Thank you so much from Korea !!!!! I totally gave up understanding Stub when I had learned it from my professor, but you save my credit !!!!! Thanks a lot ^^
I appreciate the video and it inspired me to learn more about Open and Shorted stubs! In doing so I verified the original calculation, but also first mistakenly performed a Series Stub (Open/Short) calculation instead of a Parallel Stub (Open/Short). I've include the Series below if anyone is interested. At first I thought the video was incorrect while I was still getting refreshed on Smith Charts! Normalized Impedance of Zin1norm = 1+j1.6 ohms. If one is so inclined to think of this in terms of Admittance, it can be converted to Yin1norm = 0.28-0.45j siemens. The equations for loc and lsc are only in terms of Admittance and they are missing the j term. This makes it difficult to know whether or not taking the reciprocal is going to cause a negative to appear and change the result of the arctangent. I have rederived all of the equations in terms of both Admittance and Impedance and kept the j term visible. This j that is in the numerator/denominator will algebraically cancel out (see below) when the appropriate Z or Y is inserted into the denominator/numerator. Remember that j^2 = -1 and that 1/j is -j. Any j/j can just go away and become 1. Open Circuit Stubs (derived from Zoc = -j*Z0cot(Bl) where B = 2pi/wavelength. I normalized this into Zocnorm = -j*cot(Bl) and did some Algebra/Trig to solve for the length l.) loc = 1/2pi * wavelength * arctan(-j / Zocnorm) in terms of Impedance Z loc = 1/2pi * wavelength * arctan(-j * Yocnorm) in terms of Admittance Y Short Circuit Stubs (derived from Zsc = j*Z0tan(Bl) where B = 2pi/wavelength. I normalized this into Zscnorm = j*tan(Bl) and did some Algebra/Trig to solve for the length l.) lsc = 1/2pi * wavelength * arctan(Zscnorm / j) in terms of Impedance Z lsc = 1/2pi * wavelength * arctan(-j / Yscnorm) in terms of Admittance Y Going through this example: After plotting the normalized Zlnorm = 0.5 - j1 and drawing a circle, the 2x intersection points were found where this circle hits the circle of Real Resistive part = 1.0. For a Series implentation, use Real(Z) = 1. These two intersections occurred at 0.179 wavelengths and 0.321 wavelengths just reading off the Smith Chart. We are starting at 0.365 wavelengths. That means for d1 the total wavelengths travelled is (0.50 - 0.365) + 0.179 = 0.314 wavelengths for d1. That means for d2 the total wavelengths travelled is (0.50 - 0.365) + 0.321 = 0.456 wavelengths for d2. Choosing either 0.314 for d1 or 0.456 wavelengths for d2 both will give us a Real part that is normalized to 1.0 (50 ohms resistive for most matching cases). However it will still have a j term at this point. Let's choose d1 = 0.314 wavelengths for the series transformer section since it is shorter and saves on coax. Looking at the intersection #1 on the Smith Chart, this occurs at a Normalized Impedance of Zin1norm = 1 + j1.6. This is the normalized impedance seen at the input after our new series section "d1". We could also think of this as an Admittance of Yin1norm = 0.28-0.45j, but for Series circuits, keeping things in terms of Impedance makes addition easier. In order to cancel out this j term of +j1.6, we will need the opposite which is -j1.6. Therefore our stub that we will design as an open circuit or a short circuit will have to come out to a normalized impedance of -j1.6. In other words Zscnorm = Zocnorm = -j1.6. Either design can be used! If we want to use an Open Circuit Stub to cancel out the remaining j after using d1: loc = 1/2pi * wavelength * arctan(-j / Zocnorm) loc = 1/2pi * wavelength * arctan(-j / -j1.6) loc = 1/2pi * wavelength * arctan( 1/ 1.6) loc = 0.0889 wavelengths for an open circuit stub after using a series transformer d1 = 0.314 wavelengths. This will match to 1.0 + 0j. When using Z0 = 50 ohms, it will match to 50 ohm + 0j ohm! If we want to use a Short Circuit Stub to cancel out the remaining j after using d1: lsc = 1/2pi * wavelength * arctan(Zscnorm / j) lsc = 1/2pi * wavelength * arctan(-j1.6 / j) lsc = 1/2pi * wavelength * arctan(-1.6 / 1) lsc = 1/2pi * wavelength * arctan(-1.6) lsc = -0.1611 wavelengths therefore -0.1611+0.50 = 0.3389 wavelengths. If we use 0.3389 wavelengths for a short circuit stub after using a series transformer d1 = 0.314 wavelengths, this will match to 1.0 + 0j. When using Z0 = 50 ohms, it will match to 50 ohm + 0j ohm! Thank you from N9AJD!
Thank you for commenting! Your issues with my content appear to stem from a problem converting between admittance and impedance. You might want to review manipulation of complex numbers. If you have a complex impedance Z = M + jN, the reciprocal of that impedance (the admittance) is Y = 1/Z, which is equal to 1/(M + jN), which is NOT equal to (1/M) - j (1/N) (except in the very specific case where Z = 1+j0.) To quickly address your points: 1) "the circle where Re(y) = 1 or Re(Z) = 1 should be on the right side of the Smith Chart". These are two different circles. The circle Re(Z) = 1 is on the right side. The circle Re(Y) = 1 is, as shown, on the left. 2) "the information stated is that this is the Normalized Admittance after the new series transformer "d". However this is actually the Normalized Impedance of Zin1norm = 1+j1.6. If one is so inclined to think of this in terms of Admittance, they can take the reciprocal of each term and get Yin1norm = 1 - j0.625, remembering that 1/j is -j." You are incorrect. Admittance is in this case read directly off the Smith Chart (look at the blue graph). Again, I believe your problem at least partly stems from a misunderstanding of complex numbers. 1/(1+j1.6) is not equal to (1/1) - j(1/1.6). 3) "A third issue that I see is that the equations for loc and lsc are only in terms of Admittance and they are missing the j term. This makes it difficult to know whether or not taking the reciprocal is going to cause a negative to appear and change the result of the arctangent." Nomenclature can definitely get confusing! This is why it is important to carefully define variables as they appear. In this case, at time 2:06 in this video, I define the variable Y1, (which is later used to describe loc and lsc) so that you can clearly see that Y1 = Yin,stub/(-j). I hope this helps to clarify!
@@emviso I appreciate the quick and detailed response, and you did catch a goof on my part regarding the conversion in Rectangular form between Impedance and Admittance. I had forgotten about the way I normally do it, which is to convert to Polar and take the reciprocal of the magnitude and turn the angle negative, and then convert back to Rectangular. So when I said Yin1norm = 1 - j0.625, it should have been 0.28 - 0.45j. I've now corrected this above to clarify. Showing the conversion from Zin1norm to Yin1norm: Zin1norm = 1 + j1.6 ohms MagnitudePolar = sqrt(1^2 + 1.6^2) = 1.8868 AnglePolar = arctan(1.6/1) = 58 degrees AdmitanceMagnitudePolar = 1/1.8868 = 0.53 AdmittanceAnglePolar = 0 - 58 degrees = -58 degrees Yin1norm = 0.53*cos(-58 degrees) + j*0.53*sin(-58 degrees) = 0.28 - 0.45j siemens. I've still be scratching my head as to why I think my #s are correct and even went to the extent below to check answers with the Quarter Wave transformer equation. At this point I think it looks like I am following more of an approach to a Series Stub while this video is showing the process of a Parallel Stub. I've spent all night looking at the Admittance version of the Smith Chart and I think the blue is finally starting to pop out and I'm seeing more Smith Charts within Smith Charts! I'll keep working on it and see if I can complete the Parallel Stub, as I'm thinking that I've been stuck in Impedance land which is causing me to do a process like a Series Stub from what I'm seeing on another example. Using alternate method to check answer (looks like this is verifying both my work for Series Stub and also the video's work for Parallel Stub!) Zin = Z0*(ZL + iZ0tan(Bl))/(Z0 + iZLtan(Bl)) where B = 2pi/wavelength l = length in wavelengths Z0 = 50 ohms for example ZL = 50(0.5-j1) = 25 -j50 for the example shown of ZL = 0.5-j1 In my first comment where I detailed the Impedance version of the analysis, I stated that Zin1norm = 1 + j1.6, so I would hope that we find the result is 50(1 + j1.6) = 50 + j80. When I plug in the l = d1 = 0.314 wavelengths into the above equation I get 50.875 + 79.74j out of the TI-84! This confirms that we've got the 50 ohm resistive part matched but there is still the j term. When I plug in the l = d2 = 0.456 wavelengths into the above equation, the result is then 50.66 - 79.577j. This confirms that we've got the 50 ohm resistive part matched and that the resulting j is the opposite as d1's. Trying the video's result of 0.0601 wavelengths gives a result of 14.54 - 23.66j which is clearly not matched to 50 ohm resistive. Trying the video's result of 0.207 wavelengths gives a result of 14.31 + 22.71j which confirms the process was done correctly to mirror the result across the horizontal axis, but it did not give a 50 ohm resistive match. However looking deeper into stubs and seeing examples on Series Stubs, I think I'm trying to follow that process by making the Rin = Z0, which agrees with my results of getting close to 50 ohms for the resistive part. Instead with the Parallel process it is trying to make Gin = Y0 = 1/Z0, which seems almost the same thing at first but it isn't! I definitely am agreeing that if you take for example one of the above impedances and take the reciprocal to get the admittance, 1/(14.31 + 22.71j), you get 0.01986 - 0.0315j which shows the Gin = 1/Z0 = 1/50 = 0.02 which is very close to 0.01986! _________
@@Pelnied Okay - so I might be misunderstanding what you're doing, but I think your problem is here: Yin1norm (the normalized input admittance after the addition of the series transmission line of length d1) is Yin1norm = 1 + j1.6. That comes straight from the chart, if you look at the blue lines/numbers. Impedance is shown in red; admittance is shown in blue. You don't have to convert it to admittance - it's already admittance. It's hard to read the numbers off the video - you might want to print yourself a copy of the impedance/admittance chart and work through it on paper yourself.
@@emviso I've gone back and confirmed the video has it CORRECT for the Parallel Stub implementation. I had mistakenly stuck to using Impedance and Real(Z) = 1 which caused me to do a Series Stub which was not the intention of the video as it requires the stub to be placed in series with one side of the transmission line. For Series Stub (Impedance): d1 = 0.313130325 wavelengths d2 = 0.456661387 wavelengths @d1 a series open stub would be 0.089754259 wavelengths @d1 a series shorted stub would be 0.339754259 wavelengths For Parallel Stub (Admittance): d1 = 0.0631303 wavelengths d2 = 0.20666139 wavelengths @d1 a parallel open stub would be 0.339754 wavelengths shown at 5:09 @d1 a parallel shorted stub would be 0.08975 wavelengths I'm glad I stuck to it and kept going back and forth to check answers, as I've now created an automated way to get the results using the TI-83's graphing approach and ability to find intersections so now I don't need to even draw the Smith Chart to find the intersections or to read off any new Normalized values. I can directly calculate where the intersections are at for d1 and d2 and then determine their theoretical normalized admittance or impedance. This gives more precision and accuracy than using the chart itself to eyeball the numbers. I still love the graphical approach and the visual representation that Smith Charts give in seeing how these circuits transform!
You could, and perhaps should, have used the Smith chart to calculate the length of the stubs without the use of your formula. It is easy and is one more of the magic things of the Smith chart. For those unfamiliar with the method, you just mark out the value you need for elimination of the imaginary part. For instance -j1.6 on the outer circle, draw a line from center thru the mark and out to the WTG circle, read off the WTG. For an open stub you start from where r = 0 (or g=infinity) and measure the distance needed to reach the desired imaginary part going clockwise (WTG).
Great video. So if i get this, once you know the distance from the load you can place a stub of open wire feed-line or relevant coax at ANY 1/2 wavelength plus the distance calculated from the antenna / load? The thing i would really like to know is can one keep the stub position at a constant, say any exact 1/2 wavelength and ONLY change the stub length to match the load with the rx/tx? Or is it very important to find an exact length along the feedline where to add the Tee connector / branch? I only ask because my mathematics skills are terrible, i could potentially use meters to tune the stub if location is not empirical. Thanks.
You can calculate d1 and d2 directly without needing the chart. And you can also then calculate the new normalized admittance or normalized impedance without the chart, which then allows you to calculate the length needed for the open/short stub. In a TI-83 I did the following for a Parallel Stub to find d1 and d2 directly without needing the chart by using a Characteristic Impedance, Z0 of 50 ohms and then storing that for O. I also the unnormalized the ZLnorm of 0.5 -j1 and stored it as a variable "L" as 25-j50. 50->O (25-j50)->L Go to the graphing "Y=" feature to set up two functions, y1 and y2 y1 = real(((L+i*O*tan(x))/(O+i*L*tan(x)))^-1) the inverse "^-1" turns the Normalized Impedance into Admittance for use with Parallel Stubs. Just get rid of it to have "))))" at the end if you want to use Normalized Impedance for Series Stubs instead. y2 = 1 We use this to find where the Real part of y1 is going to become 1. 2nd, calc, #5 Intersect enter, enter, enter This usually gives you the first intersection, but may give you the 2nd, depending on how it repeats. Remember if you get a negative result after dividing by 2*pi, you can add 0.5 to the result to turn it positive. Also any value you can also subtract 0.5 if you happened to have gone too far and want a shorter match closer to x = 0. Use Zoom In on the calculator to get a better view of the graph around x=0 to see the 2x intersections better. The following window works well for me: Xmin = -2.7 Xmax = 2.7 Ymin = -1.5 Ymax = 3.5 Perform the Intersection operation until you get both of the intersections close to x = 0. x1 = 0.39665953 x1/(2*pi) = 0.0631303 wavelengths for d1 x2 = 1.2984918 x2/(2*pi) = 0.20666139 wavelengths for d2. Then if you wanted to know the Normalized Admittance instead of reading it from the Smith Chart @ d1 (((L+i*O*tan(2*pi*0.0631303))/(O+i*L*tan(2*pi*0.0631303))))^-1 This gave 0.9999994908 + 1.581138428j which I rounded to about 1 + 1.58114j. The video shows around 1+1.6j from eyeballing the Smith Chart at 4:25.
Even though I didn't understand all of it I understood much more than I did going into the video, thank you for your efforts, great job.
Thank you so much from Korea !!!!! I totally gave up understanding Stub when I had learned it from my professor, but you save my credit !!!!! Thanks a lot ^^
You explain to me in five minutes what it took my professor a whole semester
I appreciate the video and it inspired me to learn more about Open and Shorted stubs!
In doing so I verified the original calculation, but also first mistakenly performed a Series Stub (Open/Short) calculation instead of a Parallel Stub (Open/Short). I've include the Series below if anyone is interested.
At first I thought the video was incorrect while I was still getting refreshed on Smith Charts!
Normalized Impedance of Zin1norm = 1+j1.6 ohms. If one is so inclined to think of this in terms of Admittance, it can be converted to Yin1norm = 0.28-0.45j siemens.
The equations for loc and lsc are only in terms of Admittance and they are missing the j term. This makes it difficult to know whether or not taking the reciprocal is going to cause a negative to appear and change the result of the arctangent. I have rederived all of the equations in terms of both Admittance and Impedance and kept the j term visible. This j that is in the numerator/denominator will algebraically cancel out (see below) when the appropriate Z or Y is inserted into the denominator/numerator. Remember that j^2 = -1 and that 1/j is -j. Any j/j can just go away and become 1.
Open Circuit Stubs (derived from Zoc = -j*Z0cot(Bl) where B = 2pi/wavelength. I normalized this into Zocnorm = -j*cot(Bl) and did some Algebra/Trig to solve for the length l.)
loc = 1/2pi * wavelength * arctan(-j / Zocnorm) in terms of Impedance Z
loc = 1/2pi * wavelength * arctan(-j * Yocnorm) in terms of Admittance Y
Short Circuit Stubs (derived from Zsc = j*Z0tan(Bl) where B = 2pi/wavelength. I normalized this into Zscnorm = j*tan(Bl) and did some Algebra/Trig to solve for the length l.)
lsc = 1/2pi * wavelength * arctan(Zscnorm / j) in terms of Impedance Z
lsc = 1/2pi * wavelength * arctan(-j / Yscnorm) in terms of Admittance Y
Going through this example:
After plotting the normalized Zlnorm = 0.5 - j1 and drawing a circle, the 2x intersection points were found where this circle hits the circle of Real Resistive part = 1.0. For a Series implentation, use Real(Z) = 1.
These two intersections occurred at 0.179 wavelengths and 0.321 wavelengths just reading off the Smith Chart. We are starting at 0.365 wavelengths.
That means for d1 the total wavelengths travelled is (0.50 - 0.365) + 0.179 = 0.314 wavelengths for d1.
That means for d2 the total wavelengths travelled is (0.50 - 0.365) + 0.321 = 0.456 wavelengths for d2.
Choosing either 0.314 for d1 or 0.456 wavelengths for d2 both will give us a Real part that is normalized to 1.0 (50 ohms resistive for most matching cases). However it will still have a j term at this point. Let's choose d1 = 0.314 wavelengths for the series transformer section since it is shorter and saves on coax.
Looking at the intersection #1 on the Smith Chart, this occurs at a Normalized Impedance of Zin1norm = 1 + j1.6. This is the normalized impedance seen at the input after our new series section "d1". We could also think of this as an Admittance of Yin1norm = 0.28-0.45j, but for Series circuits, keeping things in terms of Impedance makes addition easier.
In order to cancel out this j term of +j1.6, we will need the opposite which is -j1.6. Therefore our stub that we will design as an open circuit or a short circuit will have to come out to a normalized impedance of -j1.6. In other words Zscnorm = Zocnorm = -j1.6. Either design can be used!
If we want to use an Open Circuit Stub to cancel out the remaining j after using d1:
loc = 1/2pi * wavelength * arctan(-j / Zocnorm)
loc = 1/2pi * wavelength * arctan(-j / -j1.6)
loc = 1/2pi * wavelength * arctan( 1/ 1.6)
loc = 0.0889 wavelengths for an open circuit stub after using a series transformer d1 = 0.314 wavelengths. This will match to 1.0 + 0j. When using Z0 = 50 ohms, it will match to 50 ohm + 0j ohm!
If we want to use a Short Circuit Stub to cancel out the remaining j after using d1:
lsc = 1/2pi * wavelength * arctan(Zscnorm / j)
lsc = 1/2pi * wavelength * arctan(-j1.6 / j)
lsc = 1/2pi * wavelength * arctan(-1.6 / 1)
lsc = 1/2pi * wavelength * arctan(-1.6)
lsc = -0.1611 wavelengths therefore -0.1611+0.50 = 0.3389 wavelengths. If we use 0.3389 wavelengths for a short circuit stub after using a series transformer d1 = 0.314 wavelengths, this will match to 1.0 + 0j. When using Z0 = 50 ohms, it will match to 50 ohm + 0j ohm!
Thank you from N9AJD!
Thank you for commenting! Your issues with my content appear to stem from a problem converting between admittance and impedance. You might want to review manipulation of complex numbers.
If you have a complex impedance Z = M + jN, the reciprocal of that impedance (the admittance) is Y = 1/Z, which is equal to 1/(M + jN), which is NOT equal to (1/M) - j (1/N) (except in the very specific case where Z = 1+j0.)
To quickly address your points:
1) "the circle where Re(y) = 1 or Re(Z) = 1 should be on the right side of the Smith Chart".
These are two different circles. The circle Re(Z) = 1 is on the right side. The circle Re(Y) = 1 is, as shown, on the left.
2) "the information stated is that this is the Normalized Admittance after the new series transformer "d". However this is actually the Normalized Impedance of Zin1norm = 1+j1.6. If one is so inclined to think of this in terms of Admittance, they can take the reciprocal of each term and get Yin1norm = 1 - j0.625, remembering that 1/j is -j."
You are incorrect. Admittance is in this case read directly off the Smith Chart (look at the blue graph). Again, I believe your problem at least partly stems from a misunderstanding of complex numbers. 1/(1+j1.6) is not equal to (1/1) - j(1/1.6).
3) "A third issue that I see is that the equations for loc and lsc are only in terms of Admittance and they are missing the j term. This makes it difficult to know whether or not taking the reciprocal is going to cause a negative to appear and change the result of the arctangent."
Nomenclature can definitely get confusing! This is why it is important to carefully define variables as they appear. In this case, at time 2:06 in this video, I define the variable Y1, (which is later used to describe loc and lsc) so that you can clearly see that Y1 = Yin,stub/(-j).
I hope this helps to clarify!
@@emviso I appreciate the quick and detailed response, and you did catch a goof on my part regarding the conversion in Rectangular form between Impedance and Admittance. I had forgotten about the way I normally do it, which is to convert to Polar and take the reciprocal of the magnitude and turn the angle negative, and then convert back to Rectangular.
So when I said Yin1norm = 1 - j0.625, it should have been 0.28 - 0.45j. I've now corrected this above to clarify.
Showing the conversion from Zin1norm to Yin1norm:
Zin1norm = 1 + j1.6 ohms
MagnitudePolar = sqrt(1^2 + 1.6^2) = 1.8868
AnglePolar = arctan(1.6/1) = 58 degrees
AdmitanceMagnitudePolar = 1/1.8868 = 0.53
AdmittanceAnglePolar = 0 - 58 degrees = -58 degrees
Yin1norm = 0.53*cos(-58 degrees) + j*0.53*sin(-58 degrees) = 0.28 - 0.45j siemens.
I've still be scratching my head as to why I think my #s are correct and even went to the extent below to check answers with the Quarter Wave transformer equation. At this point I think it looks like I am following more of an approach to a Series Stub while this video is showing the process of a Parallel Stub. I've spent all night looking at the Admittance version of the Smith Chart and I think the blue is finally starting to pop out and I'm seeing more Smith Charts within Smith Charts!
I'll keep working on it and see if I can complete the Parallel Stub, as I'm thinking that I've been stuck in Impedance land which is causing me to do a process like a Series Stub from what I'm seeing on another example.
Using alternate method to check answer (looks like this is verifying both my work for Series Stub and also the video's work for Parallel Stub!)
Zin = Z0*(ZL + iZ0tan(Bl))/(Z0 + iZLtan(Bl)) where B = 2pi/wavelength
l = length in wavelengths
Z0 = 50 ohms for example
ZL = 50(0.5-j1) = 25 -j50 for the example shown of ZL = 0.5-j1 In my first comment where I detailed the Impedance version of the analysis, I stated that Zin1norm = 1 + j1.6, so I would hope that we find the result is 50(1 + j1.6) = 50 + j80. When I plug in the l = d1 = 0.314 wavelengths into the above equation I get 50.875 + 79.74j out of the TI-84! This confirms that we've got the 50 ohm resistive part matched but there is still the j term. When I plug in the l = d2 = 0.456 wavelengths into the above equation, the result is then 50.66 - 79.577j. This confirms that we've got the 50 ohm resistive part matched and that the resulting j is the opposite as d1's. Trying the video's result of 0.0601 wavelengths gives a result of 14.54 - 23.66j which is clearly not matched to 50 ohm resistive. Trying the video's result of 0.207 wavelengths gives a result of 14.31 + 22.71j which confirms the process was done correctly to mirror the result across the horizontal axis, but it did not give a 50 ohm resistive match. However looking deeper into stubs and seeing examples on Series Stubs, I think I'm trying to follow that process by making the Rin = Z0, which agrees with my results of getting close to 50 ohms for the resistive part. Instead with the Parallel process it is trying to make Gin = Y0 = 1/Z0, which seems almost the same thing at first but it isn't! I definitely am agreeing that if you take for example one of the above impedances and take the reciprocal to get the admittance, 1/(14.31 + 22.71j), you get 0.01986 - 0.0315j which shows the Gin = 1/Z0 = 1/50 = 0.02 which is very close to 0.01986!
_________
@@Pelnied Okay - so I might be misunderstanding what you're doing, but I think your problem is here: Yin1norm (the normalized input admittance after the addition of the series transmission line of length d1) is Yin1norm = 1 + j1.6. That comes straight from the chart, if you look at the blue lines/numbers. Impedance is shown in red; admittance is shown in blue. You don't have to convert it to admittance - it's already admittance. It's hard to read the numbers off the video - you might want to print yourself a copy of the impedance/admittance chart and work through it on paper yourself.
@@emviso I've gone back and confirmed the video has it CORRECT for the Parallel Stub implementation. I had mistakenly stuck to using Impedance and Real(Z) = 1 which caused me to do a Series Stub which was not the intention of the video as it requires the stub to be placed in series with one side of the transmission line.
For Series Stub (Impedance):
d1 = 0.313130325 wavelengths
d2 = 0.456661387 wavelengths
@d1 a series open stub would be 0.089754259 wavelengths
@d1 a series shorted stub would be 0.339754259 wavelengths
For Parallel Stub (Admittance):
d1 = 0.0631303 wavelengths
d2 = 0.20666139 wavelengths
@d1 a parallel open stub would be 0.339754 wavelengths shown at 5:09
@d1 a parallel shorted stub would be 0.08975 wavelengths
I'm glad I stuck to it and kept going back and forth to check answers, as I've now created an automated way to get the results using the TI-83's graphing approach and ability to find intersections so now I don't need to even draw the Smith Chart to find the intersections or to read off any new Normalized values. I can directly calculate where the intersections are at for d1 and d2 and then determine their theoretical normalized admittance or impedance. This gives more precision and accuracy than using the chart itself to eyeball the numbers. I still love the graphical approach and the visual representation that Smith Charts give in seeing how these circuits transform!
@@Pelnied Great! I'm glad it worked out. =D
Thank you very much. This has helped me a lot
I'm glad! Thanks for the feedback!
Re{y}=1 should be the other circle.
You could, and perhaps should, have used the Smith chart to calculate the length of the stubs without the use of your formula. It is easy and is one more of the magic things of the Smith chart. For those unfamiliar with the method, you just mark out the value you need for elimination of the imaginary part. For instance -j1.6 on the outer circle, draw a line from center thru the mark and out to the WTG circle, read off the WTG. For an open stub you start from where r = 0 (or g=infinity) and measure the distance needed to reach the desired imaginary part going clockwise (WTG).
excellent short and very informative... thank you!
I'm glad you found it useful - thank you for commenting!
Thanks!
Great video.
So if i get this, once you know the distance from the load you can place a stub of open wire feed-line or relevant coax at ANY 1/2 wavelength plus the distance calculated from the antenna / load?
The thing i would really like to know is can one keep the stub position at a constant, say any exact 1/2 wavelength and ONLY change the stub length to match the load with the rx/tx? Or is it very important to find an exact length along the feedline where to add the Tee connector / branch?
I only ask because my mathematics skills are terrible, i could potentially use meters to tune the stub if location is not empirical.
Thanks.
isn't there a wat to not have to calculate d2 and do it directly with the chart?
You can calculate d1 and d2 directly without needing the chart. And you can also then calculate the new normalized admittance or normalized impedance without the chart, which then allows you to calculate the length needed for the open/short stub.
In a TI-83 I did the following for a Parallel Stub to find d1 and d2 directly without needing the chart by using a Characteristic Impedance, Z0 of 50 ohms and then storing that for O. I also the unnormalized the ZLnorm of 0.5 -j1 and stored it as a variable "L" as 25-j50.
50->O
(25-j50)->L
Go to the graphing "Y=" feature to set up two functions, y1 and y2
y1 = real(((L+i*O*tan(x))/(O+i*L*tan(x)))^-1) the inverse "^-1" turns the Normalized Impedance into Admittance for use with Parallel Stubs. Just get rid of it to have "))))" at the end if you want to use Normalized Impedance for Series Stubs instead.
y2 = 1 We use this to find where the Real part of y1 is going to become 1.
2nd, calc, #5 Intersect
enter, enter, enter
This usually gives you the first intersection, but may give you the 2nd, depending on how it repeats. Remember if you get a negative result after dividing by 2*pi, you can add 0.5 to the result to turn it positive. Also any value you can also subtract 0.5 if you happened to have gone too far and want a shorter match closer to x = 0.
Use Zoom In on the calculator to get a better view of the graph around x=0 to see the 2x intersections better. The following window works well for me:
Xmin = -2.7
Xmax = 2.7
Ymin = -1.5
Ymax = 3.5
Perform the Intersection operation until you get both of the intersections close to x = 0.
x1 = 0.39665953
x1/(2*pi) = 0.0631303 wavelengths for d1
x2 = 1.2984918
x2/(2*pi) = 0.20666139 wavelengths for d2.
Then if you wanted to know the Normalized Admittance instead of reading it from the Smith Chart @ d1
(((L+i*O*tan(2*pi*0.0631303))/(O+i*L*tan(2*pi*0.0631303))))^-1
This gave 0.9999994908 + 1.581138428j which I rounded to about 1 + 1.58114j. The video shows around 1+1.6j from eyeballing the Smith Chart at 4:25.
english please 😅
Does this chick own a lot of cats?