`
`
`
`UNITED STATES PATENT AND TRADEMARK OFFICE
`__________
`
`BEFORE THE PATENT TRIAL AND APPEAL BOARD
`____________
`
`HTC CORPORATION, HTC AMERICA, Inc.,
`ZTE CORPORATION, and ZTE (USA), Inc.,
`Petitioners,
`
`v.
`
`CELLULAR COMMUNICATIONS EQUIPMENT, LLC,
`Patent Owner.
`____________
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`____________
`
`Record of Oral Hearing
`Held: July 10, 2018
`
`
`
`
`Before BRYAN F. MOORE, JENNIFER S. BISK, and
`GREGG I. ANDERSON, Administrative Patent Judges.
`
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`
`
`APPEARANCES:
`
`ON BEHALF OF THE PETITIONERS:
`
`
`BRIAN C. NASH, ESQ.
`Pillsbury Winthrop Shaw Pittman, LLP
`401 Congress Avenue, Suite 1700
`Austin, Texas 78701-3443
`(512) 580-9629
`brian.nash@pillsburylaw.com
`
`
`
`ON BEHALF OF THE PATENT OWNERS:
`
`
`BARRY J. BUMGARDNER, ESQ.
`MATTHEW JUREN, ESQ.
`Nelson Bumgardner Albritton
`3131 West 7th, Suite 300
`Fort Worth, Texas 76107
`(817) 377-3494
`barry@nbafirm.com
`
`
`
`
`The above-entitled matter came on for hearing on Tuesday, July 10,
`
`2018, commencing at 1 p.m. at the U.S. Patent and Trademark Office, 600
`Dulany Street, Alexandria, Virginia.
`
`
`
`
`2
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`P R O C E E D I N G S
`- - - - -
`
` (12:59 p.m.)
`JUDGE BISK: Good afternoon. This is the hearing for IPR2017-
`01508 and 01509 between Petitioner HTC Corporation, HTC America, ZTE
`Corporation, and ZTE (USA), Incorporated, and Cellular Communications
`Equipment, LLC, the owner of the challenged patents, U.S. Patent Number
`8,385,966 in the '1508 matter and U.S. Patent 9,037,129 in the '1509 matter.
`Just a few administrative matters before we begin. I'm Judge Bisk
`and I have Judge Moore with me here on the right. Judge Anderson is on
`video. I guess there's a few videos here today. And he's joining us
`remotely obviously. So just remember, when you're presenting, if you
`could mention the slide number you're talking about so that Judge Anderson
`can follow along. This will also make the transcript easier to read. Per
`our trial order, each party will have 30 minutes to present its argument for
`each proceeding, totaling 2 hours for this hearing. You're not required to
`take all the time. And because Petitioner has the burden to show
`unpatentability of the original claims, Petitioner will proceed first, followed
`by the Patent Owner. Petitioner may reserve time to rebut Patent Owner's
`opposition.
`And just to mention, if there are any objections that you want to
`present to any of the testimony that happens today, we'd prefer you do that
`during your own time so that we don't interrupt each other's presentations.
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`3
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`Okay. At this time, I'd like Counsel to introduce yourselves and
`who you have with you, beginning with Petitioner, please.
`MR. NASH: Your Honor, Brian Nash here on behalf of the
`Petitioners.
`JUDGE BISK: Okay, thank you. And Patent Owner?
`MR. BUMGARDNER: Your Honor, Barry Bumgardner, here on
`behalf of Patent Owner. Along with me is Matt Juren. I will argue the
`first case and Mr. Juren will argue the second case.
`JUDGE BISK: Okay, great.
`MR. BUMGARDNER: And if I may ask, Your Honor, will Judge
`Anderson be able to see what we put on the screen or should we assume that
`he will not be able to view that?
`JUDGE BISK: Well, assume that he won't be able to see the screen
`but he does have all of the slides electronically, so as long as you say the
`slide number or whatever you're looking at, if you can say the electronic
`version, he can keep up.
`MR. BUMGARDNER: Thank you, Your Honor.
`JUDGE BISK: That's correct, isn't it, Judge Anderson? You don't
`see the actual slides, right?
`
`JUDGE ANDERSON: Yeah, the hearing room personnel would
`have to turn the camera to the slides. And if you have an Elmo, I think
`somebody had an Elmo there, I can't see that, but just give me the pleading
`you're looking at. I can follow along and certainly I have the slides available.
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`4
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`MR. BUMGARDNER: Thank you, Your Honor, I just wanted to
`make sure I knew what was going on here. So, that will all be fine.
`JUDGE BISK: Okay, great. Petitioner, whenever you're ready.
`MR. NASH: Your Honor, I have a couple of hard copies. I don't
`know if you'd like this as well, but I could pass those up.
`JUDGE BISK: Yes, sure.
`MR. NASH: They're very nicely bound.
`JUDGE BISK: Okay. And I'll keep track of time for you if you let
`me know how many minutes you want to reserve for rebuttal.
`MR. NASH: Thank you, Judge. I'd like to try and reserve ten
`minutes for rebuttal.
`JUDGE BISK: Okay, so I'll put 20 minutes on the timer just so you
`can see.
`MR. NASH: On the big red clock.
`JUDGE BISK: Yes.
`MR. NASH: Thank you so much.
`JUDGE BISK: All right, whenever you're ready.
`MR. NASH: Thank you, Judge. Brian Nash on behalf of
`Petitioners. And I'm going to start on Slide 2, Judge Anderson, because I
`think the important aspect of the '966 patent that we're going to discuss
`today is that it relates to power control for Message 3.
`So we see on Slide 3 that there's Figure 1B illustrated here. We've
`got Message 1, which is your random access preamble message being sent
`from the UE to the eNodeB. And then upon the successful transmission,
`
`5
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`the eNodeB is going to reply with Message 2. And so the question is, what
`do we set the power control for Message 3?
`And what the '966 patent is teaching and relating to is utilizing
`factors and parameters that come from Messages 1 and 2 in setting that
`power control for Message 3. So we see that reflected here on Slide 3.
`In the context of Message 1, we're going to have a parameter called
`P ramp-up, delta P ramp-up. That's the incremental power change that's
`going to be applied to the RACH preamble as a method of trying to get a
`successful transmission.
`And then in Message 2 there's going to be a response, and included
`in that response from the eNodeB will be this power control command, or
`delta P PC. Those are the two factors you're going to see come into play
`here as we turn to Slide 3, because when we're setting the Message 3's initial
`power control, the '966 patent teaches these two equations that we see here
`on Slide 4, Judge Anderson, and that's equations 4A and 4B.
`And you see those factors reflected there in that equation. So
`there's the delta P PC and the delta P ramp-up factors that are included.
`And what's important, as we turn to Slide 5, is that the '966 patent
`teaches that that equation can be reduced using simple math to result in f(0)
`equal to g(0) equal to delta P PC plus delta P ramp-up. So that's a
`simplification of equations 4A and 4B from the '966 patent. So, as we
`turn to Slide 6, we see that these are the key elements. And I've tried to
`highlight the ones that we're going to be discussing today, but you're
`calculating the initial transmit power for that Message 3. And in doing so
`
`6
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`you're going to use full path loss, highlighted in purple. And then that
`calculation is going to depend on two things. It's going to depend on
`Message 1 preamble power, highlighted there in green, and it's going to
`depend on the power control adjustment state f(0), highlighted in blue.
`And again, that power control adjustment state is going to utilize a
`specific equation that was called out in the '966 patent. And we see that
`reflected there again at the bottom of Slide 6. So, as we turn into the claim
`language, which is reflected on Slide 7 and 8, we're going to see a
`corresponding highlight that's reflecting those same key elements that we
`just discussed.
`JUDGE ANDERSON: Mr. Nash, let me go back to Slide 6, please.
`So, you've got in there a bullet point of key elements, initialized power
`control, and in the '966 patent it describes the prior art as including the TS
`36.213 standard. And it says the difference that's been invented here is
`the initialization of these power control adjustment states: g(0), f(0). Tell
`me what is the initialization process that we're going through here?
`MR. NASH: Well, the initialization process in the context of the
`'966 patent is that equation that we see reflected below. That's from
`equations 4A and 4B, where we're setting f(0), a specific equation that
`utilizes f(0), and that's reduced to f(0) equal to delta P PC plus delta P ramp-
`up.
`
`JUDGE ANDERSON: So is that the first frame that's transmitted
`after Message 1 and Message 2 as part of Message 3?
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`7
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`MR. NASH: Correct, yes, that's the first successful scheduled
`transmission. So the first Message 3 that's being sent.
`JUDGE ANDERSON: Okay. Go ahead.
`MR. NASH: Thank you, Judge. I think it's helpful to take a quick
`look at Qualcomm, and so I'm going to skip to Slide 12, and actually,
`specifically Slide 13.
`So Qualcomm is obviously one of the primary issues in dispute, and
`as we look on Slide 13 we see that Qualcomm itself teaches each of those
`three key elements. And I'm using that same colored highlighting that we
`were talking about before.
`So first we see on Slide 13 that Qualcomm's got an Equation 4 that it
`discloses for PUSCH_power. And PUSCH_power is that transmit power
`for Message 3. And the first thing it depends on, you see there, is
`RACH_power. That RACH_power is the Message 1 preamble power.
`So we've got that first element in Qualcomm. And then as we turn
`to Slides 14, 15, 16, 17, and 18, we're going to see that there's some math
`that can be applied to Qualcomm's equations. And by utilizing Equations 1
`and 2 in Qualcomm, we can modify that Equation 4 to the result that we see
`here reflected on Slide 18.
`And so this is the rewritten Qualcomm Equation 4. And as we see
`there, we've got that PUSCH_power equation again. And now we've got
`these new variables in there, power_ramp_up plus PC_correction. Those
`are highlighted in blue there. And those are important provisions because
`each of those variables are the exact same as what we're talking about in the
`
`8
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`context of a '966 patent. That is your delta P PC plus your delta P ramp-up.
`So now we've shown that there's this initialization of f(0), your power
`control adjustment state, because it's teaching those provisions.
`And then finally, with respect to that third purple highlighted
`provision, which is the full path loss compensation aspect, we see on Slide
`19 that that's also reflected in Qualcomm, or one of skill in the art would
`understand that to be reflected in Qualcomm. And there's two reasons why
`that's the case; each can be done in the alternative here.
`The first reason why that's the case is because Qualcomm teaches
`that Equation 4 involves the RACH preamble, and the RACH preamble in
`the context of Qualcomm is determined using an open-loop method. The
`reason why that's important is because the prior art 3GPP specifications
`require RACH preambles to be transmitted using full path loss
`compensation.
`And the '966 patent even admits that. It admits that preamble
`power is calculated using full path loss compensation when it's being done in
`an open loop method. So we see that reflected here on Slide 19. So, that
`alone would teach one with skill in the art that we've got full path loss
`compensation in the context of Qualcomm's Equation 4.
`JUDGE ANDERSON: Okay, so let me understand. The patent,
`then, the challenged patent, shows or agrees that the prior art uses a full path
`loss in the preamble, correct?
`MR. NASH: Correct, Judge.
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`9
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`JUDGE ANDERSON: And that is the basis for you to look at
`Qualcomm and the equations on RACH preamble that are in there -- excuse
`me, yes, RACH preamble power -- and assume that those also use full path
`loss, is that right?
`MR. NASH: That's correct, Your Honor.
`JUDGE ANDERSON: Okay. What is the reason, give me the best
`explanation you can, because it's a little bit unclear at this point in my
`understanding of the case, why you necessarily would use the path loss of
`one from the prior art in the Qualcomm equations that you're relying on?
`MR. NASH: Well, I think what we're trying to point out is that
`there's a baseline understanding of those skilled in the art, and that's that the
`prior art, as the '966 patent admits, understood the RACH preamble, if
`calculated using an open-loop method, is necessarily required to use full
`path loss.
`And as a result, when somebody's reading, the skilled artisan is
`reading the Qualcomm reference and sees in Equation 4 that the
`PUSCH_power is equal to the RACH_power, it knows, well, RACH_power
`is the preamble power for Message 1 and the preamble power for Message 1
`was calculated using an open-loop method, so, as a result, I know that
`parameter is using full path loss.
`And so when you're asking the question do the claims require a
`Message 3 transmit power that's calculated and depends on full path loss
`compensation, it does, because it depends on that RACH_power or the
`preamble power. Does that answer your question, Your Honor?
`
`10
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`JUDGE ANDERSON: Yeah, I want to make sure I understand this
`as well as I can. So, first of all, Qualcomm is an open-loop system, right?
`MR. NASH: Correct.
`JUDGE ANDERSON: And so, because it's an open-loop system,
`we look at this preamble power and path loss in the preamble power based
`on what's known, and that is using full path loss. Once you do that, then
`you can fall into and look -- and you understand that in Qualcomm from
`Equation 4 because it mentions RACH power.
`MR. NASH: That's correct, Your Honor.
`JUDGE ANDERSON: Okay.
`MR. NASH: There's also a different basis that we walk through in
`the petition as well. You can see that reflected here in Slides 20, 21, and
`22. That involves Qualcomm's Equation 2, but that's a separate and
`additional basis for concluding that there's full path loss in the Qualcomm
`reference.
`I think it's helpful to quickly look at the institution decision. As we
`see on Slide 23, there's a couple of claim constructions from the institution
`decision. Those are not in dispute today. And in 24, we see the three
`grounds on which this trial was instituted. And each of those grounds
`depends on Qualcomm and then the TS 36.213 reference.
`So I think I've highlighted here on Slide 25 what I believe is not in
`dispute. And I think it's important to take a look at that quickly. The
`prior art's not in dispute, those claim constructions aren't, but, more
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`11
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`importantly, the math that we used to arrive at that simplified version of
`Equation 4 from the '966 patent is not in dispute.
`Similarly, the math that's used to modify Qualcomm's Equation 4 to
`get the result that we see there reflected at the bottom of Slide 25 is also not
`in dispute. So there's no dispute that the math that we used to get to these
`equations was correct and that there was a proper basis for doing so.
`There's only actually two issues in dispute here, and I have those
`reflected on Slide 26. First, CCE, the Patent Owner, contends that the
`Qualcomm reference doesn't utilize full path loss unless we can prove that
`Qualcomm's delta is equal to 1, it's a little delta, and that has to be 1 in order
`for there to be full path loss.
`Second, they also argue that the power control adjustment states that
`are claimed in the patent are not disclosed in Qualcomm.
`I'm going to jump into each of those issues. We've touched on it a
`little bit, Your Honor, but before I do I wanted to point out that Petitioner's
`expert testimony in this case is effectively unrebutted. CCE doesn't cite to
`explain or otherwise rely on any expert testimony of its own. So, with
`respect to experts in this case, the only testimony that's at issue is the
`testimony of the Petitioner's expert, Dr. Akl.
`So, taking a look at that first path loss provision, again, what CCE
`argues, and we see this reflected here on Slide 27, has been argued before,
`which is that delta in Qualcomm must be 1 in order for there to be full path
`loss.
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`12
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`Well, as we see in Slide 28, the first reason why that's not the case is
`exactly what we just discussed, Judge Anderson, which is that there's an
`additional basis for the reason why there's full path loss in Qualcomm that
`doesn't even look at or address Qualcomm's Equation 2. It's the fact that
`you're using that RACH preamble with an open-loop method.
`And so we see here in Slide 28, I've highlighted Paragraph 52 from
`Dr. Akl's opening testimony where he says the admitted prior art of the '966
`patent admits that prior art 3GPP specifications require RACH preambles to
`be transmitted using full path loss compensation.
`And as we turn to
`Slide 29, we see that Dr. Akl used that as the first basis for concluding that
`Qualcomm teaches the use of full path loss compensation. He says in the
`beginning of Paragraph 127 that the preamble power described in
`Qualcomm is based on the entire path loss. First -- so this is his first reason
`-- the preamble power is calculated using an open-loop method. The 966
`patent admits that preamble power is calculated using full path loss
`compensation in an open-loop method. So right there is his first basis.
`It's independent of Qualcomm's Equation 2.
`Now, with respect to that second basis that involves mathematics
`applied to --
`JUDGE BISK: Can I just ask about this expert testimony? Is he
`relying on the admitted prior art of the '966 patent for that, or is he saying
`independently that's what it is, "oh, by the way, see the '966 patent that
`basically says that as well"?
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`13
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`MR. NASH: Yes, I think I understand your question, Your Honor.
`Dr. Akl's testimony is actually very thorough on this point, and if we look
`back at Slide 28, I've included, you see there, Paragraph 52 where we first
`mentioned what a RACH preamble requires.
`JUDGE BISK: Yes.
`MR. NASH: He also walks through a number of other pieces of
`prior art that are part of this case. And that's reflected in Paragraphs 39, 73
`through 75, and 82. So he actually walks us through a number of other
`pieces of prior art to establish the fact that RACH preambles require full
`path loss compensation.
`So, when he gets to this conclusion that we see reprinted here on
`Slide 29, and that's from Paragraph 127, he does cite to the '966 patent for
`that proposition because it's kind of the easiest thing to point to. But the
`preceding analysis also addresses that point. Does that answer your
`question?
`JUDGE BISK: Yes. Thank you.
`MR. NASH: So, with respect to that second argument, the
`argument that -- actually, before we move on, I do want to point out that this
`argument has never been addressed by patent owner.
`So, they didn't cross-examine Dr. Akl on this opinion, they didn't
`raise this issue in their patent owner response, and yet this issue,
`independent of what happens with Qualcomm's Equation 2, establishes full
`path loss.
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`14
`
`
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`So I think that's an important thing to highlight because now we're
`going to jump in why they are wrong about that delta provision. But even
`if they were right -- and they're not -- you could independently determine
`that full path loss is being used here without even addressing that.
`So I'm going to jump into why they're wrong about that delta
`provision.
`JUDGE ANDERSON: Counsel, so, which equation in Qualcomm
`are you relying on to get to the initialization equation in the claims, 2 or 4?
`MR. NASH: Four, Your Honor.
`JUDGE ANDERSON: Okay, thank you.
`MR. NASH: That's the one that says PUSCH_power equal to
`various provisions. And using mathematics, we arrived at a modified
`version of Qualcomm's Equation 4 that reflects that initialization.
`JUDGE ANDERSON: Okay. And Equation 2, I've got to say,
`and maybe I'm leading you into your argument on this delta, small delta
`issue, it doesn't appear in Equation 4; it's only in Equation 2, as I read the
`record here. So, I'm going to just leave it to you to explain what this delta
`issue is about. Go ahead, please.
`MR. NASH: Well, you're right, Your Honor. I honestly don't
`know that -- and we can go back to the slide that reflects Equation 2 if that's
`more helpful. But this was just an additional analysis that was done by
`Dr. Akl to try and show how you could derive path loss using some of the
`equations that are disclosed in the context of Qualcomm, mainly the
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`
`15
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`difference between transmit power and receipt power. So, he went through
`that analysis --
`JUDGE ANDERSON: Right, and I get that, so to the extent you
`focus on this argument about delta, that would be appreciated.
`MR. NASH: You bet. So, the reason why delta isn't modifying
`the path loss is because there's nothing in Qualcomm or any other evidence
`that establishes it as a modifier of path loss.
`You have a number of variables in these equations, they're all being
`multiplied by each other, and so for it to be a modifier of path loss, there has
`to be a specific function associated with that variable that corresponds to the
`modifying path loss. And we see what Qualcomm itself says about delta
`reflected here on Slide 30.
`So, Qualcomm doesn't discuss delta much but what it does say is that
`it's a correction factor that kind of acts as a catch-all for a bunch of other
`potential parameters. And we see those reprinted here. So, you've got
`transmit power of the eNodeB reference signal, RACH slot interference
`level, and signal-to-noise ratio.
`Those are the parameters that are disclosed in Qualcomm itself,
`except those may be absorbed into delta. But there's no mention at all of
`path loss in that context or anywhere else in Qualcomm. Qualcomm
`doesn't discuss path loss itself at all.
`JUDGE ANDERSON: Well, that's sort of true, but, Qualcomm,
`I'm looking at Column 9 at Line 7, I guess, says the correction factor delta
`may be used to bias the open-loop algorithm. That's seems to walk into
`
`16
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`your argument about, well, we're using full path loss because we're in an
`open loop. So, explain that one for me.
`MR. NASH: Yes, Your Honor. I'm familiar with the provision
`that you're discussing and all that's really saying is that you can use delta to
`modify what's being disclosed in the context of Equation 2.
`Remember that Equation 2 by itself doesn't actually discuss path
`loss. There isn't a provision in Equation 2 that says "PL," and it's not like
`delta's being applied times PL. So it is a factor, what they even call a
`correction factor, meaning it can be used to bias that open-loop method.
`But the other thing that's really contemplated are these three aspects right
`here: so, transmit power to the eNodeB reference signal, RACH slot
`interference level, and signal-to-noise ratio.
`So, I think what's being taught by that is that you can bias the open-
`loop correction method by, for example, using some of these parameters as a
`potential way to correct for, or bias, that correction open-loop method.
`Does that answer your question, Your Honor? Okay.
`There's one piece of evidence the Patent Owner introduced in the
`context of this, which is Exhibit 2004. I don't want to spend a whole lot of
`time on it but, frankly, I think it's largely irrelevant. There's no connection
`between Exhibit 2004 and the Qualcomm reference we have here. There's
`no factual evidence saying that one who's skilled in the art would look to
`Exhibit 2004 to better understand the delta. And, in fact, Exhibit 2004
`doesn't even refer to a delta factor. That's not mentioned anywhere in that
`reference at all.
`
`17
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`There's an attempt to make an equivalence between delta in
`Qualcomm and alpha in the '966 patent. As we see on Slide 32, I'm just
`going to briefly mention that those equations are each solving for different
`things. So, Equation 1 in the '966 patent is solving for the PUSCH_power
`control, and the Equation 1 in the context of Qualcomm is solving for the
`RACH preamble power.
`So, these are apples and oranges, if you will, and there's no
`equivalence you can make just because they're both Equation 1 or just
`because they both happen to have Greek variables in them.
`As we look to Slide 33, another point that CCE tries to rely on is the
`testimony of Dr. Akl to try and make that equivalence. And I think it's
`very clearly from the context of Dr. Akl's testimony in cross-examination
`that he was trying to be very careful. And he was trying to say, look, if
`you're trying to say this is a factor that modifies path loss, you have to know
`something about that variable. It has to have the specific function of doing
`it. You can't just say, simply because it's 0.5 times the variable you have
`that it's there to modify path loss. It could have had some other intent.
`That 0.5 could represent some other parameter.
`And so he says that here on Slide 33, he says you have to know the
`context, what the meaning that's given to that variable is, and then, based on
`that meaning, then you can make a determination on whether or not it's there
`to result in a fractional portion of the path loss.
`And so then when we get to Slide 34, this is the quote that Patent
`Owners relied on quite a bit about delta. And as you see from the context
`
`18
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`that surrounds that, Dr. Akl was simply answering the question that was
`asked, which is delta sort of in the abstract, if it's being used to modify path
`loss and it's less than 1, wouldn't that be a fractional path loss? And he
`simply just said yes. So, nothing big to draw from that.
`If we take a look at Slide 35, I'll just briefly touch on why Dr. Akl
`does not agree that delta equals 1. CCE seems to think that's the case.
`Again, that first reason has nothing to do with delta, so he clearly doesn't
`agree on that front. He thinks path loss is there regardless of what delta
`might be.
`And then, second, with respect to the context around Equation 2, he
`made it very clear that in order for there to be modifier in Equation 2 that
`modifies path loss, delta would need to have that role. And there's no
`evidence, as we just walked through, that supports that delta has some kind
`of role for modifying path loss.
`I can briefly touch on why the disclosed power control adjustment
`states, f(0) and g(0), are disclosed. CCE has raised a number of arguments
`with respect to that claim limitation. You see them laid out here in Slide
`37. I'll note that each one of these arguments was actually raised and
`addressed before as part of the institution decision.
`But I think the one that might be helpful today is their argument that
`you see in that second bullet point, which is that TS 36.213 requires f(0)
`equal to 0, and that that somehow creates a conflict in this combination
`because one of skill in the art wouldn't set that aside and then move on to
`find something else for f(0).
`
`
`19
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`So I'm going to switch to Slide 39 where we address that argument,
`Your Honor. So I think what's important to see about that is that there's a
`starting premise that CCE uses that's got a number of flaws. And that
`starting premise is reflected here on Slide 39 at the top.
`So CCE contends you start with 36.213, which clearly requires f(0)
`equal to 0, and then the skilled artisans must recognize some kind of
`problem with that equation and then look for a different solution, and that
`would result in the combination of Qualcomm.
`Well, the first flaw in that premise, if they do start with 36.213, as
`the petition made clear and the institution decision made clear, we're starting
`with Qualcomm as the reference and Qualcomm's teaching what it's teaching
`about power control. And then you would look to 36.213 for efficiencies
`and for compliance with standards. So you don't start with 36.213 and
`move to Qualcomm, you start with Qualcomm and move to 36.213.
`The second flaw in their premise is that 36.213 requires f(0) equal to
`0. It does not. It presents f(0) equal to 0 as an example for that starting
`point of a power control adjustment state that's initialized at I equals 0.
`So that's just one example, and we have ample evidence in the record
`demonstrating that one of skill in the art would not see a conflict there
`because he would recognize that f(0) equal to 0 is just an example.
`So when he's going from Qualcomm to 36.213, he sees that f(0)
`equal to 0 -- and it's one of a couple examples that 36.213 provides -- and he
`would know that's just an example, it doesn't create a conflict, because he's
`already got a better value.
`
`20
`
`
`
`
`
`1
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
`13
`14
`15
`16
`17
`18
`19
`20
`21
`22
`23
`24
`
`Case IPR2017-01508 (Patent 8,385,966 B2)
`Case IPR2017-01509 (Patent 9,037,129 B2)
`
`So, as we see on Slide 40, there's lots of evidence in the record
`already about why that combination would be made. And to briefly touch
`on how this combination results in that claimed initialization, I'll recap this.
`I think we discussed this a little bit earlier, but as we see on Slide 42, we've
`got two things that are required for that Message 3. There's the Message 1
`preamble power and then that power control adjustment state f(0) that has a
`specific equation reflected on Slide 42.
`And as we see on 43, again, we've got that first factor Message 1
`preamble power directly from Equation 4. And then with respect to the
`other provision, we see that as our modified Equation 4. And again
Accessing this document will incur an additional charge of $.
After purchase, you can access this document again without charge.
Accept $ Charge
Still Working On It
This document is taking longer than usual to download. This can happen if we need to contact the court directly to obtain the document and their servers are running slowly.
Give it another minute or two to complete, and then try the refresh button.
A few More Minutes ... Still Working
It can take up to 5 minutes for us to download a document if the court servers are running slowly.
Thank you for your continued patience.
This document could not be displayed.
We could not find this document within its docket. Please go back to the docket page and check the link. If that does not work, go back to the docket and refresh it to pull the newest information.
Your account does not support viewing this document.
You need a Paid Account to view this document. Click here to change your account type.
Your account does not support viewing this document.
Set your membership
status to view this document.
With a Docket Alarm membership, you'll
get a whole lot more, including:
- Up-to-date information for this case.
- Email alerts whenever there is an update.
- Full text search for other cases.
- Get email alerts whenever a new case matches your search.
One Moment Please
The filing “” is large (MB) and is being downloaded.
Please refresh this page in a few minutes to see if the filing has been downloaded. The filing will also be emailed to you when the download completes.
Your document is on its way!
If you do not receive the document in five minutes, contact support at support@docketalarm.com.
Sealed Document
We are unable to display this document, it may be under a court ordered seal.
If you have proper credentials to access the file, you may proceed directly to the court's system using your government issued username and password.
Access Government Site
