Digital Electronic MCQ Set 1
1. DRAM is fabricated by using IC
a) 2114
b) 7489
c) 4116
d) 2776
Answer
Answer: c [Reason:] DRAM is fabricated by using IC 4116.
2. IC 4116 is of ______ storage memory.
a) 16 KB
b) 32 KB
c) 64 MB
d) 2 KB
Answer
Answer: a [Reason:] IC 4116 is of 16 KB storage memory.
3. How many supply voltage IC 4116 requires to operate the IC unit?
a) 3
b) 2
c) 1
d) 4
Answer
Answer: a [Reason:] IC 4116 requires three supply voltages (+5V, -5V, and +12V) to operate the IC unit.
4. The full form of PSRAM is
a) Plugged Static RAM
b) Plugged Stored RAM
c) Pseudo Stored RAM
d) Pseudo Static RAM
Answer
Answer: d [Reason:] The full form of PSRAM is Pseudo Static RAM.
5. Pseudo static RAM is a
a) Static RAM
b) Dynamic RAM
c) Cache
d) None of the Mentioned
Answer
Answer: b [Reason:] Since, it has built in refresh logic. So, it is a dynamic RAM.
6. When PSRAM is performing internal refresh,
a) The read operation is performed
b) The write operation is performed
c) It can not be accessed for read or write
d) The voltage goes HIGH
Answer
Answer: c [Reason:] Basically, it is a dynamic RAM. So, it can not be accessed for read or write during refresh operation.
7. RAMs are utilized in the computer as
a) Scratch-pad
b) Buffer
c) Main memory
d) All of the Mentioned
Answer
Answer: d [Reason:] RAMs are utilized in the computer as scratch-pad, buffer and main memories. These are the applications of RAMs.
8. The advantages of RAMs are
a) Non destructive read out
b) Fast operating speed
c) Low power dissipation
d) All of the Mentioned
Answer
Answer: d [Reason:] The above mentioned factors are the advantages of a RAM.
9. Which one is more economical?
a) ROM
b) RAM
c) EROM
d) PROM
Answer
Answer: b [Reason:] RAM is more economical than RAM because MOS memories are more economical than the magnetic core for small and medium sized systems.
10. Which one is self compatible?
a) ROM
b) RAM
c) EROM
d) PROM
Answer
Answer: b [Reason:] As semiconductor memories enjoy common interface and technology between sensing and decoding circuitry and the storage element itself, so RAMs are self compatible.
11. The memory which is used for storing programs and data currently being processed by the CPU is called
a) PROM
b) Main Memory
c) Non-volatile memory
d) Mass memory
Answer
Answer: a [Reason:] PROM has capability to store the data due to presence of MOSFET which is processed by the CPU.
12. CD-ROM is a
a) Memory register
b) Magnetic memory
c) Semiconductor memory
d) Non-volatile memory
Answer
Answer: d [Reason:] CD-ROM is a non-volatile memory. Once a program is uploaded in it then it can’t be erasable.
13. A place which is used as storage location in a computer
a) A bit
b) A record
c) An address
d) A byte
Answer
Answer: c [Reason:] A storage location of a computer is an address/memory location.
14. Which of the following is not a primary storage device?
a) Optical disk
b) Magnetic tape
c) Magnetic disk
d) None of the Mentioned
Answer
Answer: d [Reason:] The primary storage device is RAM (i.e. Random Access Memory).
Digital Electronic MCQ Set 2
1. The full form of ROM is
a) Read Outside Memory
b) Read Out Memory
c) Read Only Memory
d) None of the Mentioned
Answer
Answer: c [Reason:] The full form of ROM is Read Only Memory.
2. ROM consist of
a) NOR and OR arrays
b) NAND and NOR arrays
c) NAND and OR arrays
d) NOR and AND arrays
Answer
Answer: c [Reason:] ROM consist of NAND and OR arrays which can be programmed bythe user to implement combinational & sequential functions.
3. For reprogrammability, PLDs use
a) PROM
b) EPROM
c) CDROM
d) PLA
Answer
Answer: b [Reason:] For reprogrammability, PLDs use EPROM (i.e. Erasable PROM). It erases the previous program and starts uploading a new one.
4. The full form of PROM is
a) Previous Read Only Memory
b) Programmable Read Out Memory
c) Programmable Read Only Memory
d) None of the Mentioned
Answer
Answer: c [Reason:] The full form of PROM is Programmable Read Only Memory.
5. The full form of EPROM is
a) Easy Programmable Read Only Memory
b) Erasable Programmable Read Only Memory
c) Eradicate Programmable Read Only Memory
d) None of the Mentioned
Answer
Answer: b [Reason:] The full form of EPROM is Erasable Programmable Read Only Memory.
6. PLDs with programmable AND and fixed OR arrays are called
a) PAL
b) PLA
c) APL
d) PPL
Answer
Answer: a [Reason:] PLDs with programmable AND and fixed OR arrays are called PAL (i.e. Programmable Array Logic).
7. When both the AND and OR are programmable, such PLDs are known as
a) PAL
b) PPL
c) PLA
d) APL
Answer
Answer: c [Reason:] When both the AND and OR are programmable, such PLDs are known as PLA (i.e. Programmable Logic Array).
8. ASIC stands for
a) Application Special Integrated Circuits
b) Applied Special Integrated Circuits
c) Application Specific Integrated Circuits
d) Applied Specific Integrated Circuits
Answer
Answer: c [Reason:] In digital electronics, ASIC stands for Application Specific Integrated Circuits.
9. The programmability and high density of PLDs make them useful in the design of
a) ISAC
b) ASIC
c) SACC
d) CISF
Answer
Answer: b [Reason:] The programmability and high density of PLDs make them useful in the design of ASIC (i.e. Application Specific Integrated Circuits) where design changes can be more rapidly and inexpensively.
10. FPGA stands for
a) Full Programmable Gate Array
b) Full Programmable Genuine Array
c) First Programmable Gate Array
d) Field Programmable Gate Array
Answer
Answer: d [Reason:] In digital electronics, FPGA stands for Field Programmable Gate Array.
11. Which of the following is a reprogrammable gate array?
a) EPROM
b) FPGA
c) Both EPROM and FPGA
d) None of the Mentioned
Answer
Answer: c [Reason:] Both FPGA and EPROM are reprogrammable gate array.
12. The difference between FPGA and PLD is that
a) FPGA is slower than PLD
b) FPGA has high power dissipation
c) FPGA incorporates logic blocks
d) All of the Mentioned
Answer
Answer: c [Reason:] The differences between FPGA and PLD is that FPGA incorporates logic blocks instead of fixed AND-OR gates and is faster with low power dissipation.
Digital Electronic MCQ Set 3
1. The MOS technology based semiconductor ROMs are classified into _____ categories.
a) 2
b) 3
c) 4
d) 5
Answer
Answer: b [Reason:] The MOS technology based semiconductor ROMs are classified into three categories: Mask ROM, PROM,& EPROM.
2. MOS ROM is constructed using
a) FETs
b) Transistors
c) MOSFETs
d) BJTs
Answer
Answer: c [Reason:] MOS ROM is made up of MOSFETs.
3. The full form of EEPROM is
a) Erasable Electrically Programmable ROMs
b) Electrically Erasable Programmable ROMs
c) Electrically Erasable Programming ROMs
d) None of the Mentioned
Answer
Answer: b [Reason:] The full form of EEPROM is Electrically Erasable Programmable ROMs.
4. Which of the following best describes EPROMs?
a) EPROMs can be programmed only once
b) EPROMs can be erased by UV
c) EPROMs can be erased by shorting all inputs to the ground
d) All of the Mentioned
Answer
Answer: b [Reason:] EPROMs can be erased by UV (i.e. ultraviolet light).
5. The Width of a processor’s data path is measured in bits. Which of the following are common data paths?
a) 8 bits
b) 12 bits
c) 16 bits
d) 32 bits
Answer
Answer: a [Reason:] In generalised form, the data paths are of 8 bits.
6. What type of memory is not directly addressable by the CPU and requires special software called EMS (expanded memory specification)?
a) Extended
b) Expanded
c) Base
d) Conventional
Answer
Answer: b [Reason:] Expanded memory is not directly addressable by the CPU.
7. Which bus is used for input and output in case of microprocessor operation?
a) Address bus
b) System bus
c) Control bus
d) None of Mentioned
Answer
Answer: c [Reason:] The input and output are used to control the function of a microprocessor. Hence, the control bus is used to transfer the input and output signal from microprocessor to external peripherals and or from external peripherals to microprocessor.
8. What is the major difference between DRAM and SRAM?
a) Dynamic RAMs are always active; static RAMs must reset between data read/write cycles
b) SRAMs can hold data via a static charge, even with power off
c) The only difference is the terminal from which the data is removed—from the FET Drain or Source
d) DRAMs must be periodically refreshed
Answer
Answer: d [Reason:] DRAMs must be periodically refreshed so that it can store the new information.
9. Which of the following is not a part of Hard disk?
a) Platter
b) Read/Write
c) Valve
d) Spindle
Answer
Answer: c [Reason:] A valve is a device that regulates, directs or controls the flow of a fluid (gases, liquids, fluidized solids, or slurries) by opening, closing, or partially obstructing various passageways. So, it is not a part of hard disk.
10. Which ROM can be erased by an electrical signal?
a) ROM
b) Mask ROM
c) EPROM
d) EEPROM
Answer
Answer: d [Reason:] EEPROMs (Electrically Erased Programmable ROM) can be erased by an electrical signal.
11. In the floppy drive, data is written to and read from the disk via a magnetic _____ head mechanism.
a) Cluster
b) Read/Write
c) Cylinder
d) Recordable
Answer
Answer: b [Reason:] In the floppy drive, data is written to and read from the disk via a magnetic read/write head mechanism.
12. What does the term “random access” mean in terms of memory?
a) Any address can be accessed in systematic order
b) Any address can be accessed in any order
c) Addresses must be accessed in a specific order
d) None of the Mentioned
Answer
Answer: b [Reason:] “Random access” mean which can be accessed randomly and in other words any address can be accessed in any order.
13. Which type of ROM has to be custom built by the factory?
a) EEPROM
b) Mask ROM
c) EPROM
d) PROM
Answer
Answer: b [Reason:] All types of ROM are programmable and can be programmed as per requirement but the mask ROM is always programmed for specific application and it can’t be reprogrammed.
14. The computer’s main memory is
a) Hard drive and RAM
b) CD-ROM and hard drive
c) RAM and ROM
d) CMOS and hard drive
Answer
Answer: c [Reason:] The computer’s main memory is RAM and ROM because all the storage related operation are performed by the data present in RAM/ROM.
15. A major disadvantage of the mask ROM is that
a) It is time consuming to change the stored data when system requirements change
b) It is very expensive to change the stored data when system requirements change
c) It cannot be reprogrammed if stored data needs to be changed
d) It has an extremely short life expectancy and requires frequent replacement
Answer
Answer: c [Reason:] A major disadvantage of the mask ROM is that it cannot be reprogrammed if stored data needs to be changed.
Digital Electronic MCQ Set 4
1. Which of the following expression is used to denote spatial domain process?
a) g(x,y)=T[f(x,y)]
b) f(x+y)=T[g(x+y)]
c) g(xy)=T[f(xy)]
d) g(x-y)=T[f(x-y)]
Answer
Answer: a [Reason:] Spatial domain processes will be denoted by the expression g(x,y)=T[f(x,y)], where f(x,y) is the input image, g(x,y) is the processed image, and T is an operator on f, defined over some neighborhood of (x, y). In addition, T can operate on a set of input images, such as performing the pixel-by-pixel sum of K images for noise reduction.
2. Which of the following shows three basic types of functions used frequently for image enhancement?
a) Linear, logarithmic and inverse law
b) Power law, logarithmic and inverse law
c) Linear, logarithmic and power law
d) Linear, exponential and inverse law
Answer
Answer: b [Reason:] In introduction to gray-level transformations, which shows three basic types of functions used frequently for image enhancement: linear (negative and identity transformations), logarithmic (log and inverse-log transformations), and power-law (nth power and nth root transformations).The identity function is the trivial case in which output intensities are identical to input intensities. It is included in the graph only for completeness.
3. Which expression is obtained by performing the negative transformation on the negative of an image with gray levels in the range[0,L-1] ?
a) s=L+1-r
b) s=L+1+r
c) s=L-1-r
d) s=L-1+r
Answer
Answer: c [Reason:] The negative of an image with gray levels in the range[0,L-1] is obtained by using the negative transformation, which is given by the expression: s=L-1-r.
4. What is the general form of representation of log transformation?
a) s=clog10(1/r)
b) s=clog10(1+r)
c) s=clog10(1*r)
d) s=clog10(1-r)
Answer
Answer: b [Reason:] The general form of the log transformation: s=clog10(1+r), where c is a constant, and it is assumed that r ≥ 0.
5. What is the general form of representation of power transformation?
a) s=crγ
b) c=srγ
c) s=rc
d) s=rcγ
Answer
Answer: a [Reason:] Power-law transformations have the basic form: s=crγ where c and g are positive constants. Sometimes s=crγ is written as s=c.(r+ε)γ to account for an offset (that is, a measurable output when the input is zero).
6. What is the name of process used to correct the power-law response phenomena?
a) Beta correction
b) Alpha correction
c) Gamma correction
d) Pie correction
Answer
Answer: c [Reason:] A variety of devices used for image capture, printing, and display respond according to a power law. By convention, the exponent in the power-law equation is referred to as gamma .The process used to correct these power-law response phenomena is called gamma correction.
7. Which of the following transformation function requires much information to be specified at the time of input?
a) Log transformation
b) Power transformation
c) Piece-wise transformation
d) Linear transformation
Answer
Answer: c [Reason:] The practical implementation of some important transformations can be formulated only as piecewise functions. The principal disadvantage of piecewise functions is that their specification requires considerably more user input.
8. In contrast stretching, if r1=s1 and r2=s2 then which of the following is true?
a) The transformation is not a linear function that produces no changes in gray levels
b) The transformation is a linear function that produces no changes in gray levels
c) The transformation is a linear function that produces changes in gray levels
d) The transformation is not a linear function that produces changes in gray levels
Answer
Answer: b [Reason:] The locations of points (r1,s1) and (r2,s2) control the shape of the transformation function. If r1=s1 and r2=s2 then the transformation is a linear function that produces no changes in gray levels.
9. In contrast stretching, if r1=r2, s1=0 and s2=L-1 then which of the following is true?
a) The transformation becomes a thresholding function that creates an octal image
b) The transformation becomes a override function that creates an octal image
c) The transformation becomes a thresholding function that creates a binary image
d) The transformation becomes a thresholding function that do not create an octal image
Answer
Answer: c [Reason:] If r1=r2, s1=0 and s2=L-1,the transformation becomes a thresholding function that creates a binary image.
10. In contrast stretching, if r1≤r2 and s1≤s2 then which of the following is true?
a) The transformation function is double valued and exponentially increasing
b) The transformation function is double valued and monotonically increasing
c) The transformation function is single valued and exponentially increasing
d) The transformation function is single valued and monotonically increasing
Answer
Answer: d [Reason:] The locations of points (r1,s1) and (r2,s2) control the shape of the transformation function. If r1≤r2 and s1≤s2 then the function is single valued and monotonically increasing.
11. In which type of slicing, highlighting a specific range of gray levels in an image often is desired?
a) Gray-level slicing
b) Bit-plane slicing
c) Contrast stretching
d) Byte-level slicing
Answer
Answer: a [Reason:] Highlighting a specific range of gray levels in an image often is desired in gray-level slicing. Applications include enhancing features such as masses of water in satellite imagery and enhancing flaws in X-ray images.
12. Which of the following depicts the main functionality of the Bit-plane slicing?
a) Highlighting a specific range of gray levels in an image
b) Highlighting the contribution made to total image appearance by specific bits
c) Highlighting the contribution made to total image appearance by specific byte
d) Highlighting the contribution made to total image appearance by specific pixels
Answer
Answer: b [Reason:] Instead of highlighting gray-level ranges, highlighting the contribution made to total image appearance by specific bits might be desired. Suppose , each pixel in an image is represented by 8 bits. Imagine that the image is composed of eight 1-bit planes, ranging from bit-plane 0 for the least significant bit to bit-plane 7 for the most significant bit. In terms of 8-bit bytes, plane 0 contains all the lowest order bits in the bytes comprising the pixels in the image and plane 7 contains all the high-order bits.
Digital Electronic MCQ Set 5
1. The histogram of a digital image with gray levels in the range [0, L-1] is represented by a discrete function:
a) h(r_k)=n_k
b) h(r_k )=n/n_k
c) p(r_k )=n_k
d) h(r_k )=n_k/n
Answer
Answer: a [Reason:] The histogram of a digital image with gray levels in the range [0, L-1] is a discrete function h(rk )=nk, where rk is the kth gray level and nkis the number of pixels in the image having gray level rk.
2. How is the expression represented for the normalized histogram?
a) p(r_k )=n_k
b) p(r_k )=n_k/n
c) p(r_k)=nn_k
d) p(r_k )=n/n_k
Answer
Answer: b [Reason:] It is common practice to normalize a histogram by dividing each of its values by the total number of pixels in the image, denoted by n. Thus, a normalized histogram is given by p(rk )=nk/n, for k=0,1,2…..L-1. Loosely speaking, p(rk ) gives an estimate of the probability of occurrence of gray-level rk. Note that the sum of all components of a normalized histogram is equal to 1.
3. Which of the following conditions does the T(r) must satisfy?
a) T(r) is double-valued and monotonically decreasing in the interval 0≤r≤1; and
0≤T(r)≤1 for 0≤r≤1
b) T(r) is double-valued and monotonically increasing in the interval 0≤r≤1; and
0≤T(r)≤1 for 0≤r≤1
c) T(r) is single-valued and monotonically decreasing in the interval 0≤r≤1; and
0≤T(r)≤1 for 0≤r≤1
d) T(r) is single-valued and monotonically increasing in the interval 0≤r≤1; and
0≤T(r)≤1 for 0≤r≤1
Answer
Answer: d [Reason:] For any r satisfying the aforementioned conditions, we focus attention on transformations of the form
s=T(r) For 0≤r≤1
That produces a level s for every pixel value r in the original image.
For reasons that will become obvious shortly, we assume that the transformation function T(r) satisfies the following conditions:
T(r) is single-valued and monotonically increasing in the interval 0≤r≤1; and
0≤T(r)≤1 for 0≤r≤1.
4. The inverse transformation from s back to r is denoted as:
a) s=T-1(r) for 0≤s≤1
b) r=T-1(s) for 0≤r≤1
c) r=T-1(s) for 0≤s≤1
d) r=T-1(s) for 0≥s≥1
Answer
Answer: c [Reason:] The inverse transformation from s back to r is denoted by:
r=T-1(s) for 0≤s≤1.
5. The probability density function p_s (s) of the transformed variable s can be obtained by using which of the following formula?
a) p_s (s)=p_r (r)|dr/ds|
b) p_s (s)=p_r (r)|ds/dr|
c) p_r (r)=p_s (s)|dr/ds|
d) p_s (s)=p_r (r)|dr/dr|
Answer
Answer: a [Reason:] The probability density function p_s (s) of the transformed variable s can be obtained using a basic formula: p_s (s)=p_r (r)|dr/ds|
Thus, the probability density function of the transformed variable, s, is determined by the gray-level PDF of the input image and by the chosen transformation function.
6. A transformation function of particular importance in image processing is represented in which of the following form?
a) s=T(r)=∫0 (2r)pr (ω)dω
b) s=T(r)=∫0 (r-1)pr (ω)dω
c) s=T(r)=∫0 (r/2)pr (ω)dω
d) s=T(r)=∫0 pr (ω)dω
Answer
Answer: d [Reason:] A transformation function of particular importance in image processing has the form: s=T(r)=∫0 r pr(ω)dw, where ω is a dummy variable of integration. The right side of is recognized as the cumulative distribution function (CDF) of random variable r.
7. Histogram equalization or Histogram linearization is represented by of the following equation:
a) sk =∑k j =1 nj/n k=0,1,2,……,L-1
b) sk =∑k j =0 nj/n k=0,1,2,……,L-1
c) sk =∑k j =0 n/nj k=0,1,2,……,L-1
d) sk =∑k j =n nj/n k=0,1,2,……,L-1
Answer
Answer: b [Reason:] A plot of pk_ (rk) versus r_k is called a histogram .The transformation (mapping) given in sk =∑k j =0)k nj/n k=0,1,2,……,L-1 is called histogram equalization or histogram linearization.
8. What is the method that is used to generate a processed image that have a specified histogram?
a) Histogram linearization
b) Histogram equalization
c) Histogram matching
d) Histogram processing
Answer
Answer: c [Reason:] In particular, it is useful sometimes to be able to specify the shape of the histogram that we wish the processed image to have. The method used to generate a processed image that has a specified histogram is called histogram matching or histogram specification.
9. Histograms are the basis for numerous spatial domain processing techniques.
a) True
b) False
Answer
Answer: a [Reason:] Histograms are the basis for numerous spatial domain processing techniques. Histogram manipulation can be used effectively for image enhancement.
10. In a dark image, the components of histogram are concentrated on which side of the grey scale?
a) High
b) Medium
c) Low
d) Evenly distributed
Answer
Answer: c [Reason:] We know that in the dark image, the components of histogram are concentrated mostly on the low i.e., dark side of the grey scale. Similarly, the components of histogram of the bright image are biased towards the high side of the grey scale.
