each of the data packets, show what the 4 bits of data would be after correction (if any) [4] i) (p 1 and p 3 were wrong) ii) iii) iv) c) Using logic gates, draw a circuit that takes d[1:4] as input and generates p[1:3] as an output. [4] 1 Taken from that s a really well-written and clear definition. The rest of that article seems on-point too.

4 7) Define the following wireless terms in your own words: [4 points, 1 each] a) Keying Keying is a family of modulations where we allow only a predetermined set of values. b) FSK Use different frequencies to represent different values c) QAM Vary both the phase and amplitude to represent different values. d) PSK Change the phase to represent different values.

5 8) Consider the 4-ASK scheme shown on slide 37 of lecture 13. [10 points] a) Show how the message would be communicated as a wave. Draw all sinusoids that have a positive amplitude as starting at max value and ending at the max value while those with a negative amplitude start at the minimum value and end at that same value (just as is done in the example). [4] b) Most schemes that rely on phase or amplitude require a message start with a fixed signal to allow for synchronization. Why do you suppose this is needed? Think about both phase and amplitude issues. [6] To allow the receiving side to calibrate what phase or amplitude the sender uses for each value. In particular, amplitude might vary by distance (as the power fades) and phase could drift. 9) Consider the 4-PSK scheme as shown in the figure on the right. Show how the message would be communicated as a wave. Draw each sinusoid as lasting for exactly 3 periods and assume a sinusoid with a phase of zero degrees would be exactly a sine wave starting at zero and rising (as shown). [6 points] This is going to be really hard to grade as it s hard to read. We ll take anything that looks close. Ignore that line at the end

6 10) Consider the XBee modules as described at [10 points] a) What are the primary differences between the Zigbee PRO and Zigbee in terms of transmit power and receiver sensitivity? [2] XBEE pro use far more transmit power than a regular XBee. Receiver sensitivity also differs by 8 dbm b) Assuming they are used at the 2.4 GHz range and using a Monopole (Integrated whip) antenna on both sides i) What is the theoretic range at which a Zigbee PRO could send to a Zigbee PRO? Show your work. [6] (pt+gt+gr pr) r = f 10( ) 20 r = r = km ii) How does your answer to i) compare to the Zigbee specification? [1] The spec says outdoor with line of site is about 1.6km making the theoretical significantly higher. As mentioned in class on Tuesday, we generally expect this formula to be off by about a factor of 4 to 8. And that is the case. Switching supplies [20 points] 11) Switching supplies [12 points] a) What are the pros and cons of switching power supplies compared to linear regulators? [3] Switching: Pros: Often more power efficient (especially when dropping large voltages). Topologies are capable of stepping up a voltage or inverting it Cons: Switching induces noise on the PDN, requires tuning the capacitors and switching frequency and is therefore more difficult to get correct, and it is typically more expensive. Linear regulator: Pros: Inefficient in comparison to switching regulators if there is a large voltage drop; only capable of stepping down a voltage Cons: Voltages are more stable and noiseless, less external components and easier to set up

7 b) Explain the terms boost, buck and boost-buck as they apply to switching regulators [3] Boost: Increase the output voltage of the regulator Buck: Decrease the output voltage of the regulator Boost-Buck: Can increase or decrease input voltage to specified output voltage (or invert it) c) Find a switching power supply that can be used in place of the LM7805 you used in HW1. [6] i) Given the same problem you had in HW1, how much power would it waste doing the conversion? How does that compare to the LM7805? Switching supply: R-78E Minimum input: 8V Efficiency:91% for min Vin. 5V *.2 A/.91 = Watts W 1W (ideal 5V *.2A) = Watts of lost power. ii) How would its cost compare to the LM7805? This answer will vary. If there s something here we will give credit 12) Answer the following [8 points, -3 per wrong/blank answer, minimum 0] a. A(n) LDO / buck / boost / buck-boost converter which converts 10V to 7V can never be more than 70% efficient. b. Figure #1 is a(n) LDO / buck / boost / buck-boost converter. c. What is the purpose of the diode in figure 1? It prevents the capacitor from getting too high of a voltage. It gives the inductor someplace to draw current from when the transistor is off. It stores energy to provide power when the transistor is off. It controls the transistor s switching rate. Figure 1: A voltage converter d. An otherwise ideal buck converter which has a quiescent current of 5mA and an efficiency of 90% will pull 51.3mA ma from its 12V input if it is generating 100mA at 5 volts. (Round to the nearest ma). Needed for output=(5v*100ma)/.9=555.56mw Input current needed for output=555.56mw/12v=46.3ma. Total input current=46.3ma+5ma=51.3ma.

8 13) Consider a buck converter. For each of the following [20 points, for each of a-d (1) is worth 1 point, (2) is worth 2 points and (3) is worth 2 points] (1) Indicate if a higher value or lower value for that particular characteristic will make it more or less likely that it will enter discontinuous mode. Assume all other values are fixed. (a) Higher frequency will make it less likely. (b) Higher inductor value will make it less likely. (c) No real effect. (d) Higher output current makes it less likely. (2) Draw the picture of what the saw tooth output will look like if you double that value. Use the figure below as the baseline. (a) (b) Higher inductor value flattens the curve.

9 (c) No Change. (d) Shifts the center line up by a factor of 2. (3) Explain what the disadvantage is of changing that value in that way. (e.g. if increasing frequency helps stay out of discontinuous mode, why don t we just always use a higher frequency?) a) Frequency More power needed to switch faster. b) Inductor value Cost/size/weight increase is significant when dealing with inductors. c) Capacitor value N/A d) Output current You may not need the current, so you are just wasting power.