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These videos can be found on both their website and YouTube channel. There are also some other types of resources on their website, such as summary notes and practice questions. With a wavelength of 1. On the WTL scale, we add 0. A straight line is now drawn from the origin though the 0. A compass is then used to measure the distance between the origin and zin. With this distance set, the compass is then used to scribe off the same distance from the origin to the load impedance, along the line between the origin and the 0.

The difference in imaginary parts arises from uncertainty in reading the chart in that region. This is close to the value of the VSWR, as we found earlier. Next, yL is inverted to find zL by transforming the point halfway around the chart, using the compass and a straight edge.

This is close to the computed inverse of yL , which is 1. Now, the position of zL is read on the outer edge of the chart as 0. The point is now transformed through the line length distance of 1. The final reading on the WTG scale after the transformation is found through 0.

Drawing a line between this mark on the WTG scale and the chart center, and scribing the compass arc length on this line, yields the normalized input impedance. I will specify answers in terms of wavelength. On the WTG scale, we read the zL location as 0.

Moving from here toward the generator, we cross the positive R axis at which the impedance is purely real and greater than 1 at 0. The distance is then 0. What is s on the remainder of the line? This will be just s for the line as it was before. This would return us to the original point, requiring a complete circle around the chart one-half wavelength distance. With the aid of the Smith chart, plot a curve of Zin vs. Then, using a compass, draw a circle beginning at zL and progressing clockwise to the positive real axis.

On the chart, radial lines are drawn at positions corresponding to. The intersections of the lines and the circle give a total of 11 zin values. The table below summarizes the results. A fairly good comparison is obtained. We mark this on the positive real axis of the chart see next page. The load position is now 0.

A line is drawn from the origin through this point on the chart, as shown. We then scribe this same distance along the line drawn through the.

A line is drawn from the origin through this location on the chart. Drawing a line from the chart center through this point yields its location at 0. Alternately, use the s scale at the bottom of the chart, setting the compass point at the center, and scribing the distance on the scale to the left.

This distance in wavelengths is just the load position on the WTL scale, since the starting point for this scale is the negative real axis. So the distance is 0. Transforming the load through this distance toward the generator involves revolution once around the chart 0.

A line is drawn between this point and the chart center. This is plotted on the Smith chart below. We then set on the compass the distance between yL and the origin.

The same distance is then scribed along the positive real axis, and the value of s is read as 2. We note a reading on that scale of about 0. To this we add 0. A line drawn from the 0. This is at the zero position on the WTL scale.

The load is at the approximate 0. The wavelength on a certain lossless line is 10cm. We read the input location as slightly more than 0. The line length of 12cm corresponds to 1. Thus, to transform to the load, we go counter-clockwise twice around the chart, plus 0.

A line is drawn to the origin from that position, and the compass with its previous setting is scribed through the line. A standing wave ratio of 2. Probe measurements locate a voltage minimum on the line whose location is marked by a small scratch on the line.

When the load is replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm toward the source from the scratch. Suppose that the scratch locates the first voltage minimum. With the short in place, the first minimum occurs at the load, and the second at 25 cm in front of the load. The effect of replacing the short with the load is to move the minimum at 25 cm to a new location 7 cm toward the load, or at 18 cm.

This is a possible location for the scratch, which would otherwise occur at multiples of a half-wavelength farther away from that point, toward the generator. Using the Smith chart see below we first draw a line from the origin through the 0. As a check, I will do the problem analytically. A 2-wire line, constructed of lossless wire of circular cross-section is gradually flared into a coupling loop that looks like an egg beater.

At the point X, indicated by the arrow in Fig. A probe is moved along the line and indicates that the first voltage minimum to the left of X is 16cm from X. With the short circuit removed, a voltage minimum is found 5cm to the left of X, and a voltage maximum is located that is 3 times voltage of the minimum.

Use the Smith chart to determine: a f : No Smith chart is needed to find f , since we know that the first voltage minimum in front of a short circuit is one-half wavelength away. This point is then transformed, using the compass, to the negative real axis, which corresponds to the location of a voltage minimum. On the chart, we now move this distance from the Vmin location toward the load, using the WTL scale.

A line is drawn from the origin through the 0. With a short circuit replacing the load, a minimum is found at a point on the line marked by a small spot of puce paint.

The 1m distance is therefore 3. Therefore, with the actual load installed, the Vmin position as stated would be 3. This being the case, the normalized load impedance will lie on the positive real axis of the Smith chart, and will be equal to the standing wave ratio. This point is to be transformed to a location at which the real part of the normalized admittance is unity. The stub is connected at either of these two points. The stub input admittance must cancel the imaginary part of the line admittance at that point.

This point is marked on the outer circle and occurs at 0. The length of the stub is found by computing the distance between its input, found above, and the short-circuit position stub load end , marked as Psc.

The length of the main line between its load and the stub attachment point is found on the chart by measuring the distance between yL and yin2 , in moving clockwise toward generator. We find the stub length by moving from Poc to the point at which the admittance is j 0. This occurs at 0. The attachment point is found by transforming yL to yin1 , where the former point is located at 0.

The lossless line shown in Fig. For the line to be matched, it is required that the sum of the normalized input admittances of the shorted stub and the main line at the point where the stub is connected be unity.

So the input susceptances of the two lines must cancel. To find the stub input susceptance, use the Smith chart to transform the short circuit point 0.

This line is one-quarter wavelength long, so the normalized load impedance is equal to the normalized input admittance. To cancel the input normalized susceptance of We therefore write 2. The two-wire lines shown in Fig. In this case, we have a series combination of the loaded line section and the shorted stub, so we use impedances and the Smith chart as an impedance diagram.

The requirement for matching is that the total normalized impedance at the junction consisting of the sum of the input impedances to the stub and main loaded section is unity.

In the transmission line of Fig. Determine and plot the voltage at the load resistor and the current in the battery as functions of time by constructing appropriate voltage and current reflection diagrams: Referring to the figure, closing the switch launches a voltage wave whose value is given by Eq.

So the voltage wave traverses the line and does not reflect. The voltage reflection diagram would be that shown in Fig. Likewise, the current reflection diagram is that of Fig. First, the load voltage is found by adding voltages along the right side of the voltage diagram at the indicated times.

Second, the current through the battery is found by adding currents along the left side of the current reflection diagram. Both plots are shown below, where currents and voltages are expressed to three significant figures. Note also that when the switch is opened, the reflection coefficient at the generator end of the line becomes unity.

The reflection diagram is now constructed in the usual manner, and is shown on the next page. The load voltage as a function of time is found by accumulating voltage values as they are read moving up along the right hand boundary of the chart. The resulting function, plotted just below the reflection diagram, is found to be a sequence of pulses that alternate signs. In the charged line of Fig. This problem accompanies Example Plots of the voltage and current at the resistor are then found by accumulating values from the left sides of the two charts, producing the plots as shown.

No reflection occurs at the load end, since the load is matched to the line. The reflection diagram and load voltage plot are shown below. A simple frozen wave generator is shown in Fig. Determine and plot the load voltage as a function of time: Closing the switches sets up a total of four voltage waves as shown in the diagram below.

The reflection diagram is drawn and is used to construct the load voltage with time by accumulating voltages up the right hand vertical axis. How many modes propagate?

For the first mode, we are given 2nd 0. If the operating frequency is 32 GHz, which modes will propagate? For the guide of Problem Assume a propagation distance of 10 cm: From Problem A parallel-plate guide is partially filled with two lossless dielectrics Fig. At a certain frequency, it is found that the TM1 mode propagates through the guide without suffering any reflective loss at the dielectric interface.

The So the answer is yes. In the guide of Problem To summarize, as frequency is lowered, the ray angle in guide 1 decreases, which leads to the incident angle at the interface increasing to eventually reach and surpass the critical angle. This defines the cutoff condition in guide 2. The cutoff frequency for mode mp is, using Eq. We require that the frequency lie between the cutoff frequencies of the T E10 and T E01 modes.

We note first that f must be greater than fc01 to support both modes, but must be less than the cutoff frequency for the next higher order mode. Two rectangular waveguides are joined end-to-end.

To assure single mode operation in both guides, the operating frequency must be above cutoff for TE10 in both guides, and below cutoff for the next mode in both guides. An air-filled rectangular waveguide is to be constructed for single-mode operation at 15 GHz. Integrate the result of Problem This was considered in the discussion leading to Eq. The pulse propagates in a lossless single mode rectangular guide which is air-filled and in which the 10 GHz operating frequency is 1.

Using the result of Problem At cutoff, the mode propagates in the slab at the critical angle, which means that the phase velocity will be equal to that of a plane wave in the upper or lower media of index n2. The reasoning of part a applies to all modes, so the answer is the same, or 2. An asymmetric slab waveguide is shown in Fig. The minimum wave angle is thus determined by the greater of the two critical angles.

By what percentage must the core radius of the new fiber differ from the old one, and should it be larger or smaller? Therefore, the percentage reduction required in the core radius will be 1. The above two equations can be simplified by these substitutions, while dropping all amplitude terms that are common to both.

Prepare a curve, r vs. Both are circles. Since we are in the far zone, 84 applies. The radiated power and radiation resistance are down to a factor of 0. A dipole antenna in free space has a linear current distribution. If the length is 0. A monopole antenna in free space, extending vertically over a perfectly conducting plane, has a linear current distribution.

If the length of the antenna is 0. This reduces the radiation resistance of the equivalent dipole antenna by a factor of one-half. Additionally, the linear current distribution reduces the radiation resistance of a dipole having uniform current by a factor of one-fourth.

These lines can be considered essentially parallel, and so the difference in their lengths is. The construction and arguments are similar to those used in the discussion of the electric dipole in Sec. The electric field is now the result of part a, modified by including a shorter distance, r, in the phase term only. Again, constructing a line between B and P , we find, using the same arguments as in part b, that the length of this line is approximately 0.

Solution william h[1]. Engineering Electromagnetics - 7th Edition - William H. Rubisco, insulin, immunoglobulins, rhodopsin, collagen andspider silk as examples of the range of protein functions. Denaturation of proteins by heat or by deviation of pH from the optimum. Drawing molecular diagrams to show the formation of a peptide bond. Titin the different possible combinations of polypeptides are effectively infinite.

Also, the genes are actually longer as they contain non-coding regions that don't code for the polypeptide. Full name is ribulose bisphosphate carboxylase. It is one of the most abundant and important enzyme in the world 28 Immunoglobulins Immunity: also known as antibodies.

They are Y shaped proteinscells to identify and neutralize foreign pathogens like bacteria and viruses. They send signals to immune systems to come and destroy Rhodopsin Receptors: rhodopsin is a biological pigment in the photoreceptor cells of the retina. When the retinal absorbs light through the eye, it changes its shape and the shape of the opsin. This sends a nerve impulse through the optic nerve to the brain Collagen Tensile strengthen: main structure molecule in various connective tissues such as skin, blood vessels and ligaments.

Forms bones and teeth, to prevent crack Spider silk Tensile strengthen: spider silk consists of many different types with different functions and areused in the spokes of a web and when a spider suspends itself. Enzymes for removing stains in clothing detergent 2.

Monoclonal antibodies for pregnancy tests 3. Insulin for treating diabetics 4. A genome is unique to most individuals identical twins and clones share a genome Definition of proteome All of the proteins produced by a cell, a tissue or an organism. U2 Enzyme catalysis involves molecular motion and the collision of substrateswith the active site.

U3 U4 U5 Temperature, pH and substrate concentration affect the rate of activity ofenzymes. Enzymes can be denatured Immobilized enzymes are widely used in industry. A1 Application: Methods of production of lactose-free milk and its advantages.

A2 Skill: Design of experiments to test the effect of temperature, pH andsubstrate concentration on the activity of enzymes. A3 Skill: Experimental investigation of a factor affecting enzyme activity. Practical 3 Definitions Enzyme: A biological catalyst which speeds up the rate of a chemical reaction bylowering the activation energy.

Substrate: Reactant in a biochemical reaction. Denaturation: A structural change in a protein that results in a loss of biological properties. Most of these bonds and interactions are relatively weak and they can be disrupted or broken. This results in a change to the conformation of the protein, which is called denaturation and is permanent. Immobilized enzymes are widely used in industry. Reasons for using enzymes: 1. Convenience — only small amounts of proteins dissolve in the reactions leaving only solvent and the products.

This means the enzymes and products can be easily separated 2. Economics — The immobilized enzymes can be easily removed and recycled from the solution, saving money. Particular useful in the removal of lactase in the production of Lactose Free Milk. Stability — Immobilized enzymes generally have a greater thermal and chemical stability than the soluble form of the enzyme 4. Reaction rate is faster because substrates can be exposed to a higher concentration of enzymes 31 Individual nucleotides are referred to as monomers and always consist of three major parts: one phosphate group, one 5-carbon monosaccharide, and a single nitrogenous base.

However, there are five possible nitrogenous bases. Building models allowed them to visualize the molecule and see how well it fitted available evidence. U2 Helicase unwinds the double helix and separates the two strands by breaking hydrogen bonds.

Translation is the synthesis of polypeptides on ribosomes. The amino acid sequence of polypeptides is determined by mRNA according to the genetic code. Codons of three bases on mRNA correspond to one amino acid in a polypeptide. A2 Production of human insulin in bacteria as an example of the universality of the genetic code allowing gene transfer between species.

Helicase unwinds the double helix and separates the two strands by breaking hydrogen bonds. DNA polymerase links nucleotides together to form a new strand, using the pre-existing strand as a template. This protein family consists of multiple polypeptides sub-units. The polymerisation reaction is a condensation reaction.

DNA polymerase moves in opposite directions on each strands. It is useful when only a small amount of DNA is available for testing. These tRNA have no attached amino acid. Production of human insulin in bacteria as an example of the universality of the genetic code allowing gene transfer between species. U2 ATP from cell respiration is immediately available as a source of energy in thecell.

U3 Anaerobic cell respiration gives a small yield of ATP from glucose. U4 Aerobic cell respiration requires oxygen and gives a large yield of ATP fromglucose. A1 Use of anaerobic cell respiration in yeasts to produce ethanoland carbon dioxide in baking. A2 Lactate production in humans when anaerobic respiration isused to maximize the power of muscle contractions. S1 Analysis of results from experiments involving measurement ofrespiration rates in germinating seeds or invertebrates using a respirometer.

U2 Visible light has a range of wavelengths with violet the shortest wavelengthand red the longest. U3 Chlorophyll absorbs red and blue light most effectively and reflects greenlight more than other colours. U4 Oxygen is produced in photosynthesis from the photolysis of water. U5 Energy is needed to produce carbohydrates and other carbon compoundsfrom carbon dioxide. U6 Temperature, light intensity and carbon dioxide concentration are possiblelimiting factors on the rate of photosynthesis.

S1 S2 S3 Drawing an absorption spectrum for chlorophyll and an actionspectrum for photosynthesis. Design of experiments to investigate the effect of limiting factors onphotosynthesis. Separation of photosynthetic pigments by chromatograph. It shows which wavelength of light is most effectively used during photosynthesis. Topic 3: Genetics 3. U2 A gene occupies a specific position on a chromosome.

U3 The various specific forms of a gene are alleles. U4 Alleles differ from each other by one or only a few bases. U5 New alleles are formed by mutation.