IBHRE国际心律失常考官委员会资料： ibhre 复习资料

1、 A Summary of Standard International Units Derived Units with Special Names: Quantity Name Symbol Expressed in Terms Of Other Units Frequency Hertz Hz cycles/s * Force (exertion) Newton N kg*m/s * Pressure Pascal Pa N/m Note: -Pascal is the SI unit of pressure -mm Hg(mmHg) *, atm (atmospheres) *, an

2、d psi * (pounds per square inch) are more common pressure units 760mmHg=760 Torr=1 atm; 1 atm=14.7 psi A Summary of Standard International Units Derived Units with Special Names (contd): Quantity Name Symbol Expressed in Terms Of Other Units Energy, Work (force Joule J kg*m/s * applied over distance

3、) Power (energy/time) Watt W J/s * Quantity of electricity Coulomb C A*s * Electromotive Force Volt V W/A; J/C; V=IR * Capacitance Farad F C/V * Electrical Resistance Ohm ;R V/A; R=V/I * Conductance Siemens S A/V Inductance Henry H W*(Phi)/A Illuminance Lux lx ln/m A Summary of Standard Internationa

4、l Units Equivalents and Formulas In order to better understand the SI system and its usefulness, some helpful conversion tables are given below: Weight I = II 1 milligram 0.015432 grain 1 gram 15.432 grains 1 gram 0.03527 ounces 1 kg 2.2046 pounds * 1 grain 64.7989 milligrams 1 grain 0.0648 pounds 1

5、 ounce 28.3495 grams 1 gram 1000.0 milligrams * 1 milligram 1000.0 micrograms * A Summary of Standard International Units Equivalents and Formulas In order to better understand the SI system and its usefulness, some helpful conversion tables are given below: Volume I = II 1 cubic centimeter (cc) 1.0

6、 milliliter (ml) * 1 milliliter 0.0338 fluid ounces 1 liter 33.8148 fluid ounces 1 liter 2.1134 pints 1 liter 1.0567 quarts 1 liter 0.2642 gallons 1 fluid ounce 30.0 milliliters (or cc) * 1 pint 473.1765 milliliters 1 quart 946.353 milliliters 1 gallon 3.785 liters A Summary of Standard Internationa

7、l Units Equivalents and Formulas In order to better understand the SI system and its usefulness, some helpful conversion tables are given below: Length I = II 1 inch 2.54 centimeters * 1 french 0.335 millimeters * 1 french 0.013 inches * Time I = II 1 minute 60 seconds * 1 hour 60 minutes * A Summar

8、y of Standard International Units Equivalents and Formulas In order to better understand the SI system and its usefulness, some helpful conversion tables are given below: Temperature To Kelvin (K) Celsius (C), subtract 273.16 from K: C= K-273.16 To C K, add 273.16 to C: K= C +273.16 To C Fahrenheit

9、(F), multiply C by 9/5 then add 32: F=(9/5 * C)+32 To F C , subtract 32 from F, then multiply by 5/9 C: C=( F-32) * 5/9 In summation: C= K-273.16 K= C+273.16 F=9/5C+32 * C=(F-32) * 5/9 * 2: Relationship of Measurements A: Ohms Law Ohms Law: V=IR Ohms Law Voltage (V)=current (I)*resistance (R) or V=I

10、R Voltage “electrical pressure” or “electromotive force” causing current to flow higher voltage indicates a stronger force pushing the electrons along measured in volts (V) Ohms Law: V=IR Current flow of electrons or ions within a circuit unit of measure is the milliamp (mA) or Amp 1 mA is equal to

11、1/1000 Amp electrical symbol is I Ohms Law: V=IR Resistance opposition to current flow unit of measure is the ohm represented by Greek omega with AC, resistance is denoted as Z In terms of ohms law, voltage is directly proportional to current and resistance. Ohms Law: V=IR In summary: According to O

12、hms Law: Voltage (V)= Current (I) times resistance (R or Z) Current (I) = Voltage (V) divided by Resistance (R) Resistance (R)=Voltage (V) divided by Current (I) OR V=IR, I=V/R, and R=V/I Ohms Law: V=IR Sample questions: In a circuit with a known voltage of 12 volts (V) and a 600 ohm resistor (R), w

13、hat is the current (I) flowing through the resistor? In a lighting circuit using a 9V battery, .05A (or 50mA) of current is drawn. What is the resistance? Answer: since V=IR and I=V/R I = 12volts/600 ohms I = .02 amps (A) or 20 milliamps (mA) Answer: since V=IR and R=V/I R=9volts/.05amps R=180 ohms

14、B: Power and Energy Power and Energy Energy The ability to do work Measured in Joules 1 Joule=1 kg*m/s =amount of energy, work, or force applied over a distance of 1 m required to move a 1 kg object 1 m/s In terms of Ohms Law (V=IR), energy can be measured by the equation: energy (J) = Voltage (V) *

15、 Current (I) * time (t); substituting I=V/R for I in the previous equation, the energy equation can also be energy=(V/R) * t Example: in a pacemaker with a voltage of 2.5 Volts, lead impedance=500 and pacing pulse width of .5 ms, the energy output would be: Energy=Voltage (6.25 volts)/Resistance (50

16、0 ohms) * time (.5 milliseconds or .0005 sec) = 0.00000625 joules or .00625 millijoules or 6.3 J (microjoules) Power and Energy Power Quantity of energy (Joules) applied per unit time, usually per second Measured in watts One watt (W)=1 J/second One joule (J)=1 Ws (watt second) 3: Signal Concepts Si

17、gnal Concepts Displayed is the basic setup of an EP lab as indicated by Josephsons text Clinical Cardiac Electrophysiology: Techniques and Interpretations. Intracardiac signals are interfaced with a switch box, then conditioned by an amplifier and digitized to be analyzed for the purposes of the ele

18、ctrophysiology study. Distal Proximal 1 (-) 2 (+) 3 4 Signal Concepts Signals are transmitted via platinum electrodes to the recording system Voltage changes between each electrode pair (i.e. 1-2) sent to the viewing screen One electrode is designated as negative and the other as positive Deflection

19、 Determined by the direction of the electrical current. Upward when the current travels toward the positive pole. Downward when the current travels toward the negative pole. Deflects in both directions when the current travels perpendicular to the lead Signal Concepts Signal Concepts Filter settings

20、 High pass filters allow anything higher than them to pass and be incorporated into the recording Low pass filters allow anything lower than them to be included For surface leads filter settings should be high pass of .5 Hz and low pass of 100 Hz The His bundle and other intracardiac signals are bes

21、t visualized with filter settings of a high pass filter of 30-40 Hz and a low pass of 400-500 Hz Below, the effects of different high/low pass filters is seen in the His bundle recording. 4: Electronic Subsystems and Circuitry Electronic Subsystems and Circuitry In electronics, there are 2 basic cir

22、cuit types: series and parallel. A series circuit has one path of current flow, while a parallel circuit has 2 or more. In a series circuit (as pictured here), these rules apply: 1) The same current flows through each part of the circuit 2) The total resistance of a series circuit is equal to the su

23、m of the individual resistances. 3) Voltage applied to a series circuit is equal to the sum of the individual voltage drops 4) The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor. 5) If the circuit is broken at any point, no current will flow.

24、Each rule will now be analyzed a little more closely. Electronic Subsystems and Circuitry In a series circuit: 1) The same current flows through each part of the circuit In a series circuit, the amperage at any point in the circuit is the same. This will help in calculating circuit values using Ohms

25、 Law. You will notice from the diagram that 1 amp continually flows through the circuit. We will get to the calculations in a moment. Electronic Subsystems and Circuitry In a series circuit: 2) The total resistance of a series circuit is equal to the sum of the individual resistances. In a series ci

26、rcuit you need to calculate the total resistance to figure out the current. This is done by adding individual values of each resistor in the series. Here we have three resistors. To calculate the total resistance we use the formula: RT = R1 + R2 + R3 5 + 5 + 10 = 20 Ohms R total is 20 Ohms Electroni

27、c Subsystems and Circuitry In a series circuit: Now with these two rules we will calculate the amperage. Remember I = V / R. We will modify this slightly and say I = V / R total. Lets follow our example figure: RT = R1 + R2 + R3 RT = 20 Ohms I = V / RT I = 20V / 20 Ohms I = 1 Amp If we knew the ampe

28、rage, to calculate the voltage, we can also use Ohms Law V = I x R total V = 1 A x 20 Ohms V = 20 V Electronic Subsystems and Circuitry In a series circuit: Before going further lets define voltage drop“. amount voltage lowers when crossing a component from the negative side to the positive side in

29、a series circuit. Placing a multimeter across a resistor, the voltage drop is the voltage amount read pictured with the red arrow in the diagram. Say a battery is supplying 10 volts to a circuit of two resistors; each having a value of 5 Ohms. According to the previous rules we figure out the total

30、resistance: RT = R1 + R2 = 5 + 5 = 10 Ohms Next we calculate the amperage in the circuit: I = V / RT = 10V / 10 Ohms = 1 Amp Electronic Subsystems and Circuitry In a series circuit: Now that we know the amperage for the circuit (remember the amperage does not change in a series circuit) we can calcu

31、late what the voltage drops across each resistor are using Ohms Law (V = I x R). V1 = 1A x 5 Ohms = 5 V V2 = 1A x 5 Ohms = 5 V Now we get to the next rule. Electronic Subsystems and Circuitry In a series circuit: This simply means that the voltage drops have to add up to the voltage coming from the

32、battery or batteries. V total = V1 + V2 + V3 . In our example here, this means that 5V + 5V = 10V 3) Voltage applied to a series circuit is equal to the sum of the individual voltage drops Electronic Subsystems and Circuitry In a series circuit: 4) The voltage drop across a resistor in a series circ

33、uit is directly proportional to the size of the resistor. This is what we described in the voltage drop section. Voltage drop = Current X (times) resistor size Electronic Subsystems and Circuitry In a series circuit: 5) If the circuit is broken at any point, no current will flow. The best way to ill

34、ustrate this is with a string of light bulbs. If one is burnt out, the whole thing stops working. Electronic Subsystems and Circuitry As mentioned, a parallel circuit (as shown here) is one with several different paths. The parallel circuit has extremely different characteristics than a series circu

35、it Total resistance of a parallel circuit is NOT equal to the sum of the resistors The total resistance in a parallel circuit is always less than any of the branch resistances ALSO: adding more parallel resistances to the paths causes the total resistance to decrease. As you add more and more branch

36、es to the circuit the total current will increase. Why? Remember from Ohms Law that the lower the resistance, the higher the current. Electronic Subsystems and Circuitry In a parallel circuit, the following rules apply: 1. A parallel circuit has two or more paths for current to flow through. 2. Volt

37、age is the same across each component of the parallel circuit. 3. The sum of the currents through each path is equal to the total current that flows from the source. 4. You can find total resistance in a parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +. Rt = R (t)otal Note: T

38、he formula is not as difficult as it looks as you will see 5. If one of the parallel paths is broken, current will continue to flow in all the other paths. Lets look at each of these closer to gain an understanding of parallel circuits. Electronic Subsystems and Circuitry In a parallel circuit: 1. A

39、 parallel circuit has two or more paths for current to flow through. This is self explanatory. Simply remember that PARALLEL means two paths up to thousands of paths. The flow of electricity is divided between each according to the resistance along each route. Electronic Subsystems and Circuitry In

40、a parallel circuit: 2. Voltage is the same across each component of the parallel circuit. You may remember that in a series circuit, the voltage drops across a resistor in series. Not so with a parallel circuit. The voltage will be the same anywhere in the circuit. Electronic Subsystems and Circuitr

41、y In a parallel circuit: 3. The sum of the currents through each path is equal to the total current that flows from the source. If one path is drawing 1 amp and the other is drawing 1 amp, the total is 2 amps at the source. If there are 4 branches in this same 2 amp circuit, then one path may draw 1

42、/4A (.25A), the next 1/4A (.25), the next 1/2A (.5A) and the last 1A. Dont worry, the next rule will show you how to figure this out. Simply remember for now that the branch currents must tally to equal the source current. Electronic Subsystems and Circuitry In a parallel circuit: 1/Rt = 1/R1 + 1/R2

43、 + 1/R3 +. 4. You can find total resistance in a parallel circuit with the following formula: Before we get to the calculations, keep in mind what was said earlier: The total resistance of a parallel circuit is NOT equal to the sum of the resistors (like in a series circuit)”. That said, lets dig in

44、to the formula. We will use a parallel circuit with 3 paths as an example (it can just as easily be 2, 4 or a 1000 resistors in parallel). The power source is providing 10 volts and the value of the resistors are 4 Ohm, 4 Ohm and 2 Ohm. Electronic Subsystems and Circuitry In a parallel circuit: 1/Rt

45、 = 1/R1 + 1/R2 + 1/R3 +. 4. You can find total resistance in a parallel circuit with the following formula: Lets summarize this EXAMPLE for clarity: Voltage = 10V R1 = 4 Ohm R2 = 4 Ohm R3 = 2 Ohm Remember that Rt means Total resistance of the circuit. R1, R2, etc. are Resistor one, Resistor two, etc

46、. Electronic Subsystems and Circuitry In a parallel circuit: 1/Rt = 1/R1 + 1/R2 + 1/R3 +. 4. You can find total resistance in a parallel circuit with the following formula: Now we will apply the formula above to this example: 1 1 1 1 - = - + - + - Rt R1 R2 R3 Therefore: 1 1 1 1 - = - + - + - Rt 4 4

47、2 It is easiest to change the fractions into decimal numbers (example 1 divide by 4 equals .25): 1/Rt = .25 + .25 + .5 1/Rt = 1 Now you have to get rid of the 1 on the left side so. Rt = 1/1 Rt = 1 Ohm Electronic Subsystems and Circuitry In a parallel circuit: 5. If one of the parallel paths is brok

48、en, current will continue to flow in all the other paths. The best way to illustrate this is also with a string of light bulbs in parallel. If one is burnt out, the others stay lit. Review Questions Review Questions II: Cardiac Anatomy and Physiology 1: Cardiac Anatomy Coronary Arteries Right Corona

49、ry Artery Winds around the AV groove Supplies SA node, AV node, right atrium, right ventricle, plus inferior and posterior portion of left ventricle (PDA) Left Main Artery Left anterior descending (LAD) Travels down Inter-ventricular septum Anterior wall of LV Bundle of His Left Circumflex (LCX) Win

50、ds around left AV groove Posterior Sometimes supplies PDA Sternocostal Aspect (AP) Diaphragmatic Aspect (PA) Cardiac Veins Coronary Sinus Winds around the AV groove Parallels LCX Middle Cardiac Vein Parallels PDA Travels down Inter-ventricular septum Posterior Small Cardiac Vein -Parallels RCA Obliq

51、ue Vein of Marshall Anterior Interventricular Vein -Parallels LAD -AKA Great Cardiac Vein Sternocostal Aspect (AP) Diaphragmatic Aspect (PA) Right Atrium Right Atrial appendage Sino-atrial node IVC SVC TVA Coronary sinus AVN/HIS Right Atrium Thebesian valve Crista terminalis Eustachian ridge Fossa O

52、valis Limbus Tendon of Todaro RA and RV RAO position RV Apex SVC IVC Crista Terminalis Right Atrial Appendage Atrio-Ventricular Groove SVC IVC Appendage SA Node Right Atrial Appendage Fossa Ovalis SVC IVC Tricuspid Valve Annulus Coronary Sinus Os Limbus RA and RV RAO position AV Groove and Septum It

53、ems of importance Electrical isolation of Atrium and Ventricle Regions of breakthrough Structural relationships Triangle of Koch Sub-eustachian isthmus AV Groove and Septum AV Groove Interatrial Septum Interventricular Septum TV CT Eustachian Ridge Limbus FO CSO Triangle of Koch Subeustachian Isthmu

54、s Right Ventricle Right Ventricular outflow tract Pulmonic valve Papillary muscles TVA Septum RA and RV RAO position RV Apex SVC IVC Crista Terminalis Right Atrial Appendage Atrio-Ventricular Groove RVOT PA Pulm. Valve RVA LAD Pulmonic Valve Right Ventricular Outflow Tract TVA Papillary Muscle RV Ap

55、ex RV RAO Chordae Tendinae Left Atrium Left Atrial appendage MVA Fossa Ovalis Pulmonary veins Lt Atrial Appendage LV LAA LA LV CS LA, LV, LAA, and CS LL Position Mitral Valve Pulmonary Veins LAA Pulmonary Veins FO RSVP Lt PV Common osLAA MV Left Ventricle Left Ventricular outflow tract Aortic valve

56、Aorta Mitral valve Papillary muscles Chordae tendinae AV LVOT Papillary Muscle Chordae tendinae MVA Papillary Muscle LV Left Lateral Chordae Tendinae LVA LVOT LAALA CS Fluoro imaging 101 Anterior / Posterior Fluoro Image RA corresponds to QRS (mechanical systole of the atrium is approximately 120 ms

57、 after the electrical P wave) x wave: by atrial relaxation. c wave: caused by the closure and bulging of the tricuspid valve into the RA during RV contraction x descent: by the “pulling down” of the atrio-ventricular septum during ventricular ejection. v wave: due to RA filling with a closed tricusp

58、id valve; corresponds to the EKG T wave y descent: created by opening of the tricuspid valve with flow of blood into the RV. Right Ventricle The normal right ventricular pressure curve can be divided into 7 phases as indicated by the small letters: a: isometric contraction b: maximal ejection (measu

59、rement of peak systole usually 17-32 mmHg) c: reduced ejection d: isometric relaxation e: rapid inflow f: diastasis g: end-diastole (point at which end diastolic pressure is measured usually 5 mmHg) Normal RV: Systole: 17-32 Diastole: 0-5 Tricuspid valve closes Pulmonic valve opens Pulmonic valve cl

60、oses Tricuspid valve opens Tricuspid valve closes Pulmonary Artery PA Pressure normal pulmonary artery pressure curve displayed A steep rise in the curve to the summit follows right ventricular ejection This in turn is followed by a gradual fall (dichrotic notch), after which pressure gradually decl