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For the half-power bandwidth we have
Γ² = (V_refl/V_trans)² = P_refl/P_trans = 0.5
SWR = (1+|Γ|)/(1-|Γ|) = 5.828 => BW = 2/Q
For series and parallel RLC circuits, one mostly talkes about the half-power bandwidth, whereas for antennas, it is maybe more common to look at S-parameters. That's why...
This thread got little attention. Does anyone know of a more general expression between the impedance bandwidth and Q, i.e. I'm not looking for the -3 dB bandwidth relation BW = 1/Q (or BW = f/Q for the absolute bandwidth).
I've seen the following relation between bandwidth, Q-factor and SWR:
BW = (SWR-1)/(Q√SWR).
See for example Pozar's article, Microstrip antennas: the analysis ... - Google Books
Then for the -3 dB bandwidth (SWR = 5.85) the expression reduces to BW = 2/Q, instead of 1/Q, as I was hoping for...
If we transmit a sinusoid s(t) = cos(ωt) over a wireless fading channel, the received signal can be written
r(t) = ∑a_i(t)cos[ω(t-τ_i(t))],
where a_i(t) is the i:th attenuation constants and τ_i(t) i:th delay. In my text the above relationship is claimed to be linear. But the delays are in...
Yes, I've tried that before. However, it seems harder to obtain equation (6) (which I assume is correct) by factorizing like (s-p1)...(s-pn). Also in the text I have they factorize it like I did.
Although the derivations bother me somewhat, I'm more interested in getting answers for questions 1...
Maybe there are still some misunderstandings. Solving eq. (3) gives you the system poles. (4) is no equation but a factorization of loop poles. I've just denoted it in two ways, in case someone is not familiar with the product notation (pi). This factorization is then substituted into (3). How...
Hi LvW. You're right, poles don't necessary occur in conjugate pairs. From the little experience I have, I can only tell that loop poles are typically real (neglecting C_mu, Cgd). If I make that assumption I get eq (6). If the poles are single complex I can't cancel the (-1)^n. It is not written...
Sorry for some mistakes in the above pdf. I've corrected the errors and put in some more details.
The material I've presented is from some old lecture notes I've taken. I remember we were only provided the results. Now I want to understand the details behind all this, and I know there is almost...
compensation des poles dominants
Since nobody has answered, I'll give some background information.
The loop gain (return ratio) of an amplifier can be written as
Aβ(s) = Aβ(0)N(s)/D(s), where Aβ(0) is the DC loop gain.
Now the closed-loop gain of an amplifier is given by
A_cl = A_ideal •...
dominant closed-loop poles definition
As is well known, only the dominant loop poles can be used for compensating a negative feedback amplifier. To determine the dominant poles one then compares the sum of the loop poles with the sum of the system poles. The set of loop poles which sum is...
polynomial zeros on unit circle
I found a somewhat simpler way:
p(z) = z^n + a_n-1*z^n-1 + ... + a0 = (z-z1)*...*(z-zn) = z^n + ... + (-1)^n*z1*...*zn
So a0 = (-1)^n*z1*...*zn and we have that |a0| = |z1*...*zn| > 1. If the product of zeros is larger in magnitude than 1, then all zeros can't...
+polynomial +unit circle
I'm stuck in proving the following:
Show that if p(z) = z^n + a_n-1*z^n-1 + ... + a_0 is a polynomial of degree n ≥ 1 and |a_0| > 1, then p(z) has at least one zero outside the unit circle. Notice that the leading coefficient a_n = 1. There is also a hint given...
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