At 3:44, if z_L = 2-1j, then y_L = 1/z_L = 0.4+0.2j. Confusion arises because gamma = 0.4-0.2j . The labels on the Smith chart are correct for admittance.
For the first example, you moved the decimal point in the wrong direction: C = 9.23 * 10^-11 = 92.3 * 10 ^-12 = 92.3 pF L = 3.915 * 10 ^-6 = 3.915 uH Other than that, great explanation!
At 3:25, Why in the example 1, the normalised load impedance is inside the circle? The distance of normalised ZL (from Euclidean distance formula) is coming 1.414 which is greater than 1. So option 2 should be taken.
BTW...This is simply a video explaining how to do impedance matching on a Smith chart and does not go too deeply into the theory. If you are looking for the theory, try... Topics 7 and 8 here: empossible.net/emp3302/ Topic 2 here: empossible.net/academics/emp4301_5302/
2:28 Why can’t use type 1 circuit when the load outside 1+jx ? For example, if zl=0.55+0.7j,frequency=1 GHz I can parallel the 1.33 pF capacitance first, it will let the load on the boundary of 1+jx circle, then series 4.85 pF capacitance, the load will go to center.
Nope. ua-cam.com/video/L24aB89-m5w/v-deo.html You are correct that it removes reflection between the transmission line and the matching network (looking into the matching network it will look like 50 ohms, or whatever impedance you are designing for) ... but you are incorrect that "reflections between the load impedance and the matching network can occur" ... looking into the matching network from the load, it will match the exactly the impedance of the load, eliminating reflections. The reflection coefficient should be very low for a well deigned network. That's reslly the whol point of the matching network ... looking into the source side, it looks exactly like the source impedance (50R or whatever) and looking into the load side, it looks exactly like the load impedance.
Technically emposible is correct. Reflections between load and matching can occour, Actually this is the normal case. To see this easily, imagine the simple case of load =z0+jsL. The mathing network is a series capacitor = -jsL. Now what this means is that the energy is flowing back and forth between the L and the C. So reflection do occour always with reactive loads. But you are also correct the impedance seen from the load back towards the generator is a conjugate match _IF_ the source is z0 in the case illustrated in the video. But this may not be the case when dealing with power outputs where output impedance /= load impedance. So all in all the right answer is subtle.
Agreed. These are notes from an in-person class where we do just that. Having these animated and visualized alongside the circuit would also be great for the online content.
At 3:44, if z_L = 2-1j, then y_L = 1/z_L = 0.4+0.2j. Confusion arises because gamma = 0.4-0.2j . The labels on the Smith chart are correct for admittance.
yah
For the first example, you moved the decimal point in the wrong direction:
C = 9.23 * 10^-11 = 92.3 * 10 ^-12 = 92.3 pF
L = 3.915 * 10 ^-6 = 3.915 uH
Other than that, great explanation!
Thank you!
At 3:25, Why in the example 1, the normalised load impedance is inside the circle? The distance of normalised ZL (from Euclidean distance formula) is coming 1.414 which is greater than 1. So option 2 should be taken.
Thank you!
BTW...This is simply a video explaining how to do impedance matching on a Smith chart and does not go too deeply into the theory. If you are looking for the theory, try...
Topics 7 and 8 here:
empossible.net/emp3302/
Topic 2 here:
empossible.net/academics/emp4301_5302/
Really nice, but at 4:09, I think he means to say "we look at the admittance here..." rather than impedance.
Thank you!
2:28 Why can’t use type 1 circuit when the load outside 1+jx ?
For example, if zl=0.55+0.7j,frequency=1 GHz
I can parallel the 1.33 pF capacitance first, it will let the load on the boundary of 1+jx circle, then series 4.85 pF capacitance, the load will go to center.
at 4:28 you mentioned a positive capacative impedance. How can this be? Capacitance reactance is negative
Thanks 👍.
nice video thank u
Thanks.
About part 2: What if, hypothetically, the length of the transmission line is 50 cm (i.e. less than lA = 90.9 cm)?
👍
Nope. ua-cam.com/video/L24aB89-m5w/v-deo.html You are correct that it removes reflection between the transmission line and the matching network (looking into the matching network it will look like 50 ohms, or whatever impedance you are designing for) ... but you are incorrect that "reflections between the load impedance and the matching network can occur" ... looking into the matching network from the load, it will match the exactly the impedance of the load, eliminating reflections. The reflection coefficient should be very low for a well deigned network. That's reslly the whol point of the matching network ... looking into the source side, it looks exactly like the source impedance (50R or whatever) and looking into the load side, it looks exactly like the load impedance.
Technically emposible is correct. Reflections between load and matching can occour, Actually this is the normal case. To see this easily, imagine the simple case of load =z0+jsL. The mathing network is a series capacitor = -jsL. Now what this means is that the energy is flowing back and forth between the L and the C. So reflection do occour always with reactive loads.
But you are also correct the impedance seen from the load back towards the generator is a conjugate match _IF_ the source is z0 in the case illustrated in the video. But this may not be the case when dealing with power outputs where output impedance /= load impedance. So all in all the right answer is subtle.
it would be better if you show it by drawing on paper instead of showing it here. we could realize this better
Agreed. These are notes from an in-person class where we do just that. Having these animated and visualized alongside the circuit would also be great for the online content.