Hemodialysis Machine Block Diagram

Hemodialysis Machine Working: 

The aim of hemodialysis machine is to replace the vital kidney functions. But in most cases this ‘artificial kidney’ cannot replace the vital kidney functions. But in most cases this ‘artificial kidney’ cannot replace the total kidney operation. The hemodialysis machine propels the patient’s blood and dialysate through a dialyzer. The newest dialysis machines on the marketplace are extremely automated and constantly check different parameters including blood flow rate, dialysate flow rate, dialysis solution conductivity, blood pH, temperature etc. Any evaluation that is out of usual range, triggers an easy to hear alarm to aware the patient care technician. An extensive water filtration system is totally fundamental for hemodialysis machine. If the water utilized amid dialysis method not carefully filtered, mineral contaminations or bacterial endotoxins can go into the patient's body to cause numerous health problems. Hence water utilized as a part of hemodialysis machine is precisely purified before utilize. At first it is filtered and temperature balanced and its pH is rectified by including an acid or base. Primary purification is finished by forcing water through a penetrable with very little pores. This membrane is called reverse osmosis membrane. This gives the water pass through and holds back even very small solutes.  Once filtered water is blended with dialysate concentrate, its conductivity would increases. Amid dialysis the conductivity of dialysate solution is consistently checked to clear that the water and the dialysate concentrate are blended in proper proportions. The block diagram representation of a hemodialysis machine is shown in figure.

Hemodialysis Machine Block Diagram:

The blood is occupied from the artery of the patient and mixed with an anticoagulant such as heparin and forced into the equipment called artificial kidney or hemodialysis machine.


1. Dialysate Pail:
Dialysate pail is the storage place of the dialysate. It is through the dialysate that the blood from the patient is directed to flow through channels or tubes bounded by cellophane membrane. This membrane is permeable to small solutes and impermeable to macromolecules.

2. Dialyzer:
Dialyzer is the major part of equipment that actually filters the blood. Dialyzer membranes have dissimilar pore sizes. Those with lesser pore size are called low flux membrane and those with bigger pore size are called high flux membranes. Dialyzer membranes used in made of cellulose. Another group of membrane is made from synthetic objects using polymers. Nanotechnology is used in some high flux membranes to produce uniform pore size. Dialyzers come in numerous dissimilar sizes. A bigger dialyzer with a bigger membrane can remove more solutes.

3. Dialysate Heater:
The dialysate should be kept at a proper temperature. The temperature is controlled by the dialysate heater.

4. Check Valve:
The function of dialysate check valve is to control the flow of dialysate to the dialysate pail. If the overflow condition of dialysate occurs, it will be directed by the check valve to the dialysate overflow and drain mixer where the dialysate will be mixed with the drained dialysate. The mixed component is thrown out as exhaust.

5. Dialysate Recirculating Loop:
This loop acts in such a way that the drained dialysate is reconverted into fresh dialysate and properly heated and redirected to the dialysate pail itself. The dialysate may moreover be discarded after each action or be used again. Reused dialyzers are not shared between patients. If the reuse of dialyzers is finished cautiously and properly it will produce similar outcomes to a single use of dialyzers.

6. Dialysate Holding Tank:
The dialysate holding tank holds the fresh dialysate and whenever the dialysate of the pail has to be changed, the fresh dialysate from the holding tank is taken to the pail.

7. Flow Meter:
In order to regulate the flow of fresh dialysate from the dialysate holding tank to the dialysate pail, we normally use a flow meter in the machine set up.

Renal System Physiology

Renal System Physiology: The renal system is composed of kidneys. Kidney’s perform the functions such as excretion of waste products from metabolic processes, regulation of constancy of blood fluids, constant blood pH and blood pressure maintenance, regulation of various blood chemicals etc.

Kidney Failure:

Renal disease is classified as primary diseases and secondary diseases depending on the fact that whether the affected part is internal to kidneys or external to kidneys. Renal or kidney failure results in little or no urine formation. Due to this toxic substances such as urea accumulate in body causing symptoms such as head ache, dizziness etc. In critical cases it may also lead to death. If the kidney is not functioning enough, kidney stones will be formed due to calcium deposits and it will cause kidney blockage. When kidney failure occurs, kidney transplantation may be undertaken. But it is much expensive method. So the peritoneal dialysis or hemodialysis is recommended, than kidney transplantation.

Dialysis is broadly classified into two types, peritoneal dialysis and hemodialysis

1. Peritoneal dialysis:

This is a type of dialysis which is performed under emergency condition. This type of dialysis is done by puncturing two needles through the abdominal (peritoneum) and washing out the peritoneal cavity with a saline solution. Hence it is called peritoneal dialysis. The saline solution flows through a semi-permeable membrane outside the body and hence toxic substances are thus removed.

2. Hemodialysis:

Hemodialysis is a process for take away the waste stuff such as potassium and urea from the blood when kidneys are in failure. The chemical substances are removed from the blood by passing it through tubes surrounded by semi-permeable membrane. It is done by puncturing two needles through artery and vein and circulating the patient’s blood through a coiled plastic tube. The coiled tube is immersed in a dialysate solution (solution containing salts of Ca, Mg, K, Na etc). The dialysed solution utilized is a pure solution of mineral ions. Sodium bicarbonate is mixed in a more concentration compared to plasma to exact blood acidity. A little quantity of glucose is too added to the dialysate. As told, hemodialysis contains diffusion of solutes a semi-permeable membrane. It uses counter current pass where the dialysate is passing in the opposite direction to blood flow. Fluid removal is attained by varying the hydrostatic pressure of the dialysate section reasoning free water and various melt solutes to go across the membrane along a produced pressure gradient. Toxins such as urea, uric acid gradually pass through plastic tube into dialysate solution. The semi-permeable membrane does not allow blood cells and large protein molecules.

Heart Lung Machine Block Diagram

The main components of a heart-lung machine and their functions are briefly described below.

1. Roller Pump:

The pump comfort frequently consists of numerous revolving motor pumps that peristaltically ‘massage’ tubing. Pumping action occurs because the rollers on the rotating arm comprises the tubings carrying the blood and forces the blood ahead of the compressed section. This action is called peristaltic action and this leads to a pulsatile flow of blood through the tubings. The heart-lung machine uses five pump heads normally.

In some cases we use centrifugal pumps for the safeguarding and control of blood flow during the surgical procedure. By changing the RPM of pump head, blood flow is maintained. This kind of pumping action is measured to be better to the action of roller pumps because it produces less damage to the blood.

2. Oxygenator: 

The oxygenator is planned to move oxygen to the infused blood and take away carbon dioxide from venous blood. So simply the oxygenator assembly allows to oxygenate the blood. Bubble oxygenators, membrane oxygenators and heparin coated blood oxygenators are the different types of oxygenators. Of them, heparin coated blood oxygenators are the commonly used one as they prevent blood coating.

3. Heat Exchanger:

Heat exchangers permit body and organ temperatures to be attuned. The uncomplicated heat exchange design is a container of water. As the blood transfers through the tubing positioned in the bath, the blood temperature will alter. A more complex system separated the blood and water interface with a metallic barrier. As the water temperature is changed, the blood temperature also changes, which in turn changes the tissue temperature. Once the tissue temperature reaches the desired level, the water temperature is maintained.

Heat exchangers can be of two types namely plate type and coil type. In plate type of heat exchangers, two sheets of metals are used and they are coated with a non – thrombogenic substance. This configuration offers good efficient transfer of heat. Coil configuration consists of a hollow coil through which water circulates. The disadvantage of this type is it is a non disposable unit and is very difficult to clean.

4. Temperature Controller:

During the time the patient uses a heart-lung machine, body temperature is to be maintained at a normal range. This unit allows the pump operator to keep the blood at a proper temperature.

5. Blood and Oxygen Dispersion Plates:

In a heart – lung machine, the gas bubbles are dispersed into the venous blood in the bubble column and acts as vehicles for both oxygen and carbon dioxide. Here oxygen diffuses from the bubble into the blood film surrounding the bubble. Various types of dispersion plates are used.

6. Reservoir:

A reservoir gather blood drained from the venous flow. Reservoir arranges include open or closed systems. The open system displays demarcations corresponding to blood volume in the container. The design is open to environment permitting blood to cross with environmental gases. The closed system contains a bag and it eliminates the air-blood interface. Volume is measured by weight or change in radius of the container.

7. Nylon Mesh Filters:

The nylon mesh filters used in heart-lung machines are attuned with a broad choice of solvents. They are particularly used for gathering of algae and cells, atom analysis, big particulate filtration, prefiltration of solvents and so on.

8. Cannulae: 

Numerous cannulae are sewn into the suffering patient’s body in a diversity of position depending on the kind of surgery. A venous cannula takes away the oxygen destitute blood from the patient’s body and an arterial cannula is sewn into the suffering patient’s body to impart oxygen rich blood. The cannula utilized to return oxygenated blood is frequently inserted in the rising aorta.

Applications of Bioinformatics

Bioinformatics are applicable in the following fields.

1. Personalised Medicine
2. Gene therapy
3. Microbial genome applications
4. Biotechnology
5. Forensic study of microbes
6. Insect Resistance
7. Growth of drought resistance varieties
8. Molecular medicine
9. Drug development
10. Climate Change studies
11. Antibiotic resistance
12. Crop development
13. Veterinary Science

Applications of Bioinformatics:

a. Prediction of Protein Structure:
It is a further vital application of bioinformatics. The amino acid series of a protein is called its primary structure. The primary structure of a protein is of vital importance in understanding the function of protein. The feeling of homology is also important in bioinformatics. It means homology can be used to find out the function of a gene. So in bioinformatics, if a sequence and its function are known, we usually assign a similar function for another gene with almost similar sequence.

b. Computational evolutionary biology:
Bioinformatics helps scientists to find out the evolution of huge number of organisms by determining changes in their DNA. The scientists can find out the evolution of different organism by tracing the changes in their DNA. In that sense, it is the field of bioinformatics that contributed substantially for the study of evolutionary biology.

c. Measuring Biodiversity:
Bioinformatic databases can be used to collect size of populations and to study how each organisms interact with other species. By using specialized software programs the collected information can be visualized and analysed for the purpose of analyzing biodiversity.

d. Analysis of mutations in cancer:
As we know, diseases such as cancer affect the cells well. In cancer, the genomes of influenced cells are reorganized in a complex way. Massive sequencing efforts are used to identify the sequence so as to make diagnosis of the diseases easy. Also the various changes that occur in the cells at the onset of such diseases can be studied deeply.

e. Sequence analysis:
Here an assessment of genes within a species or amid dissimilar species can show similarity between protein functions or relation between species. Now a day, computer programs are utilized to look for the genome of thousands of organisms. We can use sequence analysis in usual search for genes. Also we can use different computational tools to search the genomes of a vast number of organisms. By using these programs, we can identify the related and identical sequences. With the help of bioinformatics, automatic search tools for genes and regulatory sequences within a gene can be implemented.

f. Genome annotation:
Annotation is the method of spotting the genes and other biological characteristics in a DNA sequence. Even though the systems for the annotation of genes are almost similar, the programs for the analysis of genomic DNA are changing day by day.

g. High information image analysis:
Biomedical imaging is a core for the research and diagnosis of diseases. So with the help of bioinformatics, computational technology can be used to accelerate the analysis of large amount of high information content biomedical image clusters. Bioinformatics can also be used in clinical image analysis and visualization.

h. Modelling biological systems:
Bioinformatics is applied in modelling biological systems. It means that computer simulation software can be used to study the different systems of body such as cells and their structures. Also artificial models can be used to study the complex systems of various living beings. Virtual evolution is another field where we can understand the evolutionary processes via the computer simulation.

i. Comparative genomics:
The comparison between different genes and their structural and functional details are of very importance for the advancement of evolutionary biology. The main principle of comparative genome study is to find out the connection among genes or other genomic characteristics in different organisms.

j. Crop improvement:
This information enables new methods to learn gene expression patterns in plants. Comparative genomics of the plant genomes has exposed that the organisation of their genes has remained more preserved over growth time.

k. Study of climate change:
Study of genomics of microbes can be used to decrease atmospheric carbon dioxide levels. Increasing levels of carbon dioxide emission are thought to contribute to global climate change. Bioinformatics study has proved that certain bacteria converts sunlight to cellular energy by absorbing atmospheric carbon dioxide and converting it to biomass.

Micro and Macro Shock Hazards

Patient Safety against Electricity: Hospitals are facing the problem of creating a safe electric environment for the care and comfort the patients. The main problems encountered are electric shock, burns and fire hazards, which results from the care less use of electricity. Shock resulting from electric power is a common experience. Electric current can flow through the human body either accidentally or intentionally. Electric currents are used deliberately in the subsequent cases.

a. For the dimension of respiration rate by impedence method
b. For therapeutic and surgical purpose
c. When taping signals like ECG and EEG

Accidentally electric current can pass through the human body, are due to defect of the equipment or due to defect in design, and are due to defect of the equipment or due to defect in design.

Electric Shock Hazards:

Hazards because of electric shock are additionally connected with types of equipments other than that used for healing. There are two circumstances which represent dangers from electric shock. They are gross shock and micro current shock. An electric shock in the person with any cause of voltage is sufficiently high to cause adequate current passing through the skin, muscles or hair. The base current a person can sense is believed to be around 1 mA. The current may harm tissue or fibrillation if it is adequately high. Death caused by an electric shock is alluded to as electrocution. For the most part, currents close to 100 mA are dangerous if they go through vital parts of the body. Micro and Macro Shock Hazards are explained below.

Macro Shock:

In the case of gross shock, the current flows through the body of the subject. In the case of micro current shock, the current passes directly through the heart wall. Gross shock is experienced by the person by an accidental contact with the electrical wiring at any point on the surface of the body. Macro shock hazards are usually caused by electrical wiring failures. This type of hazard is dangerous to the patient as well as to the medical and attending staffs. Voltages of more than 50 Volts connected crosswise over dry unbroken human skin are fit for delivering heart fibrillation if they create electric streams in the body tissues which happen to go through the chest zone. Hazards because of lightning incorporate an immediate strike on people or property.

Micro Shock:

Micro shock is a more serious hazard than macro/gross shock in which the current flows directly to the heart wall and requires only less current to produce ventricular defibrillation.
When current enter the body at one point and leaves some other points, some physiological effects takes place,

a. Electrical stimulation of the excitable tissue (nerves and muscules).
b. Resistive heating of tissue
c. Electro-chemical blazes and tissue harm for direct current and very high voltage.

Type of Current
Current Rate (mA)
Physiological effect
Threshold
1 - 5
Tingling sensation
Pain
5 - 8
Intense or painful sensation
Let go
8 - 20
Threshold of involuntary muscle
Paralysis
> 20
Respiration paralysis and pain
Fibrillation
80 - 100
Ventricular and heart fibrillation
Defibrillation
1000 - 10000
-



EMG Block Diagram Explanation

Electromyography is the technique for calculating and recording the action potential of muscles. EMG is taken using a device called electromyography and the record obtained is known as electromyogram. The electrical activity of muscle cells when they are active and at rest can be analysed using an EMG. The measured EMG potentials range from 50 µ Volt to 30 millivolts.

Mainly there are two kinds of EMG measurements. The first method is using surface electrodes and the second one is using needle electrodes. The Surface EMG electrodes are used to monitor the electrical activity of muscles generally whereas the needle electrodes are used to monitor the electrical activity of muscles generally whereas the needle electrodes are used to observe the electrical activity of only few fibers.

A trained expert can observe the electrical activity of muscles when the needle is inserted. There is a normal electrical activity for the muscle fibers at rest. The physician or concerned expert examines the normal activity of muscles when the needle is inserted. There is a normal electrical activity for the muscle fibers at rest. The physician or concerned expert examines the normal activity of muscles. The abnormal spontaneous activity indicates that some nerve or muscle cells are damaged. At a time potentials from different electrodes are taken. So the needle electrodes have to be placed at different locations to obtain an accurate EMG. So the intramuscular EMG is considered to be too invasive.

So in order to obtain the general activity of the muscle cells we use surface electrodes which need to be placed only on the concerned area. So no insertion is required. This technique is commonly used in many applications such as in a physiotherapy clinic where the muscle activity is monitored by the surface EMG electrodes and the patients can have visual stimulus when they activating the muscles.

So when the motor neuron or muscle fiber is stimulated, the action potential is transmitted across the muscle it is passed to the connected nerve fibers. Actually during EMG we are evaluating this bioelectric potential from different cells. This potential is collectively called motor unit action potential (MUAP). EMG signals are made up of superimposed MUAPs. Hence the shape of the electromyogram is affected by factors such as number of muscle fibers under consideration, the metabolic type of muscle fibers etc.

EMG Block Diagram

EMG Block Diagram Explanation:

1. EMG electrode: As told earlier, the electrode used for EMG recording can be of surface type or needle type depending on the area from which the EMG is to be obtained and the type of measurement. If we need to have EMGs from many individual muscle cells rather than from the surface as a group, needle electrodes are the best choice. But if the general activity of a muscle is to be analysed the surface electrodes can give the accurate values. 

2. Bioelectric amplifier: As the name implies, the bioelectric amplifiers are used to amplify the bioelectric signals obtained from EMG electrodes.

3. AF amplifier: At rest under normal condition the sound does not undergo large variations. But abnormal and spontaneous activity may be distinguished by the sudden change in sound and this can be analyzed by the physician. The abnormal activity usually indicates muscle damage and they can easily find out the nerve or muscle damage. So physicians normally use AF amplifiers during EMG measurement so as to distinguish there sounds clearly.

4. Oscilloscope: The measured EMG can be connected to the oscilloscope to visualize the EMG. The abnormalities in the working of nerves and muscle  cells can be identified by a physician by analyzing the EMG waveform. The EMG can also be stored using special oscilloscopes such as DSO (Digital Storage Oscilloscope) for future analysis.

5. EMG recorder: Unlike ECG, EMG cannot be recorded in a low speed chart paper recorders because of its extreme low frequencies. So it will be less useful. Normally we use the photographic recording of EMG. For this a light sensitive paper is moved over the recording CRT.

Frequency Limitations:

1. The normal EMG is in the range of 60 – 70 Hz. The EMG appear as a random noise signal the shape of the waveform may vary based on the part of the body from which EMG is taken.

2. Due to the low frequency limitation, EMG cannot be recorded on a strip chart recorder because it cannot give a clear idea of the waveform.

Applications of EMG:
1. EMG can be used to diagnose two main categories of diseases. They are neuropathies and myopathies. In EMG representation, an increase in duration of action potential is an indication of neuropathic disease and a decrease in duration of action potential is an indication of myopathic disease.

2. EMG is utilized as a diagnostic instrument for detecting neuromuscular diseases.

3. EMG can be utilized for silent speech detection which identifying the speech by examining the EMG activity of muscles related with speech.


Reduction of Product Of Sums (POS) form using K Map

So far, we have discussed only the sum-of-products expressions. We now discuss the reduction process in the POS form. It may be noted that, because of the dual nature of binary systems, any procedure or rule that is valid in the SOP form is equally valid in the POS form also.
      Just like SOP functions, POS functions can also be represented in a K-map and the elimination of variables can be done as in the case of SOP. However, the K-map designations of cells differ and the entries are 0s instead of 1s. The four-variable K-map off a POS system is shown in Fig. 2.32.
      The cells are numbered from 0 to 15 as in the case of SOP form. However, the cell designations have been changed. In SOP form, the individual cells were designated as product functions like a′b′, a′b, etc.  In the case of POS form, multiplication gets changed to summation For example, the Cell 0 is designated as A+B+C+D, Cell 1 is designated as A+B+C+D’, and so on. In this text, the cells are designated using capital letters to distinguish them from SOP form (not as a rule, but as a convenient arrangement). Further, for the same reason, we have replaced the number designation of 00, 01, etc. with A+B, A+B’, etc. on the top and sides of the map.


Example 9: Simplify P = Π (0, 1, 2, 3, 8, 9, 10, 13)

Solution: The Greek uppercase letter Π represents a POS expression, just as S represents an SOP expression. The entries in the POS cells must be 0s. The cells in which there are no entries must be filled with 1s. Just as in the case of the SOP form, here also we can have don’t-care terms. Reduction process is similar in both the SOP and POS forms. Notice that the POS terms are called maxterms. The K-map for Example 9 is shown in Fig. 2.33.


The groupings are as shown in Fig. 2.33. These are:

                                                                     Group 1:  A+B
                                                            Group 2: B+D
                                                            Group 3: A′+C+D′

The final POS result is
                            P = (A+B)(B+D)(A′+C+D’)           (2.33)

Simplification of POS Expressions with Don’t-Care Terms:

Example 10: Let us now introduce don’t care terms in cells 4, 5, 11, 14, and 15 of Example 9.  The new function is P = Π (0, 1, 2, 3, 8, 9, 10, 13) + d (4, 5, 11, 14, 15). Obtain its reduction.

Solution: The K-map for this is shown in Fig. 2.34. The entries are made as shown in the figure.  We find that there are only two groups, viz., an octet and a quad. The final POS expression is:

                                               P = B (C+D’)              (2.34)

It can be seen alternate groups using don’t-care terms also exist in this case. Then the final expression will be different from that given in Eq. (2.34).

Author Name: B. SOMANATHAN NAIR
Author Website: www.randomelectronicdreams.com

Six Variable K Map Examples

A six-variable K-map is shown in Fig. 2.30, which also shows cell designations. This consists of  a group of 4 four-variable K-maps. It can be seen that the reduction process becomes highly complicated with six variables because locating  similar positions in the individual maps is a very difficult task. Entries are made as per the designations of the cells shown in Fig. 2.30.  It may be noted that the individual charts are designated as 00, 01, 11, and 10, respectively. The cell designations begin from 000000 (corresponding to decimal 0) to 111111 (corresponding to decimal 63) as shown. As in the case of the five-variable K-map, entries in similar cells can be combined to eliminate the involve variable (or variables) and thus simplify the given logic function. Example 8 will illustrate this.


Example 8: Simplify S(12, 13, 14, 15, 28, 29, 30, 31, 44, 45, 46, 47, 48, 49, 50, 51, 60, 61, 62, 63).

Solution: Figure 2.31 shows the entries for Example 8.
The encircled five quads can be combined to form into two main groups. Group 1 consists of  four quads whose individual members are (12, 13, 14, 15), (28, 29, 39, 31), (44, 45, 46, 47), and (60, 61, 62, 63), respectively. Group 2 consists of just one quad whose members are (48, 49, 50, 51). The final solution is:
                                                            S = cd + abc′d′                 (2.32)


Author Name: B. SOMANATHAN NAIR
Author Website: www.randomelectronicdreams.com

Five Variable K Map Examples

Figure 2.27 shows a five-variable K-map. The variables may be designated as a, b, c, d, and e, respectively.
The five-variable K-map is in effect two four-variable K-maps drawn horizontally to form an extension of each other. The two four-variable maps are designated as a = 0 and a = 1, respectively. Thus, the map a = 0 represents cells designated from 00000 (≡decimal 0) to 01111 (≡decimal 15) and the map a = 1 represents cells designated from 10000 (≡decimal 16) to 11111 (≡decimal 31). The simplification using the five-variable K-map is slightly complex and requires a lot of careful inspection of the two constituent maps. We shall now consider an example to illustrate the simplification process in a five-variable K-map.


Example 6: Simplify S = S (0, 1, 2, 3, 8, 12, 15, 16, 17, 18, 19, 22, 28, 31)

Solution: We know that adjacent-cell entries can be reduced by grouping them. In Fig. 2.27, we have two separate charts or maps. How do we go about to find the adjacencies? There is no real difficulty in this if we consider the fact that the two charts themselves are adjacent to one another with the main element a = 0 for the first map and a = 1 for the second map. For example, the entries in the zeroth cell and the sixteenth cells are adjacent to each other because the zeroth cell represents 00000 and the sixteenth cell 10000. The two cells differ in the variable ‘a’ only and this gets eliminated if we group the entries in cells 0 and 16 together.  In fact, a gets eliminated whenever the entries in corresponding cells of both the charts can be grouped together. Based on this principle, the entries of the given example are grouped, as shown in Fig. 2.28.


Now, consider Group 1 in the figure. We have a quad in chart a = 0 and a similar quad in chart a = 1. Since these two quads occupy similar positions in the two charts, we group them together to form an octet. The resultant terms for Group 1 is b′c′.  Group 2 has entries in chart 1 only, and reduces to  abde.  Group 3 has a single entry in chart 1, and a single entry in chart 2. The reduced term then is bcde. Group 4 is similar to group 3 and reduces to  bcde. Finally group 5 gives abde. The solution, therefore, is:

              S = b′c′ + a′bd′e′+  bcd′e′ +  bcde +  ab′d′e′        (2.30)

      It may be noted that don’t-care terms can appear in five-variable maps also. The following example illustrates this.

Example 7: Reduce S = S (1, 2, 3, 8, 9, 10, 12, 15, 16, 17, 18, 22, 26, 27) + d (0, 4, 11, 19, 28, 29, 30, 31)

Solution:    The K-maps for Example 7 is shown in Fig. 2.29.


From Fig. 2.29, we obtain the final expression as:

                              S = a′c′ + b′c′+  ade′ +  bde +  a′d′e′    (2.31)

Author Name: B. SOMANATHAN NAIR

Author Website: www.randomelectronicdreams.com

Boolean Algebra and Logic Simplification Examples

Boolean Algebra and Logic Simplification Worked Exercises:


LOGIC SIMPLIFICATION USING ALGEBRAIC METHODS

Example 1:  Prove that AB + AB′ = A.
Proof:                                            LHS = A(B + B′) = A = RHS
Example 2: Prove that (A + B′) B = AB.
Proof:                                    LHS = (A + B′) B = AB +BB = AB = RHS               

Example 3:  Prove that (A + B′) A = A,  

Proof:                            LHS = (AA + AB′) = A (1+B′) = A = RHS               

Example 4: Prove that A + A′B = A + B
Proof:         LHS = A(1+B) + A′B = A + AB + A′B = A+ B(1+ A′) = A + B

In the above proof, we have used the relation= A(1+B) = A.

Example 5: Prove that  ABAB = A + B.
Proof: We first expand the term 
                                               BAB = B′AB + B(AB)′= B(AB)′
where we have used B′B = 0. Now we expand using De Morgan the relation
               
                                                           B(AB)′ = B(A′+B′) = BA

Then we find that    A BA′ = ABA′+ A(BA′)= AB + A(B′ + A)

Further reducing, we find that          AB + A + AB′ = AB + A(1+ B′)

which reduces to       AB + A(1+ B′) = AB + A = A + B = RHS
                                                                                                                          
Example 6: Prove that B + A B′C = B + AC
Proof: We have              LHS  = B + A B′C = B + B′ (AC) = B + AC = RHS
where we have used a relation similar to B + B′C = B + C.                                                       
                                               
Example 7: Prove that  A + A B + A B′C + A B′ C′D + … = A + B + C +…
Proof: We know that             A + A B = A + B

Extending the above relation, we find
                                     A + B + A B′C = A + (B + A B′C) = A + B + A C             
where we have used the relation  (B + A B′C) = B + A C

But, we know that A + A C = A + C.  Thus we obtain the relation

                              A + B + A B′C = A + (B + A B′C) = A + B + C
Carrying out this operation and mathematical induction, we obtain the final relation:                                           A + A B + A B′C + A B′C′D + … = A + B + C + D +…

Hint: As stated earlier, logic simplification through algebraic methods is very tedious and requires the knowledge of many fundamental theorems and laws, especially when the RHS part is not given. To keep these theorems in memory is not that easy since there exist several of them.
      
To simplify the procedure, we suggest that the student (especially one who is writing an examination) first find the correct solution using an appropriate K-map. Once we have the answer with us, we can proceed to solve the problem algebraically. As an example, we solve Example 6 (assuming the RHS part is not given) using this method. For this, let us assume that the given problem is stated as  
A + A B + A B′C + A B′C′D + … = ?
     
In the above problem, since the RHS is not given, we are not sure what the answer (RHS) would be. To find the answer (i.e., RHS), we first draw the three-variable map M1 as shown in Fig. E6a and enter the implicants A, A’B, and A’B′C in it as shown.




Next we draw K-map M2 as shown in Fig. E5b and remove the first encirclements. Then we regroup them as shown in M2 to find what A + A B + A B′C would be. 



From map M, we get the answer as:

A + A B + A B′C = A + B + C

Now we are sure what the RHS would be in this case. Knowing the answer in advance, we can prepare our strategy accordingly to solve the problem.

Example 8: Prove that (i) AA A… = 0, if the number of A’s is even
                                       (ii) A A A… = A, if the number of A’s is odd
Proof: We know that  
             (i) A A = A A + A A = 0
This means that if the number of A’s is even, then the sum will be 0. Now, if the number is odd (say, three), we get            
                                       (ii)  AA A = 0 + A = 0. A+ 1. A = A

      Extending the above two results, using mathematical induction, we get the desired results.



Example 9: Prove that AB AB′ = A
Proof:          LHS = AB AB′ = AB(AB′)′ + (AB)′ AB′
                           = AB(A′+B) + (A′+B′) AB′
                           = AB + AB′ = A =RHS


Example 10: Prove that A′ + B′ + ABC′ = A′ + B′ + C′
Proof: We know that     A′ + ABC′ = A′ + BC′                            
Using the above relation we find LHS = A′ + B′ + BC′ = A′ + B′ + C′ = RHS
                                                
Example 11: Prove that     x′y′z +yz + xz = ?                                 
Proof:                                  LHS = z(y + y′x′) + xz = z(y + x′) + xz
                                                                   = zy + zx′ + xz = zy + z = z = RHS

Example 12: Reduce the expression 
(Bars denote negation).                                     

Proof:  Applying De Morgan on the barred term in square brackets yields
Example 13: Prove that xy + x′y′ + yz = xy + x′y′ + x′z                                  
Proof:                                      LHSxy + x′y′ + yz (1 + x′) = xy + x′y′ + yz +yzx′
                                                          = xy + x′(y′ + yz) + yz = xy + x′y′ + x′z + yz
                                                          = (xy + yz + x′z) + x′y′ = xy + x′z +x′y′ = RHS
where we have applied the consensus theorem on the bracketed terms.
                          
Example 14: Prove thatA′C′ + ABD + AB′D′ + ABCD′ = ?
                                         
Proof: Since the RHS is not given, we use a K-map and find the RHS                               


Example 15: Prove the De Morgan’s laws, viz.,(A + B)′ = A′B′ and (AB)′ = A′ + B′.
Proof:  These laws were enunciated by Augustus De Morgan (to be pronounced as da morgan), a nineteenth century British mathematician. To prove these laws, we make use of truth tables: Table E14a is used to prove the first law. It can be observed that the entries in the rightmost two columns are the same; this proves the first law.

                          Truth Table E14a Proof of (A + B)′ = A′B′

A
A′
B
B′
A+B
(A+B)
A′ B′
0
1
1
0
0
1
1
0

0
1
1
0
1
0


      Table E14b proves the second law. The entries related to the second law are as shown in the table. As in the first case, in this case also the entries in the rightmost two columns are the same, which proves the second law.

                            Truth Table E14b Proof of (AB)′ = A′ + B′


A
A′
B
B′
AB
(AB)
A′+ B′
0
1
1
0
0
1
1
0

0
1
1
0
1
0


Example 16: Prove the transposition theorem, which states that
 AB + A′C = (A′ + B)(A + C)

Solution: Consider the RHS. We find
RHS(A′ + B)(A + C) = A′A + A′C + BA + BC = A′C + BA + BC
where we have used the basic law A′A = 0.  Now, the remaining expression A′C+BA+BC can be seen
to be a statement of the consensus theorem, which reduces to A′C+BA. That is,
AB + BC + A′C = AB +A′C = LHS

Example 17: Involution law: This law states that (A′)′ = A
                                                
Proof: Involution law becomes highly useful in solving problems applying De Morgan’s laws. The law can be proved using the truth table E16. We find that the first and last columns agree with each other, which proves the law.

Truth Table E16 Involution Law
A
A’
(A′)′= A
0
1

1
0

0
1 


Example 18: Define the EXOR function.

Solution: The logic function Z = AB′ + A′B  is called the EXCLUSIVE-OR (EXOR or XOR) function. This relation can be derived from Table E17a, the truth table for the OR function.

Truth Table E17a The OR function
A
B
Z = A+B
0
0
1
1
0
1
0
1
0
1
1
1


In Table E17a, if we change the last row as shown in Table E17b, we get the XOR function.

Truth Table E17b The XOR function
A
B
Z = A ⊕ B
0
0
1
1
0
1
0
1
0
1
1
0


From Table E17b, we get the XOR relationA B = AB′ + A′B
where the symbol ‘’ represents the XOR operation. It is to be noted that it is the XOR operation (and not the OR operation) that really represents the algebraic addition of two bits.

Example 19: Define the EXNOR function.
Solution: The logic statement (AB+AB′)′ is called the EXCLUSIVE-NOR (EXNOR or XNOR) function. This relation is derived from the truth table for the OR function as shown below. EXNOR represents the complementary operation of the algebraic addition of two bits. The EXNOR operation in the last column may be read as “NOT of either A OR B”.

Truth Table E18 The XNOR function
A
B
   A B
(AB′ + A′B)
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
1


From Table E18, we get the EXNOR relation as (A B)′ = A′B′ + AB

Example 20: Use De Morgan’s laws to expand the XNOR relation.
Solution: The XOR operation given by
                                                            Z = (AB′ + A′B)  
Since XNOR operation is the complement of , by applying De Morgan on the RHS, we get
                                                           Z = (AB′)′.(A′B)′

Applying De Morgan on the two RHS terms individually yields

Z = [A′ + (B′)′][(A′)′+ B′]
                                                                         = (A′ + B)(A + B′) = AA′ + BB′ + AB + A′B′
                                                                         = AB + A′B′
Thus we get
                                         (AB′ + A′B)′ = AB + A′B′



Example 21: Explain the principle of duality.
Solution: Consider the function
f(x) = x1 + x2x3 + x4x5x6                                            (1)
Replacing the pluses with dots and dots with pluses in the RHS of Eq. (1), we get a new function
                                                               F(x) = x1 (x2 + x3) ∙ (x4 + x5 + x6)                                 (2)
The function F(x) defined in Eq. (2) is called the dual of the function f(x). We find that f(x) and F(x) are equally valid functions and duality is a special property of Boolean (binary) algebra. The property of duality exists in every stage of Boolean algebra. For example, positive and negative logic schemes are dual schemes. Similarly, AND is the dual of OR, NAND is the dual of NOR, and so on. To illustrate further, consider the De Morgan’s law

                                                                    (A + B)′ = A′.B′                                                       (1)
Now, in Eq. (1), replace the plus with a dot and the dot with a plus; this action yields the expression
                                                                    (A.B)′ = A′ + B′                                                        (2)
We find that Eq. (2) is the complementary De Morgan’s law. This suggests that the De Morgan’s laws form a dual pair.
      We now state that every rule and law applicable to a positive-logic scheme is applicable to its corresponding negative- (or, complementary-) logic scheme also. We now formally define the dual of a given function as follows:                                                                                         
      A function F(x) is said to be the dual of another function f(x), provided that F(x) is obtained from  f(x) by replacing the multiplication signs (∙) and summation signs (+) in f(x) with summation signs and multiplication signs, respectively, and with no change in the variables of f(x).
      The definition given above may also be considered as the duality theorem. In this context, we may define duality as the state of being dual

Example 22: Find (ABC′)+(ABC′)′
Solution: Let us assume that ABC′ = P. Then we find that (ABC′)′ = P′. Now the given function can be written as:
                                                       (ABC′) + (ABC′)′  = P + P′ = 1

Example 23: Find (ABC′)+(ABC′)′ using a different method.
Solution: Applying De Morgan on the bracketed term yields
                                                           (ABC′) + (ABC′)′  = ABC′+A′+B′+C
But we know that                                               ABC′+A′ = BC′+A
and                                                            A′+BC′+B′ = C′+B′+A
Finally,                                                   C′+B′+C + A′= (C′+ C) +B′ + A′ = 1
Example 24: Find w(x + y′z) + x + y′z.
Solution: Expanding the given expression yields
                                                wx + wy′z + x + y′z = wx + x + wy′z + y′z
                                                                              = x + y′z
Example 25: Find (w + x′ + yz)(w′ + x′  + yz).
Solution: Expanding the given expression yields
              (w + x′ + yz)(w′ + x′  + yz) = wx′ + wyz + x′ w′ + x′ + x′yz + yzw′ + yzx′ + yz
                                                      = wx′ + wyz + x′ w′ + x′ + x′yz + yzw′ + yzx′ + yz
                                                       = x′ (w +w′)+ yz (w +w′)+ x′ (1 + yz) +  yz (x′ + 1)
                                                       = x′ + yz + x′  + yz = x′ + yz

Example 26: Find (a + b′)(a + c + d)(a + c + d′).
Solution: Performing the first multiplication in the given expression yields
                                (a + b′) (a + c + d) = (a + ac + ad) + b′c + b′d
                                                           =a + b′c + b′d
Now, we perform the second multiplication, which gives
                   (a + b′c + b′d) (a + c + d′) = (aa + ac+ ad′ + ab′c + ab′d) + (b′c + b′cd′ + b′cd)
                                                            = a + b′c  

Example 27: Find ab′(c + d) + (c + d).
Solution: Performing the first multiplication and applying De Morgan to the complemented (third) term in the given expression yields

                                          (ab′)(c + d) + (c + d) = ab′c + ab′d + c′d′
Now, we are in confusion regarding the route through which we have to move to reach the destiny. As the first step, we try expanding the term c′d′. This is because the term c′d′ contains four 4-variable terms, given within brackets below, and contains components related to ab′c and ab′d. Thus we find

                                 RHS = ab′c + ab′d + (a′b′c′d′ + a′bc′d′ + abc′d′ + ab′c′d′)    
Now, we find that terms
 ab′d + ab′c′d′ = ab′(d + c′d′) = ab′d + ab′c′   
Using the above two factors, the RHS may be expressed as
                                RHS = ab′c + ab′d + ab′c′ + a′b′c′d′ + a′b c′d′ + ab c′d′       
Combining factors ab′c and ab′c yields
                                 RHS = ab′ + ab′d + a′b′c′d′ + a′bc′d′ + abc′d′       
Combining factors ab′ and ab′d, we get
                                    RHS = ab′ + a′b′c′d′ + a′bc′d′ + abc′d′       
Combining factors a′b′c′d′ and a′bc′d′
                                 RHS = ab′ + a′c′d′ + abc′d′       
Finally, combining factors a′c′d′ and a′bc′d′, we obtain the reduced expression as
                                 RHS = ab′ + c′d′       
      It can be seen that the reduction process is quite laborious and lengthy. Further, the reduction has been performed based on hunches and previous experience. Experiences and difficulties of this kind led to the development of the K-map and QM methods of Boolean reduction.

Example 28: Find a + b + c′d(a + b).
Solution: Performing the first multiplication and applying De Morgan to the complemented (third) term in the given expression yields
 
a + b + c′d (a + b) = a + b + c′d (a′b′) = a + b + a′b′c′d
                                                    = (a + a′b′c′d) + b  = a + b′c′d + b
                                                    = a + b + c′d

Example 29: Find [(ab)′ + a′ + ab].
Solution: Applying De Morgan to the terms within the square brackets yields
                                    [(ab)′ + a′ + ab] = [(ab)′]′ ∙ a (ab)′
Applying De Morgan to RHS terms, we get
                                  [(ab)′]′ ∙ a (ab)′ = aba ∙ (a′ + b′) = ab (a′ + b′) = 0
[Note: This may also be reduced at the second stage itself (without second the demorganization. For this observe that
                                              [(ab)′]′ ∙ a (ab)′ = (ab) ∙ (ab)′  = 0

Example 30: Prove that (a + b′)(b + c′)(c + d′)(d + a′) = (a′ + b)(b′ + c)(c′ + d)(d′ + a)
Solution: Multiplying the individual terms in LHS yields
                                           LHS = (ab + ac′ + b′c′)(cd + ca′ + d′a′) =  abcd + b′c′d′a′        (1)
Similarly, multiplying the RHS terms yields                                                                                    
                                        RHS =  (a′b′ + a′c + bc)(c′d′ + c′a + da) =  a′b′c′d′ + bcda     (2)

We find that
                                                Equation (1) = Equation (2)
which is the desired result.

Author Name: B. SOMANATHAN NAIR

Author Website: www.randomelectronicdreams.com

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