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Hello LvW,
thank you very much. I will study your documents and try to prepare some others for a discussion :roll:.
FvM is right that we rather sheered off the original problem. I know that SpectreRF from Cadence can solve frequency responses of SC circuits, see for example here...
LwH,
OK, let us make the things clear.
After studying your comments, I (perhaps) found the reason why you defend the mistaken thesis
(T)"Replacement of a resistor by a switched cap is allowed only if BOTH sides of the resistor are connected to a voltage source or ground"
which was completed by...
For low frequencies, you can use power OpAmp, for example L165 by SGS-Thomson (output current up to 3A). Vpp is typically 27V (decreases for higher output currents).
Mashnayn,
in RC cell, the capacitance is 150nF, which corresponds to the time constant RC=150us and the cutoff frequency of 1.06 kHz.
In SC cell, the capacitance is 15nF, which corresponds to the time constant RC=15us and the cutoff frequency of 10.6 kHz. Since the switching frequency is only...
Hi Zorx,
you are right:-D.
Your result
Hn(w) = [H(w)]^N = 1 / [1+b^2 * w^2]^N
is computed on the assumption that the impulse response is even function of time, i.e. the response to the Dirac impulse precedes this impulse. In other words, the corresponding linear system is not causal.
Based on...
I think that such standard built-in function is not available. Of course, this function can be easily compiled from basic Matlab matrix functions.
See for example here:
MATLAB Teaching Codes
and read the M-file projmat.m
I suppose that delta(t) is the Dirac impulse.
Then the impulse response is
h(t)=3*exp(-2*t)-2*exp(-3*t)
and the transfer function H(s) is its Laplace transform, thus
H(s)=3/(s+2)-2/(s+3)=(s+5)/(s^2+5*s+6).
Frequency response can be obtained after the substitution
s=jw.
One note should be mentioned: multiplication in the frequency domain is equivalent to convolution, not correlation, in the frequency domain.
I am afraid that Mamdulahad is interested in the analytical, not numeric form of the result. Any way, he should clearly state if his formula for h(t)...
Yes, but there are two points which should be cleared up:
1) The gain is Av=-RF/R1=-1.5
2) \[ Fo = \frac{R1}{ 2 *\pi * C1 (R1*R2 + R1*RF + R2*RF) } = 9MHz\]
The formula can be simplified to
\[ Fo = \frac{1}{ 2 *\pi * C1 (R2 + RF + R2*RF/R1) } = 9MHz\]
The "resistive" term is Rb (the well-known...
Try to decrease the step ceiling. Note that your command
.TRAN 1m 1 uic
is not optimal: the first number 1m has no effect to the simulation run; you have not set the step ceiling, thus it is defined automatically as 1/50=20ms and it is too high.
Try the following:
.TRAN 0 1 0 1m uic
Then it...
I think that it is a way how to do frequency compensation of this amplifier in order to provide its stable behavior.
Try the following:
1) Make the wye-delta transformation of R1, Rf, and R2. You obtain resistors Ra, Rb, Rc; Ra=11.666kOhms between Vin and - nodes, Rb=17.5kOhms between - and Vout...
Hello Goldsmith,
LvW tried to help you such that you would be able to find your own path to the aim. It is the best method of understanding.
However, I see that you lose the way.
Please follow the points below and think over them:
1] OpAmp is connected in negative feedback loop, thus its input...
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