We will find out later why a slower input results in less glitches.
It turns out that circuits can be used to model many real-life situations - the syringe and the drug can be modeled as a current source that is going to be dumping some charge. So it is an injection. And then the body can be modeled as a capacitor. So this is the body storage. So the body can be modeled as a capacitor where, when the syringe injects some drug, and in the electrical analogy when
the current source provides some current, that will then go and charge the capacitor and put the
charge in the capacitor. In the body situation, the injection will deposit some drug in the body.| Once a drug is deposited in the body, the way in which the drug is dissipated through the body is modeled with a little resistor where the drug is then finally dissipated. So it can all be modeled as a first-order RC circuit.
It is possible to model the input as a short pulse as input time duration does not matter. What matters is how much charge is deposited by the pulse. This type of pulse is called as an impulse.
There are several kinds of input-step, impulse, pulse and ramp.
A step input is one in which the signal has been 0 for some time and at a time 0, the signal goes to some d.c.value, say a 1. Then if the signal is V_I(t), it can be written as V_I(t) = u ( t).
If the signal has been 0 for some time and at a time 0, the signal goes to some d.c.value, say V_I, then V_I(t) = V_I*u ( t).
If the signal has been V_I for some time and at a time 0, the signal goes to some 0, then
V_I(t) = V_I - V_I*u ( t). This is a falling step.
If the signal has been 0 for some time and at a time T_0, the signal goes to some some dc value, say a 1, then V_I(t) = u (t - T_0). This is a translated step.
When analyzing RC response to a rising step, we need to find out the voltage of the capacitor just before the step rises, v_c(0-).
It's been 0 for a long period of time- if there had been any voltage across the capacitor, it would
have discharged through the resistor, so therefore the initial condition on the capacitor is v_c(0-), is 0 volts.
After a long period of time, the input takes a step, the voltage will be V_I . The capacitor will charge up fully to V_I. The output is going to be of the form v_c= V_I(1-e^(-t/RC)) .
The circuit can be transformed into a Norton form. After a long period of time, the input takes a step, the current will be I. The capacitor will charge up fully to IR. Then the current, I, will begin flowing to the resistor and cause a voltage IR across the resistor. The output across the capacitor is going to be of the form v_c= I R(1-e^(-t/RC)) .
For the RC response to a falling step, the Norton style of the circuit is used. The reason a current source is used is that, for impulses, it makes it a little easier to think in terms of depositing charge on the capacitor.
When analyzing RC response to a falling step, we will also need to find out the voltage of the capacitor just before the step falls, v_c(0-).
After a long period of time, the capacitor looks like an open, so v_c(0-)=IR.
Then at time t equal to 0, the input goes down to 0. The capacitor that has an initial voltage IR across it, will now discharge through the resistor. The output across the capacitor is going to be of the form v_c= I R(e^(-t/RC)) .
For the RC response to a pulse input which starts at t=0 and has a long pulse width (more than time constant RC), the output across the capacitor is going to be of the form v_c= I R(1-e^(-t/RC)) in the beginning and then later v_c= I R(e^(-(t-T)/RC)) where T is the pulse width.
For the RC response to a pulse input which starts at t=0 and has a shorter pulse width (less than time constant RC), the output across the capacitor is going to be of the form v_c= I R(1-e^(-t/RC)) in the beginning and then later v_c= I R(1-e^(-T/RC)) *(e^(-(t-T)/RC)) where T is the pulse width.
(the point where the waveform starts to discharge is I R(1-e^(-T/RC)) )
Looking at the current pulse, we can see that the area of the pulse is Q. It is possible to make a pulse deliver the same charge when it is made more narrow, to the point that it becomes an impulse.
For the RC response to an impulse input, the waveform is v_c= Q/C(e^(-t/RC)). At t=0, v_c is at Q/C and analysis is done with Taylor series.
v_c(t) is largely independent of T. The capacitor behaves like a short circuit for changes that are much faster than the time constant. All charge goes to the capacitor V=Q/C. The impulse sets up initial conditions for the capacitor.
Current impulse delivers charge while voltage impulse delivers flux linkage. For the RL response to an impulse input, the waveform is i_L= λ/L(e^(-Rt/L)). In this case, the inductor behaves like an instantaneous open.
Superposition does not apply when initial conditions exist. It does apply when initial conditions are zero. One way of solving circuits with with multiple sources and
multiple initial conditions is the following:-
(1) Treat initial conditions like sources.
(2) Find the response to each source or each initial condition acting alone and then sum the partial response.
The notation for an impulse is δt. The notation for unit pulse is u(t) and the notation for ramp is tu(t). The other way of representing them are u_0(t) for impulse, u_-1(t) for unit pulse and u_-2(t) for ramp.
Integrating an impulse produces a step. Integrating a step produces a ramp.
For a linear system, differentiating the input will get the same output differentiated. If the input is integrated, then the corresponding output is integrated as well. This provides another way of finding the output, especially for ramps.
If there is an input that looks like a step and a ramp on top of the step, first find the response to a step. and integrate it to get that of a ramp. Then I add the two up. So using superposition, it is possible get the outputs for all kinds of inputs.
Going back to the slower may be better topic, one can see how parasitic capacitance between two pins of a chip can affect a circuit. Consider R_1 to be the resistance of the wire for pin l and R_0 to be the load resistance and that they are connected to a near ideal voltage source. Also consider pin 2 having a system consisting of an ideal voltage source with a much higher frequency waveform and a resistance R_2. The wires of pin 1 and pin 2 are so close together that they form a parasitic capacitance which causes crosstalk.
Crosstalk is when a signal on one wire influences the signal on another wire when it's not supposed to do so. By superposition, R_1 is in parallel with R_0 and is in series with the parasitic capacitance and the source with a much higher frequency waveform. Taking R_1 in parallel with R_0 to be equal to R. The voltage across the resistor is V_R=V_I*e^(-t/RC) which is an exponential decaying form. The positive spike corresponds to the rising edges of the source waveform while the negative spikes correspond to the falling part of it. These spikes are superimposed on top of the waveform due to the other voltage source.
Slowing down the edges of waveforms to that of triangular waveforms will produce a waveform with less glitches. ( no change in frequency)
Taking 1/A as a constant value to control the slope, integrating the step input will result in a ramp of the form (V_I*t)/A. Then by integrating the output of the step, the output of the ramp will be V_R= ((V_I*R*C_p)/A)*(1-e^(-t/RC_p)). If RC_p is much less than A then the output of the ramp will be a very small blip.
When analyzing RC response to a rising step, we need to find out the voltage of the capacitor just before the step rises, v_c(0-).
It's been 0 for a long period of time- if there had been any voltage across the capacitor, it would
have discharged through the resistor, so therefore the initial condition on the capacitor is v_c(0-), is 0 volts.
After a long period of time, the input takes a step, the voltage will be V_I . The capacitor will charge up fully to V_I. The output is going to be of the form v_c= V_I(1-e^(-t/RC)) .
The circuit can be transformed into a Norton form. After a long period of time, the input takes a step, the current will be I. The capacitor will charge up fully to IR. Then the current, I, will begin flowing to the resistor and cause a voltage IR across the resistor. The output across the capacitor is going to be of the form v_c= I R(1-e^(-t/RC)) .
For the RC response to a falling step, the Norton style of the circuit is used. The reason a current source is used is that, for impulses, it makes it a little easier to think in terms of depositing charge on the capacitor.
When analyzing RC response to a falling step, we will also need to find out the voltage of the capacitor just before the step falls, v_c(0-).
After a long period of time, the capacitor looks like an open, so v_c(0-)=IR.
Then at time t equal to 0, the input goes down to 0. The capacitor that has an initial voltage IR across it, will now discharge through the resistor. The output across the capacitor is going to be of the form v_c= I R(e^(-t/RC)) .
For the RC response to a pulse input which starts at t=0 and has a long pulse width (more than time constant RC), the output across the capacitor is going to be of the form v_c= I R(1-e^(-t/RC)) in the beginning and then later v_c= I R(e^(-(t-T)/RC)) where T is the pulse width.
For the RC response to a pulse input which starts at t=0 and has a shorter pulse width (less than time constant RC), the output across the capacitor is going to be of the form v_c= I R(1-e^(-t/RC)) in the beginning and then later v_c= I R(1-e^(-T/RC)) *(e^(-(t-T)/RC)) where T is the pulse width.
(the point where the waveform starts to discharge is I R(1-e^(-T/RC)) )
Looking at the current pulse, we can see that the area of the pulse is Q. It is possible to make a pulse deliver the same charge when it is made more narrow, to the point that it becomes an impulse.
For the RC response to an impulse input, the waveform is v_c= Q/C(e^(-t/RC)). At t=0, v_c is at Q/C and analysis is done with Taylor series.
v_c(t) is largely independent of T. The capacitor behaves like a short circuit for changes that are much faster than the time constant. All charge goes to the capacitor V=Q/C. The impulse sets up initial conditions for the capacitor.
Current impulse delivers charge while voltage impulse delivers flux linkage. For the RL response to an impulse input, the waveform is i_L= λ/L(e^(-Rt/L)). In this case, the inductor behaves like an instantaneous open.
Superposition does not apply when initial conditions exist. It does apply when initial conditions are zero. One way of solving circuits with with multiple sources and
multiple initial conditions is the following:-
(1) Treat initial conditions like sources.
(2) Find the response to each source or each initial condition acting alone and then sum the partial response.
The notation for an impulse is δt. The notation for unit pulse is u(t) and the notation for ramp is tu(t). The other way of representing them are u_0(t) for impulse, u_-1(t) for unit pulse and u_-2(t) for ramp.
Integrating an impulse produces a step. Integrating a step produces a ramp.
For a linear system, differentiating the input will get the same output differentiated. If the input is integrated, then the corresponding output is integrated as well. This provides another way of finding the output, especially for ramps.
If there is an input that looks like a step and a ramp on top of the step, first find the response to a step. and integrate it to get that of a ramp. Then I add the two up. So using superposition, it is possible get the outputs for all kinds of inputs.
Going back to the slower may be better topic, one can see how parasitic capacitance between two pins of a chip can affect a circuit. Consider R_1 to be the resistance of the wire for pin l and R_0 to be the load resistance and that they are connected to a near ideal voltage source. Also consider pin 2 having a system consisting of an ideal voltage source with a much higher frequency waveform and a resistance R_2. The wires of pin 1 and pin 2 are so close together that they form a parasitic capacitance which causes crosstalk.
Crosstalk is when a signal on one wire influences the signal on another wire when it's not supposed to do so. By superposition, R_1 is in parallel with R_0 and is in series with the parasitic capacitance and the source with a much higher frequency waveform. Taking R_1 in parallel with R_0 to be equal to R. The voltage across the resistor is V_R=V_I*e^(-t/RC) which is an exponential decaying form. The positive spike corresponds to the rising edges of the source waveform while the negative spikes correspond to the falling part of it. These spikes are superimposed on top of the waveform due to the other voltage source.
Slowing down the edges of waveforms to that of triangular waveforms will produce a waveform with less glitches. ( no change in frequency)
Taking 1/A as a constant value to control the slope, integrating the step input will result in a ramp of the form (V_I*t)/A. Then by integrating the output of the step, the output of the ramp will be V_R= ((V_I*R*C_p)/A)*(1-e^(-t/RC_p)). If RC_p is much less than A then the output of the ramp will be a very small blip.
No comments:
Post a Comment