UNIVERSITY QUESTION PAPER SOLUTION - E-Learning

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DSPAA10EC751UNIVERSITY QUESTION PAPER SOLUTIONUNIT - 1INTRODUCTION TO DIGITAL SIGNAL PROCESSINGQuestion No. 11. Explain the decimation and interpolation process with an example.(DEC’09-JAN’10) (Dec.12, 6m).(6marks)Solution:Decimation and Interpolation are two techniques used to alter the sampling rate of asequence. Decimation involves decreasing the sampling rate without violating the samplingtheorem whereas interpolation increases the sampling rate of a sequence appropriately byconsidering its neighboring samples.Decimation is a process of dropping the samples without violating sampling theorem. The factorby which the signal is decimated is called as decimation factor and it is denoted by M. It is givenby,Ex: Let x(n) [3 2 2 4 1 0 –3 –2 –1 0 2 3] be decimated with a factor of 2. Let the filteredsequence be w(n) [2.1 2 3.9 1.5 0.1 –2.9 –2 –1.1 0.1 1.9 2.9]. Decimated sequence y (m) can beobtained by dropping every alternative sample of w (n). y (m) [2 1.5 -2.9 -1.1 1.9]Interpolation is a process of increasing the sampling rate by inserting new samples in between.The input output relation for the interpolation, where the sampling rate is increased by a factor L,is given as,Let x(n) [0 3 6 9 12] be interpolated with L 3. If the filter coefficients of the filters are bk [1/32/3 1 2/3 1/3], the interpolated sequence is After inserting zeros,Dept ECE, SJBITPage 1

DSPAA10EC751W(m) [012]00300600900Question No. 22 The sequence x(n) [0,3,6,9] is interpolated using interpolation sequence bk [1/3,2/3,1,2/3,1/3] and the interpolation factor of 3. Find the interpolated sequence y (m).(6marks) (DEC’09-JAN’10)Solution:After inserting zeros,w (m) [0 0 0 3 0 0 6 0 0 9 ]bk [1/3 2/3 1 2/3 1/3]We have,y (n) [ 0 1 2 3 4 5 6 7 8 9]Question No. 33 Describe the basic features that should be provided in thr DSP architecture to be used toimplement the Nth order FIR filter, y(n) x(i) h(n-i) where x(n) denotes the inputsample, y(n) the output sample and h(i) denotes ith filter coefficient.(8marks)(DEC’09-JAN’10)Solution:In order to implement the above operation in a DSP, the architecture requires thefollowing features:Dept ECE, SJBITPage 2

DSPAA10EC751i. A RAM to store the signal samples x (n)ii. A ROM to store the filter coefficients h (n)iii. An MAC unit to perform Multiply and Accumulate operationiv. An accumulator to store the result immediatelyv. A signal pointer to point the signal sample in the memoryvi. A coefficient pointer to point the filter coefficient in the memoryvii. A counter to keep track of the countviii. A shifter to shift the input samples appropriatelyQuestion No. 44 Explain the issues to be considered in designing and implementing a DSP system, with thehelp of a neat block diagram. (6 marks) (MAY-JUNE)Solution:The Block Diagram of a DSP SystemDesign issues:Complexity of algorithmSample rateSpeedData representationQuestion No. 55 Briefly explain the major features of programmable DSPs.(6 marks) (MAY-JUNE 10)Solution:Although addition and multiplication are two different operations, they can be performed inparallel. By the time the multiplier is computing the product, accumulator can accumulate theDept ECE, SJBITPage 3

DSPAA10EC751product of the previous multiplications. Thus if N products are to be accumulated, N-1multiplications can overlap with N-1 additions. During the very first multiplication, accumulatorwill be idle and during the last accumulation, multiplier will be idle. Thus N 1 clock cycles arerequired to compute the sum of N products. DSPs should have zero overhead looping &specialized addressing capability.Question No. 61c.Explain the operation used in DSP to increase the sampling rate. The sequencex(n) [0,2,4,6,8] is interpolated using interpolation sequence bk [1/2,1,1/2] and the interpolationfactor is 2.find the interpolated sequence y(m).(8 marks) (MAY-JUNE)Solution:Interpolation factor L 2W(m) {0,0,2,0,4,0,6,08}Y(m) {0,0,1,2,3,4,5,6,7,8}There are four special cases in this addressing mode. They area. SAR EAR & updated PNTR EARb. SAR EAR & updated PNTR SARc. SAR EAR & updated PNTR SARd. SAR EAR & updated PNTR EARThe buffer length in the first two case will be (EAR-SAR 1) whereas for the next lowcases (SAR-EAR 1)Question No. 7Why signal sampling is required? Explain the sampling process. (Dec.12, 5m)Solution:The Sampling Process:ADC process involves sampling the signal and then quantizing the same to a digitalvalue. In order to avoid Aliasing effect, the signal has to be sampled at a rate at least equal to theNyquist rate. The condition for Nyquist Criterion is as given below, fs 1/T 2 fmWhere, fs is the sampling frequency, fm is the maximum frequency component in themessage signal. If the sampling of the signal is carried out with a rate less than the Nyquist rate,the higher frequency components of the signal cannot be reconstructed properly. The plots of thereconstructed outputs for various conditions are as shown in figure 1.4.Dept ECE, SJBITPage 4

DSPAA10EC751Question No. 8List the major architectural features used in DSP system to achieve high speed programexecution. (Dec.12, 6m).Solution:A programmable DSP device should provide instructions similar to aconventional microprocessor. The instruction set of a typical DSP device should includethe following,a. Arithmetic operations such as ADD, SUBTRACT, MULTIPLY etcb. Logical operations such as AND, OR, NOT, XOR etcc. Multiply and Accumulate (MAC) operationd. Signal scaling operationDept ECE, SJBITPage 5

DSPAA10EC751In addition to the above provisions, the architecture should also include,a. On chip registers to store immediate resultsb. On chip memories to store signal samples (RAM)c. On chip memories to store filter coefficients (ROM)Question No. 9Explain how to simulate the impulse responses of FIR and IIR filters. (Dec.11, 6m).Solution:FIR Filter DesignFrequency response of an FIR filter is given by the following expression,Design procedure of an FIR filter involves the determination of the filter coefficients bk.IIR Filter DesignIIR filters can be designed using two methods viz using windows and direct method. Inthis approach, a digital filter can be designed based on its equivalent analog filter. An analogfilter is designed first for the equivalent analog specifications for the given digital specifications.Then using appropriate frequency transformations, a digital filter can be obtained. The filterspecifications consist of passband and stopband ripples in dB and Passband and Stopbandfrequencies in rad/sec.Fig Lowpass Filter SpecificationsDirect IIR filter design methods are based on least squares fit to a desired frequency response.These methods allow arbitrary frequency response specifications.Dept ECE, SJBITPage 6

DSPAA10EC751UNIT – 2ARCHITECTURES FOR PROGRAMMABLE DIGITAL SIGNAL-PROCESSORSQuestion No.11. What is the role of a shifter in DSP? Explain the implementation of 4-bit shift right barrelshifter, with a diagram. (DEC’09-JAN’10)6MSolution:Shifters are used to either scale down or scale up operands or the results. The following scenariosgive the necessity of a shiftera. While performing the addition of N numbers each of n bits long, the sum cangrow up to n log2 N bits long. If the accumulator is of n bits long, then an overflow errorwill occur. This can be overcome by using a shifter to scale down the operand by anamount of log2N.b. Similarly while calculating the product of two n bit numbers, the product cangrow up to 2n bits long. Generally the lower n bits get neglected and the sign bitis shiftedto save the sign of the product.c. Finally in case of addition of two floating-point numbers, one of the operands has to beshifted appropriately to make the exponents of two numbers equal.d. From the above cases it is clear that, a shifter is required in the architecture of a DSP.Dept ECE, SJBITPage 7

DSPAA10EC751Question No.22. Identify the addressing modes of the operands in each of the following instructions & theiroperation. (DEC’09-JAN’10)Solution:i) ADD B :REGISTER addressing modeii) ADD #1234H :Immediateiii) ADD 5678H : directiv) ADD *ADDRREG : pre incrementDept ECE, SJBIT8mPage 8

DSPAA10EC751Question No.33. Explain the features of a program sequencer unit of a programmable DSP with a neat blockdiagram(DEC’09-JAN’10) (Dec.10-Jan.11, 6m)6mSolution:It is a part of the control unit used to generate instruction addresses in sequence needed to accessinstructions. It calculates the address of the next instruction to befetched. The next address can be from one of the following sources.a. Program Counterb. Instruction register in case of branching, looping and subroutine callsc. Interrupt Vector tabled. Stack which holds the return addressQuestion No.44. Explain Baugh-wooley multiplier for signed numbers. Show the multiplication operation for4x4 signed multiplication(6 marks)Solution: Consider two signed numbers A and B,Dept ECE, SJBITPage 9

DSPAA10EC751Question No.55. What is meant by circular addressing mode? Write pointer updating algorithm for the Circularaddressing mode and show different cases that encounter during the updating process of thepointer. (MAY-JUNE 10Solution:Circular Addressing Mode:While processing the data samples coming continuously in a sequential manner, circular buffersare used. In a circular buffer the data samples are stored sequentially from the initial location tillthe buffer gets filled up. Once the buffer gets filled up, the next data samples will get storedonce again from the initial location. This process can go forever as long as the data samples areprocessed in a rate faster than the incoming data rate.The pointer updating algorithm for the circular addressing mode is as shown below.Circular Addressing mode requires three registers viza. Pointer register to hold the current location (PNTR)b. Start Address Register to hold the starting address of the buffer (SAR)c. End Address Register to hold the ending address of the buffer (EAR)Dept ECE, SJBITPage 10

DSPAA10EC751Question No.66. Explain implementation of 8- tap FIR filter, (i) pipelined using MAC units and (ii) parallelusing two MAC units. Draw block diagrams.(MAY-JUNE 10)(8marks)Question No.77. With a neat block diagram explain ALU of DSP system. (Dec.11, 6m)Solution:Arithmetic and Logic UnitA typical DSP device should be capable of handling arithmetic instructions like ADD,SUB, INC, DEC etc and logical operations like AND, OR , NOT, XOR etc. The blockdiagram of a typical ALU for a DSP is as shown in the figure 2.8.It consists of status flag register, register file and multiplexers.Fig Arithmetic Logic Unit of a DSPDept ECE, SJBITPage 11

DSPAA10EC751Status FlagsALU includes circuitry to generate status flags after arithmetic and logic operations.These flags include sign, zero, carry and overflow.Overflow ManagementDepending on the status of overflow and sign flags, the saturation logic can be used tolimit the accumulator content.Register FileInstead of moving data in and out of the memory during the operation, for better speed, alarge set of general purpose registers are provided to store the intermediate results.Question No.88.Give the structure of a 4X4 Braun multiplier, Explain its concept. Whatmodification is required to carry out multiplication of signed numbers? Comment on thespeed of the multiplier.(Dec.12, 10m)Solution:Parallel MultipliersConsider the multiplication of two unsigned numbers A and B. Let A be representedusing m bits as (Am-1 Am-2 . A1 A0) and B be represented using n bits as (Bn-1Bn-2 . B1 B0). Then the product of these two numbers is given by,This operation can be implemented paralleling using Braun multiplier whose hardwarestructure is as shown in the figure 2.1.Dept ECE, SJBITPage 12

DSPAA10EC751Fig 2.1 Braun Multiplier for a 4X4 MultiplicationQuestion No.99. Explain guard bits in a MAC unit of DSP. Consider a MAC unit whose inputs are 24-bitnumbers. How many guard bits should be provided if 512 products have to be added inthe accumulator to prevent overflow condition? What is the overall size of theaccumulator required? (Dec.12, 10m)Solution:Multiply and Accumulate UnitMost of the DSP applications require the computation of the sum of the products of aseries of successive multiplications. In order to implement such functions a special unitcalled a multiply and Accumulate (MAC) unit is required. A MAC consists of amultiplier and a special register called Accumulator. MACs are used to implement thefunctions of the type A BC. A typical MAC unit is as shown in the figure 2.5.Dept ECE, SJBITPage 13

DSPAA10EC751Fig 2.5 A MAC UnitAlthough addition and multiplication are two different operations, they can be performedin parallel. By the time the multiplier is computing the product, accumulator canaccumulate the product of the previous multiplications. Thus if N products are to beaccumulated, N-1 multiplications can overlap with N-1 additions. During the very firstmultiplication, accumulator will be idle and during the last accumulation, multiplier willbe idle. Thus N 1 clock cycles are required to compute the sum of N products.Question No.1010. Explain circular buffer addressing mode ii) Parallelism iii) Guard bits. (Dec.11, 9m)Solution: (July.11, 9m).i) Circular Addressing ModeWhile processing the data samples coming continuously in a sequential manner, circularbuffers are used. In a circular buffer the data samples are stored sequentially from the initiallocation till the buffer gets filled up. Once the buffer gets filled up, the next data samples will getstored once again from the initial location. This process can go forever as long as the datasamples are processed in a rate faster than the incoming data rate.Circular Addressing mode requires three registers viza. Pointer register to hold the current location (PNTR)b. Start Address Register to hold the starting address of the buffer (SAR)c. End Address Register to hold the ending address of the buffer (EAR)There are four special cases in this addressing mode. They area. SAR EAR & updated PNTR EARb. SAR EAR & updated PNTR SARDept ECE, SJBITPage 14

DSPAA10EC751c. SAR EAR & updated PNTR SARd. SAR EAR & updated PNTR EARThe buffer length in the first two case will be (EAR-SAR 1) whereas for the next tow cases(SAR-EAR 1)iii) Guard bitsAs the normalization process does not yield accurate result, it is not desirable for someapplications. In such cases we have another alternative by providing additional bits called guardbits in the accumulator so that there will not be any overflow error. Here the add/subtract unitalso has to be modified appropriately to manage the additional bits of the accumulator.Question No.1111. Consider a MAC unit whose inputs are 16 bit numbers. If 256 products are to besummed up in this MAC, how many guard bits should be provided for theaccumulator to prevent overflow condition from occurring? (Dec 08, 9m).Solution:As it is required to calculate the sum of 256, 16 bit numbers, the sum can be aslong as (16 log2 256) 24 bits. Hence the accumulator should be capable of handlingthese 22 bits. Thus the guard bits required will be (24-16) 8 bits.The block diagram of the modified MAC after considering the guard or extension bits is asshown in the figureQuestion No.1212. What are the memory addresses of the operands in each of the following cases of indirectaddressing modes? In each case, what will be the content of the addreg after the memoryaccess? Assume that the initial contents of the addreg and the offsetreg are 0200h and0010h, respectively. (July.09, 10m).a. ADD *addregb.ADD *addregc. ADD offsetreg ,*addregDept ECE, SJBITPage 15

DSPAA10EC751d. ADD *addreg,offsetregSolution:Question No.1313. A DSP has a circular buffer with the start and the end addresses as 0200h and 020Fhrespectively. What would be the new values of the address pointer of the buffer if, in thecourse of address computation, it gets updated toa. 0212hb. 01FCh(July.09, 10m).Buffer Length (EAR-SAR 1) 020F-0200 1 10ha. New Address Pointer Updated Pointer-buffer length 0212-10 0202hb. New Address Pointer Updated Pointer buffer length 01FC 10 020ChDept ECE, SJBITPage 16

DSPAA10EC751UNIT - 3PROGRAMMABLE DIGITAL SIGNAL PROCESSORSQuestion No.11. Describe the multiplier/adder unit of TMS320c54xx processor with a neat block diagram.Solution:The multiplier/adder unit of TMS320C54xx devices performs 17 x 17 2’s complementmultiplication with a 40-bit addition effectively in a single instruction cycle. In addition to themultiplier and adder, the unit consists of control logic for integer and fractional computationsand a 16-bit temporary storage register, T.The compare, select, and store unit (CSSU) is a hardware unit specifically incorporated toaccelerate the add/compare/select operation. This operation is essential to implement the Viterbialgorithm used in many signal-processing applications.The exponent encoder unit supports the EXP instructions, which stores in the T register thenumber of leading redundant bits of the accumulator content. This information is useful whileshifting the accumulator content for the purpose of scaling.Question No.22. Describe any four data addressing modes of TMS320c54xx processorSolution:1. Immediate addressing:The instruction contains the specific value of the operand. The operand can be short (3,5,8 or 9bit in length) or long (16 bits in length). The instruction syntax for short operands occupies onememory location,Example: LD #20, DP.RPT #0FFFFh.2. Absolute Addressing:The instruction contains a specified address in the operand.i) Dmad addressing.MVDK Smem,dmadMVDM dmad,MMRii) Pmad addressing.MVDP Smem,pmadMVPD pmem,Smadiii) PA addressing.PORTR PA, Smem,iv)*(lk) addressing .Example:MVKP 1000h, *AR5 ; 1000 H *AR5 (dmad addressing)MVPD 1000h, *AR7 ; 1000h *AR7 (pmad addressing)PORTR 05h, *AR3 ; 05h *AR3 (PA addressing)LD *(1000h), A ; *(1000h) A (*(lk) addressing)3. Accumulator Addressing:Dept ECE, SJBITPage 17

DSPAA10EC751Accumulator content is used as address to transfer data between Program and Datamemory.Ex: READA *AR24. Direct Addressing:Base address 7 bits of value contained in instruction 16 bit address. A page of 128 locationscan be accessed without change in DP or SP. Compiler mode bit (CPL) in ST1 register is used.If CPL 0 selects DPCPL 1 selects SP,It should be remembered that when SP is used instead of DP, the effective address is computedby adding the 7-bit offset to SP.Question No.33. assume that the current content of AR3 is 400h, what will be its contents after each of thefollowing. Assume that the content of AR0 is 40h.Solution:i) *AR3 0: AR3 AR3 AR0; AR3 400h 40 440hii) *AR3 :AR3 401hiii) *AR3 0b: AR3 AR3 AR0 440h (reverse carry propagation).Question No.44. Compareprocessors.architectural features of TMS320C25 and DSP6000 fixed point digital(6marks)signalSolution:Dept ECE, SJBITPage 18

DSPAA10EC751Question No.55. Write an explanatory note on direct addressing mode of TMS320C54XX processors.Giveexample. (MAY-JUNE 10)(6marks)Solution:Direct Addressing:Base address 7 bits of value contained in instruction 16 bit address. A page of 128locations can be accessed without change in DP or SP.Compiler mode bit (CPL) in ST1register is used.If CPL 0 selects DPCPL 1 selects SP, when SP is used instead of DP, the effective address is computed byadding the 7-bit offset to SP.Dept ECE, SJBITPage 19

DSPAA10EC751Block diagram of the direct addressing mode for TMS320C54xx ProcessorsEx:- when CPL 0, to add the contents of the memory location 0 on page 4 in the datamemory to accumulator B, the instruction sequence is :LD #4, DPADD#0, BThe contents of the first locations on data page 4 are added to accumulator B.Question No.66. With a block diagram explain the indirect addressing mode of TMS320C54XX processorusing dual data memory operand. (June.12, 6m)Solution:o Data space is accessed by address present in an auxili

UNIVERSITY QUESTION PAPER SOLUTION UNIT - 1 INTRODUCTION TO DIGITAL SIGNAL PROCESSING Question No. 1 1. Explain the decimation and interpolation process with an example. (6marks) (DEC’09-JAN’10) (Dec.12, 6m). Solution: Decimation and Interpolation are two techniques used to alter the sampling rate of a sequence.

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