Information about RC and LC circuits
Thanking In advance,
Cheers,
Pankaj
usually you can read them on books discussing filters.
for low pass filters:
Low pass RC filter.
The two circuits to the right behave the same way for the most part, but there are some important differences. Both circuits operate as low-pass filters. That is, they will readily transmit signals below a certain frequency from input to output, with no appreciable loss in signal amplitude.
The first circuit shows a resistor and a capacitor, connected as a voltage divider. We have already looked at the behavior of this circuit with a single input frequency. The question now is, how does it behave for a range of frequencies?
One thing we can note at once: at very low frequencies, XC will be quite large, so C will have no practical effect on the signal. In essence, there will only be R in series with the signal, between vIN and vOUT. At the same time, C does provide dc isolation from ground. Sometimes this is important.
Low pass RL filter.
Our second circuit on the right is a low-pass RL filter. At low frequencies, the series inductance has negligible effect on the signal, and we essentially have a resistor connecting the signal line to ground. This is often designated a shunt resistance, indicating that it is in parallel with the signal.
As the frequency increases, XL also increases, thus increasing the total impedance of the filter. Since R doesn't change, more and more of the signal is dropped across L at higher frequencies, leaving less and less across R and available at vOUT.
At audio frequencies, the RC filter is generally preferred. This has nothing to do with any inherent benefit or lack of one circuit over the other, but rather is strictly a question of cost. Inductors for the audio frequency band are generally physically large and have heavy iron cores. This makes them expensive, and also causes them to take up a lot of space inside the cabinet of the device using it. In addition, inductors can be affected by external magnetic fields, such as the fields surrounding all power lines in homes and offices. Capacitors are much smaller, lighter, cheaper, and less susceptible to external energies, so they are used far more often.
The Frequency Response Curve
Normalized frequency response of a low pass filter.
Regardless of which circuit you choose for your filter, its frequency response will match the curve to the right. Note that the frequency scale is logarithmic rather than linear. In addition, the attenuation of the filter is shown in units called decibels (db). This is also a logarithmic unit used to show ratios. Mathematically, voltage gain or loss is described as:
Gain or Attenuation (db) = 20 log10 vOUT
vIN
Because for these filters vOUT is always less than (or equal to) vIN, the ratio is a fraction no higher than 1, so the logarithm of this fraction will be negative or zero. Thus, a voltage loss is expressed as negative decibels. A voltage gain would be expressed in positive decibels.
Regardless of the component values we use in our filter, there will be one particular frequency at which R = XC (for the RC filter), or R = XL (for the RL filter). At this frequency, vOUT/vIN = 0.707, and the calculation above will indicate a ratio of -3 db. This point is shown clearly on the graph. At lower frequencies, attenuation is reduced and rapidly becomes 0 db. At higher frequencies, attenuation increases rapidly until it falls off at a constant slope of 20 db per frequency decade as shown.
The frequency at which the two components have equal individual impedances (without regard to phase shift) is called the cut-off frequency of that filter, designated fco or ωco, according to whether we have specified frequency in Hz or radians/second. Since this graph is normalized, that frequency is shown as 1 (actually 1 times the actual cut-off frequency). The Frequency axis on the graph thus indicates frequencies in powers of 10 above and below the cut-off frequency. The shape of the graph will remain the same for any first-order RC or RL low-pass filter of the types described on this page.
Frequencies below the cut-off frequency will be passed through without significant loss, while frequencies above the cut-off frequency will be attenuated significantly. This in fact is how the filter gets its name: it passes low frequencies but not high frequencies.
If the constant slope is extended up to the 0 db line as shown in green in the graph, the intersection point is once again the cut-off frequency. In many circumstances it is useful and easier to simply use the straight lines to describe a filter's performance. This is an acceptable approximation, and the actual curve is understood to be present.
Phase Response
Voltage vectors for first-order low-pass filters.
For a first-order low-pass filter, vOUT always lags vIN by some phase angle betweeen 0 and 90°. The vector diagrams to the right show why.
The RL vector diagram at the top properly shows that vL leads vR by 90°. However, the input voltage is applied to the series combination of the two components. Thus, vL leads vIN, while vR lags vIN. Since the output is taken from across R, vOUT = vR and lags behind vIN by some phase angle that depends on frequency.
In the case of the RC low-pass filter, vC lags vR, and again vIN is applied to the series combination of the two components. This time, however, the output is taken from across C, so vOUT = vC, which again lags vIN by some angle in the range 0 to 90°, according to the signal frequency.
Phase response of a first-order low-pass filter.
Most of the variation in phase angle occurs within one decade of the cutoff frequency, as shown in the figure to the right. Note that at frequencies well below cutoff, there is essentially no phase shift. The phase lag begins to become significant about a decade below ωco, and reaches 45° (or π/4 radians) at ωco itself. Above ωco, the output phase continues to change rapidly during the first decade, by which time the phase lag is close to 90° (π/2).
Because of this shifting phase lag, non-sinusoidal signals with frequency components near ωco will be distorted by this filter. Such distortion must be taken into account when designing filters of this type for some kinds of applications.
The Calculations
We perform our calculations without considering the loading effects of any circuit connected to vOUT. Any such loading effects are calculated separately. Therefore, signal current is the same through both components in the filter, and the ratio vOUT/vIN is equal to the ratio ZOUT/ZIN.
In the case of the RC filter, ZOUT = XC. For the RL filter, ZOUT = R. On the input side, Z is always the series combination of the two elements. We'll use the RC filter as our main example here; you'll find that the RL filter works out to the same result.
We are using normalized values for these calculations. This means that we arbitrarily assign values as follows:
* R = 1Ω.
* C = 1f.
* L = 1H.
For convenience, we will use frequency in radians/second, symbolized by the Greek letter ω (omega), rather than frequency in Hz. You need only remember when we're done with our calculations that ω = 2πf. But for the calculations themselves, this allows us to use XC = 1/ωC and avoid performing lots of calculations using π. This simplifies many calculations, and still allows the circuit to be scaled to any impedance and any frequency.
We begin at the cut-off frequency, since that is what controls our component values. We know that R = XC at this frequency, so we have:
R = XC
= 1
ωC
ω = 1
RC
Since R = 1Ω and C = 1f, we can immediately calculate that ω = 1 radian/second. Then, at the cut-off frequency we have:
XC = 1
ωC
= 1
1
= 1 Ω
Z = (R2 + XC2)?
= (12 + 12)?
= Ω
Thus, at the cut-off frequency (generally designated ωco), XC/Z = 1/ = 0.707 as we determined above. To calculate the vOUT/vIN ratio at all frequencies, we leave ω as a variable. We can also make it complete with R and C accounted for, as follows:
vOUT = XC
vIN Z
= 1/ωC
(R2 + 1/(ωC)2)?
= 1
ωC(R2 + 1/(ωC)2)?
= 1
((RωC)2 + 1)?
If we use the values R = 1Ω and C = 1f as we did above, and let ω vary over a range from 0.01 to 100 radians/second, and then convert the resulting ratios to db, we will get exactly the curve shown in the graph above, in blue. In fact, that's how this graph was computed. Many years of experiments have shown that it is valid, and accurately portrays the frequency response of a real low-pass filter.
The RL low-pass filter may be computed in the same way, except that here the ratio vOUT/vIN = R/Z, and Z = (R2 + (ωL)2)?. The final formula is very similar, except that we must replace (RωC)2 with (ωL/R)2. For the RL low-pass filter, ωco = R/L. Thus, if we pick the normalized values of R = 1Ω and L = 1H, we will get exactly the same curve as we plotted above for the RC filter.
The phase response for both filters near ωco is exactly the same, and may be calculated as:
φ = arctan - R = arctan - XL
XC R
= arctan - ωRC = arctan - ωL
R
for high pass filters:
High-Pass Filters
The Circuits
High pass RC filter.
The basic first-order high-pass filters use the same components as the low-pass filters we just studied. However, their positions are swapped. Thus, the RC high-pass filter has the capacitor in series with the signal and the resistor across the output, as shown in the first diagram to the right. At high frequencies, C has very low impedance, and the signal passes through unhindered. As the frequency decreases, however, XC becomes significant, until at the cutoff frequency, XC = R, just as with the low-pass filter. At still lower frequencies, XC increases, and less of the signal reaches the output.
One useful feature of the RC high-pass filter is that the capacitor serves to block direct current between vIN and vOUT. Thus, two circuits that operate at different DC voltages can be connected by this type of high-pass filter without encountering any problems with dc component bias voltages as a consequence.
High pass RL filter.
The RL version of the high-pass filter uses a series resistor and a shunt inductor to accomplish its purpose. At high frequencies, XL is large, so the inductor is nearly an open circuit for such signals. At low frequencies, XL is very small, and effectively connects those signals directly to ground. As before, the cutoff frequency occurs where R = XL, so that L is just beginning to have a significant effect on the signal.
The Frequency Response Curve
Normalized frequency response of a high pass filter.
As shown to the right, the frequency response of a basic high-pass filter is actually a mirror image of its low-pass counterpart. At the cutoff frequency where R = XL or R = XC, the attenuation is only 3 db, so the signal voltage is still 70.7% of its higher-frequency value.
Below the cutoff frequency, attenuation increases at the rate of 20 db per decade, which is the same roll-off as for the low-pass filter. Above the cutoff frequency, attenuation rapidly decreases to nothing, and all higher frequencies pass with ease.
The green line in the graph is the straight line extension of the constant slopes of the actual frequency response. As with the low-pass filter, the intersection point is the cutoff frequency. This straight-line approximation of the real frequency response curve is very easy to draw, and is sufficiently accurate for some kinds of applications. Of course, the actual curve near the cutoff frequency is understood.
Phase Response
Voltage vectors for first-order high-pass filters.
As we have already seen, in a first-order low-pass filter, vOUT always lags vIN by some phase angle betweeen 0 and 90°. Exactly the reverse is true for a first-order high-pass filter: as shown in the vector diagrams to the right, vOUT is always taken from across the component whose voltage leads vIN by some phase angle, φ.
For the RL filter, vOUT is taken from across L, so its phase angle necessarily leads vIN as shown in the upper vector diagram. For the RC filter, vOUT is taken from across R, which again leads vIN as shown in the lower vector diagram.
Phase response of a first-order high-pass filter.
Of course, the actual phase angle by which vOUT leads vIN depends on the specific frequency of the signal, as compared to the cutoff frequency of the filter. As shown in the phase diagram to the right, signals more than 10 times the cutoff frequency show little or no appreciable phase shift, while signals less than 0.1 times the cutoff frequency are shifted close to 90°. Most of the change in phase occurs within a factor of 0.1 to 10 times ωCO.
As with the low-pass filter, non-sinusoidal signals with frequency components at or near ωCO will be distorted when passing through the high-pass filter.
The Equations
The equations for the high-pass filter are very similar to the ones for the low-pass filter, as you might expect. Certainly the basic comparisons are the same. Thus, for the RC circuit at the cutoff frequency,
R = XC
= 1
ωCOC
ωCO = 1
RC
This much is the same as for the RC low-pass filter. However, because the components have been swapped, the equation for attenuation over a frequency range has become:
vOUT = R
vIN Z
= R
(R2 + XC2)?
= R
(R2 + (1/ωC)2)?
= 1
1/R × (R2 + (1/ωC)2)?
= 1
(1 + (1/RωC)2)?
In this equation, the higher the value of ω, the less effect it has on vOUT. This is exactly the reverse of the low-pass filter, where high values of ω would seriously reduce vOUT. This is in fact the essential difference between the low-pass filter and the high-pass filter.
The RL filter behaves the same way, except that at the cutoff frequency:
XL = R
ωCOL = R
ωCO = R
L
Then, over the frequency spectrum,
vOUT = XL
vIN Z
= XL
(R2 + XL2)?
= ωL
(R2 + (ωL)2)?
= 1
1/ωL × (R2 + (ωL)2)?
= 1
((R/ωL)2 + 1)?
At high frequencies, the fraction R/ωL becomes very small and has negligible effect on the signal. At very low frequencies, this fraction becomes very large and blocks nearly all of the signal. And at cutoff, ω = ωCO = R/L, so that vOUT/vIN = 1/ = 0.707 as expected. Thus, the frequency response of the RL filter is exactly the same as the frequency response of the RC filter.
As with the low-pass filters, the phase response for both high-pass filters near ωco is exactly the same, and may be calculated as:
φ = arctan XC = arctan R
R XL
= arctan 1 = arctan R
ωRC ωL
and for the bandpass filters:
Band-Pass Filters
The Circuit
RC band-pass filter.
We've looked at low-pass filters and high-pass filters individually. But what happens if we combine them into a single circuit, as shown to the right?
In this case, R1 and C1 form a high-pass filter, while R2 and C2 form a low-pass filter. For the purpose of discussion, we arbitrarily assign a cutoff frequency ωCO1 = 10 radians/sec for R1 and C1, and a higher cutoff frequency ωCO2 = 10,000 radians/sec for R2 and C2. The actual frequencies don't matter, so long as ωCO1 is less than ωCO2. That way, R1 and C1 pass signals that will also be passed by R2 and C2.
It is equally possible to swap the two filter sections, putting the low-pass filter first. However, if we use the circuit shown here, the dc resistance between vOUT and ground is R1 + R2. If we swap the two filters, R1 will be the only resistance from vOUT to ground. In addition, the second filter section will present a load to the first section. Since the low-pass section has a higher cutoff frequency (ωCO2), R2 and C2 have higher impedances and constitute less of a load on R1 and C1 than would be true if the sections were swapped. Therefore the two filters operate pretty much independently, even though they are electrically connected.
The Frequency Response
Band-pass frequency response curve.
If we apply the cutoff frequencies assumed above, the frequency response curve for our filter will appear as shown to the right. R1 and C1 govern the low-frequency cutoff, and will block signals at lower frequencies while passing higher-frequency signals.
These signals will also be passed by R2 and C2, so long as their frequency doesn't get too high. Frequencies above ωCO2 pass through C2 to ground, and therefore are kept away from vOUT.
The actual band or range of frequencies passed by this type of filter does not have to cover three decades as shown here. The two parts of the band-pass filter can be adjusted independently of each other to widen or narrow the pass band as much as you like. The minimum effective pass band occurs when ωCO1 is set equal to ωCO2. Then the response curve peaks at the mutual cutoff frequency and rolls off immediately on either side.
If you attempt to set ωCO1 to a higher frequency than ωCO2, the band-pass filter will block all frequencies, and no signal will get through.
Phase Response
Phase response of a first-order bandpass filter.
Because the band-pass filter is actually two independent first-order filters, the phase response of the entire circuit is simply the combination of the phase responses of the two separate sections. This combined phase response is shown in the graph to the right.
In this case, the pass band is only three decades wide, so the output phase shift is zero only for a very narrow range of frequencies. A wider pass band would mean a correspondingly wider frequency range with no phase shift. A narrower pass band results in a narrower frequency range with no phase shift.
In the case where the two filter sections have the same cutoff frequency, the phase lead from the high-pass section cancels the phase lag from the low-pass section at the cutoff frequency only, so that is the only frequency with no phase shift.
The Calculations
There are no special extra calculations required for the band-pass filter. In our example circuit, the high-pass filter comes first, and has its effect on the signal. vOUT from the high-pass filter becomes vIN for the low-pass filter, which then has its effect on the signal. The two filters don't really interact with each other, beyond the fact that the second filter, depending on component values, may act as a load on the first one. To minimize this effect, we generally put the high-pass filter first, since the low-pass filter, with a higher cutoff frequency, will have higher relative impedance values for its components.