Implantable Telemetry System

Sometimes it is wanted to implant the telemetry or receiver subcutaneously. Implantable telemetry systems allow the measurement of physiological variables over long periods of time. Here no sensors need to be attached to the body surface. Implantable telemetry is mainly used in animal research and the equipment must be sheltered from the animals. In simple conditions the implantable telemetry system consists of a power supply, transducer which is connected to the transmitter. For long term telemetry, implant telemetry is more helpful. The whole electronic circuit is fully packed as a capsule and then implanted subcutaneously (under skin) or deep in the body. This is called encapsulation unit. Silicon encapsulation is commonly used.

The telemetry transmitters are as small as possible. The weight and size of the batteries must be minimal and the operating lights must be maximal. Once a unit is implanted its service depends on the battery capacity. So the power source is of great importance. Mercury and silver oxide main batteries are used commonly. Depending on the application, the size and capacity of implantable unit battery varies. To save the power, radio frequency switches can be used to turn the system ON and OFF on depending on need. The use of implanted unit also limits the distance of transmission of the signal.

Why PWM is used in multichannel telemetry system?

PWM systems has a high noise immunity.
Techniques are cheaper and less complex than FM system.
PWM is lower sensitive to temperature changes.

Disadvantages of Implant Telemetry

1. One disadvantage of implantable telemetry unit is that the range of the signal is affected by the body fluids and skin which greatly attenuate the signal. For implant telemetry the size and weight limitations are much more serious than other types as heavy units can make the subject comfortless.

2. Some patients may be allergic to external materials and this can also make problems with implantable telemetry.

3. Proper protection of the patients is also a vital factor to be considered. In any means the patient will not be affected by the battery power supply.


Single Channel Telemetry System

Single Channel Telemetry System:

Here the physiological variables from a solitary patient are calculated using appropriate electrodes and sensors and fed to the radio transmitter after appropriate signal conditioning. In a single channel system a battery functioned radio transmitter is linked to the electrodes of the patients. This transmitter transmits the bio-potential over a restricted range to a distantly situated receiver. The receiver senses the radio signals and after good signal conditioning, recovers the original signals for additional processing. The basic block diagram of a single channel telemetry system is parallel to that of a basic biotelemetry system.

block diagram of biotelemetry system
Figure shows a block diagram of a distinctive radio telemetry single channel system. From the block diagram, the Biological signal from patient is transmitted to Transducer and sensor, There it is transmitted to Signal conditioner and then to the Transmitter. This transmitted original signal is transferred by Signal conditioner and then to the Display and readout. The radio frequency utilized in the system differs from a hardly any hundred KHz to some MHz. Further than this frequency range the reduction due to subject body turn out to be excessive. AM is not adopted for biotelemetry system since the presence of noise can affect the measured variables severely. That is why we usually adopt either frequency modulation or pulse modulation techniques to transmit the bio-signals.

Block Diagram of Biotelemetry system

Block Diagram of Biotelemetry system:


The transducer converts the biological signal from Patient into electrical signal. The signal conditioner amplifies and alters this signal for efficient transmission. The transmission link connects the signal conditioner to the read out and monitoring device by wire or wireless means.

Major Specifications of a Bio Telemetry System:

a. The telemetering system must be selected to send out the bio-electric signals with maximum reliability and ease and directivity.
b. There would not be any discomfort for living system due to these telemetry systems. In any condition the natural living environment of the living system should not be affected.
c. The size and weight of telemetry system should be small and should not be too costly.
d. It should have more stability and reliability and efficiency.
e. The power consumption should be moderate.
f. The presence of common mode noise will affect the measured physiological variable. So at the transmitter side differential amplifier must be present to reject this component.
g. For wire transmission the cable should be shielded well to avoid noise and Electro Magnetic Interference (EMI).
h. In the case of long term units or implant units the weight and of the telemetry system must be designed in a way such that the system should be comfortable to the subject.
i. There should be any interference with living system.

Biotelemetry Transmitter Block Diagram:



Mobile units close to the patient comprises of a transmitter. The physiological variables are measured using a transducer (ex: Temperature, pressure etc) and the bioelectric signals are directly measured using suitable electrodes. The measured variables are then suitably amplified and processed and given to the modulator circuit. The physiological signal modulates a sub-carrier in audio frequency range. The sub-carrier in turn adapts the RF signal to be spread and sent to the transmitting antenna. The sub carrier is modulated moreover by AM or by FM techniques. But usually FM technique is used.

Biotelemetry Receiver Block Diagram:


The receiver consists of a tuner, demodulator, and a display unit. The receiving antenna receives the multiplexed RF carrier which was transmitted by the transmitting antenna. The tuning circuit is properly tuned to receive the appropriate signals and the unwanted signals are rejected. The receiving antenna must have high selectivity. The multiplexed RF carrier is demodulated to improve the individual sub carriers from which original physiological signals can be decoded. The recording system is used to record the physiological variables either in a chart paper (ECG, EEG etc) or in a display unit (Computer, monitor, oscilloscope etc.). The signals can be stored on any secondary storage media such as magnetic disk.

Branching method using QM Method and K Map

Branching method is a technique used when a given function does not contain any essential prime implicant in it. The solution to such a problem contains more than one set of solutions. There may be two independent sets of solutions, which may be regarded as two branches of solutions; hence the name branching method.

Example 17: Reduce the expression S = S(0, 1, 5, 7, 8, 10, 14, 15).

Solution: We solve the problem using the QM branching method. For this, we prepare the QM charts as shown below. It can be seen, we can form only two charts. Also we notice from Chart 2 that only eight pairs are obtained and that no quads can be formed from these pairs.

Now, from Chart 2, we obtain the pairs as A(0, 1), B(0, 8), C(1, 5), D(8, 10), E(5, 7), F(10, 14), G(7, 15), and H(14, 15). Table 2.14 shows the desired QM cyclic chart showing entries of these groups.


From the table, we obtain the four selected groups (encircled) as A(0, 1), D(8, 10),, E(5, 7), and H(14, 15). We may also obtain a second set of groups containing pairs B(0, 8), C(1, 5), F(10, 14), and

G(7, 15). Thus in this problem using a cyclic chart (see problem S2.4), we find two independent sets of pairs, both pairs being equally valid and minimum. As stated above, since there are two independent branches of pairs in this problem, this method of solving the problem is called as the branching method. The reduced function using the first branch of pairs mentioned above is:

S = a′b′c′ + ab′d′ + a′bd + abc                                  (2.37a)                                                                                           
and the reduced function using the second branch of pairs mentioned above is:

S = a′c′d + acd′ + bcd +b′c′d                                    (2.37b)                                                                                    

a.   Solution using the Modified QM method

In this scheme, we delete the terms covered by other implicants. For this, let us first choose A(0, 1). Then we strike out B since it contains 0. We next select E(5, 7) and strike out C(1, 5) as  1 and 5  are covered by A and E, respectively. Then we select D(8, 10) and H(14, 15) and delete F (10, 14) and G(7, 15) since the elements in them  are already covered. The finally selected groups are A, D, E, and H, respectively. These are indicated in bold letters.

The selection shown above is quite random, which suggests that there are other possibilities in this case. For example, we can choose B, C, F and G to form a second set of implicants. It can be clearly observed that the minimum number selected is half of the number of implicants present in this case.

It can be seen that the modified QM technique is a general technique and can be easily applied to all types of logic reduction including the problems which use the branching method.

b. Solving the Problem using the K-map Method

We first solve the problem using the K-map method. We prepare the K-map in this case as shown in Fig. 2.28(a). Notice that in this map, we have shown entries only in the cells mentioned in the given problem, i.e., we have shown 1 as the entry in the cells 0, 1, 5, 7, 8, 10, 14, and 15. The remaining cells are left vacant instead of filling them with 0s. This operation will help in identifying the pairs, quartets etc. without confusion. The cells left unfilled may be filled with 0s after the grouping operation is completed. This technique will be followed in this text for solving problem.

From Fig. 2.28(a), we find that the reduced function is the same as that given by Eq. (2.37-b). Similarly, from Fig. 2.28(b), we get a second reduced function, which is the same as that given by Eq. (2.37-a).  
              



QM Method with Don't Care

Simplification When Don’t-Care Terms are Present:

Example 16: We now use an example to illustrate the reduction process in the QM when don’t-care terms appear in the given logic function. Reduce S = Sm(1, 2, 4, 5, 6, 8, 9, 12) + d(3, 10, 13, 15).

Solution: Chart 1 shows the initial grouping, Chart 2 shows the pairs, and Chart 3 shows the quads. In the case of the K-map reduction technique, don’t-care terms were treated as valid terms if it were helpful in the reduction process. The same assumption is brought into the QM reduction process also. The QM charts of Example 16 are drawn as shown below using the principles discussed in the previous sections.


It may be noticed that the don’t-care terms are identified with the letter d attached to them in Chart 1. However, once they are identified, the letter d is dropped from respective terms, as shown in Chart 3. Now, the quads and pairs are designated using bold letters digits as shown. The reduced terms represented using bold letters are:

                          A :  1, 5, 9, 13                                  
                          B :  4, 5, 12, 13                                
                          C :  8, 9, 12, 13                                
                          D :  1, 3                                            
                          E :  2, 3                                            
                          F :  2, 6
                                                             G :  2, 10
                                                             H :  4, 6
                                                              I  :  8, 10
                                                              J :  13, 15

The above groups are now entered as shown in Table 2.18 for reduction by the modified QM technique. In Table 2.18, we represent the selected groups by bold digits. The groups that are not selected are represented by conventional letters and digits. In Table 2.18, we have shown a third column with entries Y/N. Y represents yes, indicating the selection of that group, while N represents no indicating the rejection of that group.
The finally selected reduced Groups are A, B, C, and F. In selecting these groups, we have discarded Groups D, E, H, I, and J, as each one of them contains at least one don’t-care term, which has no relevance at all in the final groups. For example, Groups D (2, 3d) and G (2, 10d) have no relevance at all since the only valid member in these groups, viz., 2, has already been selected in Group F. Hence they are discarded. The same is the case with E, H, I, and J. The reduction process to get the final SOP expression is then carried out, as shown in Table 2.19.
From Table 2.19, we obtain the resultant expression as

                              S = c d + bc  + ac + a b d +a cd             (2.40)

We now solve Example 15 using the K-map shown in Fig. 2.40. From the K-map, we obtain the resultant expression as the same one given by Eq. (2.40).

Table 2.19

Group
Group members
Elimination
Reduced Function


A


1, 5, 9, 13,

0  0  0  1
0  1  0  1
1  0  0  1
1  1  0  1

c′d


B


4, 5, 12, 13
1  0  0
1  0  1
1  0  0
            1  0  1

b c′


C

8, 9, 12, 13
1  0  0  0
1  0  0  1
1  1  0  0
1  1  0  1


a c′
F

2,  6
0  0 1  0
          0  1 1  0
a′ c d′



Quine McCluskey (Tabular) Method Example


Quine McCluskey (Tabular) Method Examples

Example 12: Reduce S = S(0, 1, 2, 3, 6, 7, 8, 12, 13, 15).

Solution: To solve the given problem, we follow the steps given below:

Step 1: The first step in the QM method is to separate the minterms into specific groups, as shown in Table 2.12. These groups are formed on the basis of the number of 1s in their binary form.  For example, the binary number 0000 has no 1 in it and hence forms the first group.  Binary numbers 0001, 0010, 1000, 10000, etc. have one 1 in them and are put together to form the second group. This process is continued to form groups with two 1s, three 1s, etc., as shown in Table 2.12.

      We shall make use of Table 2.12 to proceed further with our example. By comparison of the minterms in our example with the tabulation given in Table 2.12, we form the groups belonging to the first chart of the QM scheme. It may be noted that the members in each group differ in one-bit position only from the members of the adjacent group below (or above) that group. For example, ‘0’ in Group P and 1, 2 or 8 in Group Q differ from each other in one bit position only. This rule follows for all the subsequent groups in the first as well as other charts. It may also be noted that the groups are named as P, Q, R, etc. for the convenience of their identification. Once we are familiar with the procedure of groupings, this designation is not required as will be clear from the examples to follow. 

      Now, the first column of the first chart shows the names of the groups as P, Q, R, S, etc. and the second column shows the member(s) in them. Thus we find that Group P has only ‘0’ as its member. Below this, in Group Q, we have numbers 1, 2, and 8 as its members. Similarly, every member of Q is different from every member of R in only one position. Thus, we conclude that the groups P, Q, R, etc. are chosen such that every member in a group differs in only one bit position from every other member in the group immediately above or below it. It may also be noted that this is not true for distant groups. Hence grouping and pairing is not possible among distant groups.

Step 2: In this, as stated above, we first form the groups, as shown in Chart 1. From this chart, we prepare Chart 2, which shows the members of each group that can be paired together, these pairings being indicated by tick marks, as shown in Chart 2. For this, we inspect adjacent groups and collect members that differ from one another by a power of 2. For example, ‘0’ of Group P differs by a power of 2 from the members of Group Q. Thus, 0 and 1 differ by 20, 0 and 2 differ by 21, and 0 and 8 differ by 23. Hence, they can be paired together to form the members of Chart 3. The principle behind these pairings is that each of these pairs can be combined to eliminate one bit, as shown:
Table 2.12

Minterm
Number of 1s in the binary equivalent number
Decimal number
Binary equivalent number
0
0000
0
1
2
4
8
0001
0010
0100
1000

1
3
5
6
9
10
12
0011
0101
0110
1001
1010
1100


2
7
11
13
14
0111
1011
1101
1110

3

15
1111
4

Illustration 1: We can combine decimal numbers 0 and 1 as shown below to eliminate D.

As stated, in the pairing shown above, we find that variable D is eliminated.



Illustration 2: As in Illustration 1, we find that decimal numbers 0 and 8 can be combined to eliminate A, shown below. Thus we find that appropriate pairing eliminates the variables appearing in the complemented and uncomplemented forms.

      In this way, one pairing operation between adjacent groups results in logic reduction by one bit. As stated previously, it may be noted that pairing must be done strictly between adjacent groups, i.e., between P and Q, Q and R, R and S, and so on.  Pairing cannot be done between P and R, P and S, etc., as they do not obey the rule of difference-in-one-bit-position-only.  The results of the pairing are tabulated, as shown in Chart 3. It may also be noted that when a pair is chosen, the corresponding numbers (in decimal form) must be marked with a tick (ΓΌ) sign so as to indicate that these numbers have already been selected.  This is shown in Column 2 of Chart 3. Column 3 shows the difference between the decimal numbers in each pair; it must be remembered that this difference must be exactly a power of 2, as indicated in Column 4.  Usually, Columns 3 and 4 need not be shown in the charts; these can be easily avoided in practical solutions using the QM method. Here, they are shown only for illustration. It also may be noted that to find the difference between two numbers, decimal numbers are easier to remember and handle than binary numbers. For example, it is easy to subtract (mentally) 7 from 12 than subtracting 0111 from 1100. Also, it is easier to find that 12 –7 = 5 is not a power of 2 and hence pairing of 12 and 7 is impossible.

                                               Chart 1                                                           Chart 2

Group
name
Members of the group

Group
name
Members of the group
P
0
P
0b
Q
1
2
8
     Q
1b
2b
8b
R
 3
 6
12
             R
  3b
  6b
12b
       S
7
13
             S
              7b
             13b
T
15

     T
15b

                                                                               Chart 3                                     

Group
Name
1
Members of the group
2
Difference in the pairs in  decimal numbers
3
Difference in the pairs in the power of 2
4
P1
0, 1
0, 2
0, 8
1 – 0 = 1
2 – 0 = 2
8 – 0 = 8
20
21
23
Q1
1, 3
2, 3
2, 6
8, 12
3 – 1 = 2
3 – 2 = 1
             6 – 2 = 4
           12 – 8 = 4
21
20
                   22
                   22
R1
3, 7
6, 7
12, 13
7 – 3 = 4
7 – 6 = 1
           13 – 12 = 1
22
20
                   20
S1
7, 15
13, 15
15 – 7 = 8
15 – 13 = 2
23
21

Step 3: In Step 3, we make still lager pairings. For example, a pair from group P1 can be combined with a pair from the adjacent group Q1 if they differ by a power of 2.  Thus, in Chart 3, the pair (0, 1) of Group P1 can be combined with the pair (2, 3) from Group Q1, as they differ by the same power of 2 (i.e., 20 = 1) between individual members and by 21 between the pairs. That is, between 0 and 1, the difference is 20  (= 1); between 2 and 3, difference is 20  (= 1).

      Now, to find the second condition, i.e., the difference between pairs, take any member in pair P1, say 0. We now find the difference between 0 and 2, its corresponding member in pair Q1.  The difference can be seen to be equal to 2 (=21).  Since the pairs (0, 1) and (2, 3) have satisfied our two conditions, we can combine them to form a quad P2.  Once 0, 1, 2, and 3 are grouped, the grouping 0, 2, 1, and 3 can also be ticked off, as this quad has already been selected once.  By checking out for groupings in this fashion, we find that we have one more quartet (i.e., 2, 3, 6, 7) available to us.  No further pairings are possible in this group. The second set of groupings is shown in Chart 4.

      Now, we notice that some pairs such as (0, 8), (8, 12), (12, 13), (7, 15) and (13, 15) are left behind in Chart 3 which cannot be combined further.  We encircle each one of these groups and keep them as independent groups with names A, B, C, etc. as shown in Chart 5. According to this scheme, the pair (0, 8) is designated as A, the pair (8, 12) is designated as B, and so on. The pairs and quads show reduction in variables, which are then checked for repetition; if a  pair or quad is covered in other pairs, quads etc., then such pairings will be discarded. The remaining SOP expression will be used for logic implementation


Step 4: The next step in the QM reduction process is to find whether pairing is possible from groups P2 and Q2 to form an octet.  For this, we must see that individual members differ by the same pattern in both groups, and any member in one group differ by a power of 2 with the corresponding member in the other group.

In our example, adjacent numbers of Group P2 differ by a factor of 1 (= 20) between them. Thus we observe that 0, 1, 2, and 3 differ by 1 from each other.  Now, consider the members in Group Q2. These are, respectively, 2, 3, 6 and 7.  Even though 2 and 4, and 6 and 7 differ by 1 from each other, the adjacent members 3 and 6 differ by 3 and hence the pattern of 1-difference-between-adjacent members is not followed here.  So, in the first instance itself, these groups cannot be combined to form an octet. Thus, we stop at this point, and name the quad (0, 1, 2, 3) as final Group F and the quad (2, 3, 6, 7) as final Group G and encircle them, as shown in Chart 5.

Chart 5

Final groups 
Members of the final groups
A
B
C
D
E
F
G
0, 8
8, 12
12, 13
7, 15
13, 15
0, 1, 2, 3
2, 3, 6, 7
                                     
Step 5: The last step is to remove the common (or, redundant) terms, which have been selected more than once from different groups. In the QM method, Table 2.13 is drawn as shown in for the reduction purposes.

As shown in Table 2.13, the first column represents the names of the groups. The remaining columns are drawn to enter all the minterms in the given problem. In our problem, we have the minterms 0, 1, 2, 3, 6, 7, 8, 12, 13 and 15. So, we must draw tabular columns to accommodate all of these minterms.

After drawing the rows and columns of Table 2.12 as shown, we mark ‘X’ in cells where there is an entry. For example, final Group A has entries in the cells under the columns designated 0 and 8. Similarly, Group F has entries under columns 0, 1, 2 and 3. In this way, we fill the relevant cells with entries from groups A, B, C, D, E, F, and G as shown.
      
After having completed the filling of the cells, as explained above, we start the elimination process. In this elimination process, we have to remove the redundant terms (terms that have already been selected in various groups, and whose presence in the final logic expression will only add unnecessary extra gates). This is done as follows.
      
Let us first consider the larger groups. In the given problem, groups E and F are larger groups (quartets). Each group has two common entries in 2 and 3. But E and F differ in the remaining two entries. So, we choose them as essential and encircle all their members.
      
Next, we look for common terms in the paired groups of A, B, C, D, and E. We notice that, if we choose Group B with members 8 and 12, and Group F, we can eliminate Group A whose members are 0 and 8. This is because 0 is included in group F and 8 is included in Group B. We then score out the Group A with elements of 0 and 8. This process is repeated until all the redundant terms are eliminated, by visual inspection of the table. We find that groups A, C, and D are redundant. The selected groups are F, G, B, and E and we encircle them as selected groups. The final SOP expression can be obtained from these selected groups by the elimination process illustrated in the following


Example 13: Consider group F = 0, 1, 2, 3 (designated in decimal numbers). We now find the binary equivalent numbers of these decimal numbers and enter them one below the other, as shown in Table 2.14.

Table 2.14

Decimal number

Equivalent bits

Equivalent variables

0
1
2
3
0  0  0  0
0  0  0  1
0  0  1  0
0  0  1  1
a′  b′  c′  d′
a′  b′  c′  d
a′  b′  c  d′
a′  b′  c  d
Reduced solution
0  0  – –
a′  b′ – –

          In the above example, we have written the binary and algebraic equivalents of decimal numbers 0, 1, 2, and 3, as shown. We then encircle the member elements in the vertical columnar groups, and eliminate those variables, which appear in complemented and uncomplemented forms. The resultant SOP expression is given by: 
                                                             S = a′b′

This process is continued, as shown in Table 2.15 and the final SOP expression is obtained by combining the terms left in the right-most column of the table.
       From Table 2.15, we obtain the reduced final SOP expression as:

                                               S = a′ b′ + a′ c + a c′ d′ + a b d                (2.38)

Table 2.15 Reduction Method

Final group

Members (digits)

Members (bits)
Members  (variables)

Reduced function


F

0, 1, 2, 3

0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
a′b′c′d′
a′b′c′d
a′b′cd′
a′b′cd
a′b′

G
2, 3, 6, 7
0 0 1 0
0 0 1 1
0 1 1 0
0 1 1 1
a′b′cd′
a′b′cd
a′bcd′
a′bcd

a′c

B
8, 12
1 0 0 0
1 1 0 0
ab′c′d′
abc′d′
ac′d′

E
13, 15
1 1 0 1
1 1 1 1
abc′d
abcd
abd