# How To Calculate Pulse Width From Frequency?

This is the pulse width, or PW, of the signal. Calculate the period, or ‘T’, of the frequency, or ‘f,’ using the formula: T = 1/f. For example, if the frequency is 20 hz, then T = 1/20, with a result of 0.05 seconds.

#### How is pulse width calculated?

Impulse (abbrev. IMPL) Alternate term for Upper Level System and Shortwave ; a general term for any large-scale or mesoscale disturbance capable of producing upward motion (lift) in the middle or upper parts of the atmosphere. Peak Pulse The amount of power transmitted by a radar during a given pulse.

1. Note that because these pulses are widely spaced, the average power will be much smaller.
2. Pulse A short burst of electromagnetic energy that a radar sends out in a straight line to detect a precipitation target.
3. The straight line that this pulse travels along is called a radar beam.
4. Pulse Duration The time over which a radar pulse lasts.

The pulse duration can be multiplied by the speed of light to determine the pulse length or pulse width. Pulse Length The linear distance in range occupied by an individual pulse from a radar. h = c * t, where t is the duration of the transmitted pulse, c is the speed of light, h is the length of the pulse in space.

• Note, in the radar equation, the length h/2 is actually used for calculating pulse volume because we are only interested in signals that arrive back at the radar simultaneously.
• This is also called a pulse width.
• Pulse Radar A type of radar, designed to facilitate range (distance) measurements, in which are transmitted energy emitted in periodic, brief transmission.

Pulse Repetition Frequency (PRF) The amount of time between successive pulses, or bursts, of electromagnetic energy that is transmitted by a radar. The PRF determines the maximum range at which echoes can be detected and also the maximum radial velocity that can be detected by a Doppler radar.

• Pulse Repetition Time (PRT) The time elapsed between pulses by the radar.
• This is also called the pulse interval.
• Pulse Resolution Volume A discrete radar sampling volume, of dimensions (horizontal beamwidth * vertical beamwidth * 1 range gate).
• Pulse Severe Thunderstorms Single cell thunderstorms which produce brief periods of severe weather (3/4 inch hail, wind gusts in the excess of 58 miles an hour, or a tornado).

Pulse Storm A thunderstorm within which a brief period (pulse) of strong updraft occurs, during and immediately after which the storm produces a short episode of severe weather. These storms generally are not tornado producers, but often produce large hail and/or damaging winds.

See also overshooting top. Pulse Width Same as Pulse Length ; the linear distance in range occupied by an individual pulse from a radar. h = c * t, where t is the duration of the transmitted pulse, c is the speed of light, h is the length of the pulse in space. Note, in the radar equation, the length h/2 is actually used for calculating pulse volume because we are only interested in signals that arrive back at the radar simultaneously.

Pulse-Pair Processing Nickname for the technique of mean velocity estimation by calculation of the signal complex covariance argument. The calculation requires two consecutive pulses, hence “pulse-pair”. Sudden Impulse (SI+ or SI-) In solar-terrestrial terms, a sudden perturbation of several gammas in the northward component of the low-latitude geomagnetic field, not associated with a following geomagnetic storm.

#### How is pulse width related to frequency?

The pulse width is a measure of the elapsed time between the leading and trailing edges of a single pulse of energy. The measure is typically used with electrical signals and is widely used in the fields of radar and power supplies, There are two closely related measures.

The pulse repetition interval measures the time between the leading edges of two pulses but is normally expressed as the pulse repetition frequency (PRF), the number of pulses in a given time, typically a second. The duty cycle expresses the pulse width as a fraction or percentage of one complete cycle.

Pulse width is an important measure in radar systems. Radars transmit pulses of radio frequency energy out of an antenna and then listen for their reflection off of target objects. The amount of energy that is returned to the radar receiver is a function of the peak energy of the pulse, the pulse width, and the pulse repetition frequency.

• Increasing the pulse width increases the amount of energy reflected off the target and thereby increases the range at which an object can be detected.
• Radars measure range based on the time between transmission and reception, and the resolution of that measurement is a function of the length of the received pulse.

This leads to the basic outcome that increasing the pulse width allows the radar to detect objects at longer range but at the cost of decreasing the accuracy of that range measurement. This can be addressed by encoding the pulse with additional information, as is the case in pulse compression systems.

### How do you calculate pulse frequency?

Pulse frequency is calculated by dividing 1000 by the total cycle time (on-time + off-time) in microseconds (44).

#### Does frequency affect pulse width?

PWM Characteristics Continued – We discussed the vast array of applications that ideally suit PWM’s functionality, including LEDs and motors (servo). Since frequency is a primary component of the PWM technique, it is understandable that frequency affects PWM’s ability to exert control within an application.

• Therefore, the square wave frequency does need to be sufficiently high enough if controlling LEDs, for example, to get the proper dimming effect.
• As an example, a duty cycle of 20% at 1 Hz will be noticeable to the human eye that an LED is turning OFF and ON.
• Whereas, a duty cycle of 20% at 100 Hz or higher will merely exhibit a slightly less dim light output.

As I am sure you are aware, we can utilize PWM to control motors (servo). We can also use it to control a servo motor’s angle. In terms of applications, this is beneficial when we attach it to a mechanical device like a robotic arm in an assembly or manufacturing environment.

## What is PWM value?

Learn how PWM works and how to use it in a sketch. The Fading example demonstrates the use of analog output (PWM) to fade an LED. It is available in the File->Sketchbook->Examples->Analog menu of the Arduino software. Pulse Width Modulation, or PWM, is a technique for getting analog results with digital means. Digital control is used to create a square wave, a signal switched between on and off. This on-off pattern can simulate voltages in between the full Vcc of the board (e.g., 5 V on UNO, 3.3 V on a MKR board) and off (0 Volts) by changing the portion of the time the signal spends on versus the time that the signal spends off. The duration of “on time” is called the pulse width. To get varying analog values, you change, or modulate, that pulse width. If you repeat this on-off pattern fast enough with an LED for example, the result is as if the signal is a steady voltage between 0 and Vcc controlling the brightness of the LED. In the graphic below, the green lines represent a regular time period. This duration or period is the inverse of the PWM frequency. In other words, with Arduino’s PWM frequency at about 500Hz, the green lines would measure 2 milliseconds each. A call to analogWrite () is on a scale of 0 – 255, such that analogWrite ( 255 ) requests a 100% duty cycle (always on), and analogWrite ( 127 ) is a 50% duty cycle (on half the time) for example. Once you get this example running, grab your Arduino and shake it back and forth. What you are doing here is essentially mapping time across the space. To our eyes, the movement blurs each LED blink into a line. As the LED fades in and out, those little lines will grow and shrink in length. Now you are seeing the pulse width. Written by Timothy Hirzel

#### What is bandwidth pulse width?

Width is the duty cycle of a pulse in time domain, which is inversely proportional to the pulse bandwidth in frequency domain. The pulse width of a Gaussian pulse is defined as the pulse’s temporal width at half of the maximum amplitude.

## What is PWM output frequency?

The PWM output provides a so-called pulse width modulated voltage with a fixed high frequency ( 53.6 kHz ) to the motor by the ESCON’s power stage. This motor voltage adjusts the motor’s speed and current.

## Does PWM change frequency?

PWM (Pulse width modulation) is one of the most useful feature used in many applications. PWM is used by using function like ‘analog Write’. With this function although width of the PWM cycle(Duty Cycle) can be changes but frequency remains constant.

#### Is bandwidth the same as pulse width?

• Frequency (MHz)
• B = 1.6 MHz
• Figure 1: A graph of a bandpass filter’s gain magnitude, illustrating the concept of -3 dB (or half-power) bandwidth

1. Frequency (MHz)
2. B = 1.6 MHz
3. Figure 1: A graph of a bandpass filter’s gain magnitude, illustrating the concept of -3 dB (or half-power) bandwidth
• Frequency (MHz)
• B = 1.6 MHz
• Figure 1: A graph of a bandpass filter’s gain magnitude, illustrating the concept of -3 dB (or half-power) bandwidth

Bandwidth B, BW or Δf is the difference between the upper and lower cut-off frequencies of radar receiver, and is typically measured in hertz. In case of a baseband channel or video signal, the bandwidth is equal to its upper cut-off frequency. In a Radar receiver the bandwidth is mostly determined by the IF filter stages.

1. The receiver must be able to process the signal bandwidth of the backscattered pulse.
2. The wider the bandwidth, the greater the degree of that will be input to the receiver.
3. Since noise exists at all frequencies, the broader the frequency range to which the receiver bandpass filters are tuned, then the higher the intensity level of the noise and the lower the, and so the receivers sensibility.

The bandwidth is roughly proportional to the amount of information carried by the signal. To detect a rectangle pulse with the (FFT) the bandwidth of the receiver is equal to the highest sine wave frequency component that is significant. The larger the bandwidth of the receiver, the shorter the rise time of the edges of the rectangular pulse.

• Generally the necessary bandwidth B of a pulse in form of a half wave sine signal of duration τ is: The influence of the will change the signal duration and bandwidth of the backscattered pulse.
• To obtain the Doppler information the installed bandwidth of the radar receiver must be higher than the signal bandwidth of the transmitted pulse.

In radar system using the of the transmitted pulse, the necessary bandwidth of radar receiver is much higher than the reciprocal of their pulse width. In this case the necessary bandwidth of radar receiver depends on the internal modulation of the signal, the compressed pulse width and a weighting function, to achieve the required,

### Is pulses per second frequency?

A pulse per second ( PPS or 1PPS ) is an electrical signal that has a width of less than one second and a sharply rising or abruptly falling edge that accurately repeats once per second, PPS signals are output by radio beacons, frequency standards, other types of precision oscillators and some GPS receivers.

## What is the formula for pulse repetition frequency?

• Principle of Operation
• Radar is an acronym for Radio Detection and Ranging. The term “radio” refers to the use of electromagnetic waves with wavelengths in the so-called radio
• wave portion of the spectrum, which covers a wide range from 10 4 km to 1 cm. Radar
• return. If the time delay is D t, then the range may be determined by the simple formula:

systems typically use wavelengths on the order of 10 cm, corresponding to frequencies of about 3 GHz. The detection and ranging part of the acronym is accomplished by timing the delay between transmission of a pulse of radio energy and its subsequent R = c D t/2 where c = 3 x 10 8 m/s, the speed of light at which all electromagnetic waves propagate.

The factor of two in the formula comes from the observation that the radar pulse must travel to the target and back before detection, or twice the range. A radar pulse train is a type of amplitude modulation of the radar frequency carrier wave, similar to how carrier waves are modulated in communication systems.

In this case, the information signal is quite simple: a single pulse repeated at regular intervals. The common radar carrier modulation, known as the pulse train is shown below. The common parameters of radar as defined by referring to Figure 1. PW = pulse width. PW has units of time and is commonly expressed in m s. PW is the duration of the pulse. RT = rest time. RT is the interval between pulses. It is measured in m s. PRT = pulse repetition time. PRT has units of time and is commonly expressed in ms.

PRT is the interval between the start of one pulse and the start of another. PRT is also equal to the sum, PRT = PW+RT. PRF = pulse repetition frequency. PRF has units of time -1 and is commonly expressed in Hz (1 Hz = 1/s) or as pulses per second (pps). PRF is the number of pulses transmitted per second and is equal to the inverse of PRT.

RF = radio frequency. RF has units of time -1 or Hz and is commonly expressed in GHz or MHz. RF is the frequency of the carrier wave which is being modulated to form the pulse train.

1. Mechanization
2. A practical radar system requires seven basic components as illustrated below:
3. Figure 3
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Transmitter, The transmitter creates the radio wave to be sent and modulates it to form the pulse train. The transmitter must also amplify the signal to a high power level to provide adequate range. The source of the carrier wave could be a Klystron, Traveling Wave Tube (TWT) or Magnetron. Each has its own characteristics and limitations.2. Receiver, The receiver is sensitive to the range of frequencies being transmitted and provides amplification of the returned signal. In order to provide the greatest range, the receiver must be very sensitive without introducing excessive noise. The ability to discern a received signal from background noise depends on the signal-to-noise ratio (S/N). The background noise is specified by an average value, called the noise-equivalent-power (NEP). This directly equates the noise to a detected power level so that it may be compared to the return. Using these definitions, the criterion for successful detection of a target is P r > (S/N) NEP, where P r is the power of the return signal. Since this is a significant quantity in determining radar system performance, it is given a unique designation, S min, and is called the Minimum Signal for Detection,

• S min = (S/N) NEP
• Since S min, expressed in Watts, is usually a small number, it has proven useful to define the decibel equivalent, MDS, which stands for Minimum Discernible Signal,
• MDS = 10 Log (S min /1 mW)

When using decibels, the quantity inside the brackets of the logarithm must be a number without units. I the definition of MDS, this number is the fraction S min /1 mW. As a reminder, we use the special notation dBm for the units of MDS, where the “m” stands for 1 mW.

This is shorthand for decibels referenced to 1 mW, which is sometimes written as dB//1mW. In the receiver, S/N sets a threshold for detection which determines what will be displayed and what will not. In theory, if S/N = 1, then only returns with power equal to or greater than the background noise will be displayed.

However, the noise is a statistical process and varies randomly. The NEP is just the average value of the noise. There will be times when the noise exceeds the threshold that is set by the receiver. Since this will be displayed and appear to be a legitimate target, it is called a false alarm,

If the SNR is set too high, then there will be few false alarms, but some actual targets may not be displayed known as a miss). If SNR is set too low, then there will be many false alarms, or a high false alarm rate (FAR). Some receivers monitor the background and constantly adjust the SNR to maintain a constant false alarm rate, and therefore all called CFAR receivers.

Some common receiver features are: 1.) Pulse Integration. The receiver takes an average return strength over many pulses. Random events like noise will not occur in every pulse and therefore, when averaged, will have a reduced effect as compared to actual targets that will be in every pulse.2.) Sensitivity Time Control (STC).

This feature reduces the impact of returns from sea state. It reduces the minimum SNR of the receiver for a short duration immediately after each pulse is transmitted. The effect of adjusting the STC is to reduce the clutter on the display in the region directly around the transmitter. The greater the value of STC, the greater the range from the transmitter in which clutter will be removed.

However, an excessive STC will blank out potential returns close to the transmitter.3.) Fast Time Constant (FTC). This feature is designed to reduce the effect of long duration returns that come from rain. This processing requires that strength of the return signal must change quickly over it duration.

Since rain occurs over and extended area, it will produce a long, steady return. The FTC processing will filter these returns out of the display. Only pulses that rise and fall quickly will be displayed. In technical terms, FTC is a differentiator, meaning it determines the rate of change in the signal, which it then uses to discriminate pulses which are not changing rapidly.3.

Power Supply, The power supply provides the electrical power for all the components. The largest consumer of power is the transmitter which may require several kW of average power. The actually power transmitted in the pulse may be much greater than 1 kW.

1. The power supply only needs to be able to provide the average amount of power consumed, not the high power level during the actual pulse transmission.
2. Energy can be stored, in a capacitor bank for instance, during the rest time.
3. The stored energy then can be put into the pulse when transmitted, increasing the peak power.

The peak power and the average power are related by the quantity called duty cycle, DC. Duty cycle is the fraction of each transmission cycle that the radar is actually transmitting. Referring to the pulse train in Figure 2, the duty cycle can be seen to be: DC = PW / PRF Synchronizer,

• The synchronizer coordinates the timing for range determination.
• It regulates that rate at which pulses are sent (i.e.
• Sets PRF) and resets the timing clock for range determination for each pulse.
• Signals from the synchronizer are sent simultaneously to the transmitter, which sends a new pulse, and to the display, which resets the return sweep.

Duplexer, This is a switch which alternately connects the transmitter or receiver to the antenna. Its purpose is to protect the receiver from the high power output of the transmitter. During the transmission of an outgoing pulse, the duplexer will be aligned to the transmitter for the duration of the pulse, PW.

• After the pulse has been sent, the duplexer will align the antenna to the receiver.
• When the next pulse is sent, the duplexer will shift back to the transmitter.
• A duplexer is not required if the transmitted power is low. Antenna,
• The antenna takes the radar pulse from the transmitter and puts it into the air.

Furthermore, the antenna must focus the energy into a well-defined beam which increases the power and permits a determination of the direction of the target. The antenna must keep track of its own orientation which can be accomplished by a synchro-transmitter.

There are also antenna systems which do not physically move but are steered electronically (in these cases, the orientation of the radar beam is already known a priori ). The beam-width of an antenna is a measure of the angular extent of the most powerful portion of the radiated energy. For our purposes the main portion, called the main lobe, will be all angles from the perpendicular where the power is not less than ½ of the peak power, or, in decibels, -3 dB.

The beam-width is the range of angles in the main lobe, so defined. Usually this is resolved into a plane of interest, such as the horizontal or vertical plane. The antenna will have a separate horizontal and vertical beam-width. For a radar antenna, the beam-width can be predicted from the dimension of the antenna in the plane of interest by q = l /L where: q is the beam-width in radians, l is the wavelength of the radar, and L is the dimension of the antenna, in the direction of interest (i.e.

1. Width or height).
2. In the discussion of communications antennas, it was stated that the beam-width for an antenna could be found using q = 2 l /L.
3. So it appears that radar antennas have one-half of the beam-width as communications antennas.
4. The difference is that radar antennas are used both to transmit and receive the signal.

The interference effects from each direction combine, which has the effect of reducing the beam-width. Therefore when describing two-way systems (like radar) it is appropriate to reduce the beam-width by a factor of ½ in the beam-width approximation formula.

1. The directional gain of an antenna is a measure of how well the beam is
2. gain would merely be the ratio 2 p / q, Since the same power is distributed over a smaller range of angles, directional gain represents the amount by which the power
3. G dir = 4 p/q f
4. since there are 4 p steradians corresponding to all directions (solid angle, measured in steradians, is defined to be the area of the beam front divided by the range squared, therefore a non-directional beam would cover an area of 4 p R 2 at distance R, therefore 4 p steradians).
5. Here we used:

focused in all angles. If we were restricted to a single plane, the directional in the beam is increased. In both angles, then directional gain would be given by: q = horizontal beam-width (radians) f = vertical beam-width (radians) Sometimes directional gain is measured in decibels, namely 10 log (G dir ).

As an example, an antenna with a horizontal beam-width of 1.5 0 (0.025 radians) and vertical beam-width of 20 o (0.33 radians) will have: directional gain(dB) = 10 log (4 p / 0.025 0.333) = 30.9 dB Example: find the horizontal and vertical beam-width of the AN/SPS-49 long range radar system, and the directional gain in dB.

The antenna is 7.3 m wide by 4.3 m tall, and operates at 900 MHz. The wavelength, l =c/f = 0.33 m. Given that L= 7.3 m, then q = l /L = 0.33/7.3 = 0.045 radians, or q = 3 0, The antenna is 4.3 m tall, so a similar calculation gives f = 0.076 radians f = 4 0, The directional gain, G dir = 4p /(0.045 0.076) = 3638. Expressed in decibels, directional gain = 10 Log(3638) = 35.6 dB.

Display, The display unit may take a variety of forms but in general is designed to present the received information to an operator. The most basic display type is called an A-scan (amplitude vs. Time delay). The vertical axis is the strength of the return and the horizontal axis is the time delay, or range.

The A-scan provides no information about the direction of the target. Figure 4 The most common display is the PPI (plan position indicator). The A-scan information is converted into brightness and then displayed in the same relative direction as the antenna orientation. The result is a top-down view of the situation where range is the distance from the origin.

• Figure 5
• In this example, the use of increased STC to suppress the sea clutter would be helpful.

All of the parameters of the basic pulsed radar system will affect the performance in some way. Here we find specific examples and quantify this dependence where possible. Pulse Width The duration of the pulse and the length of the target along the radial direction determine the duration of the returned pulse.

In most cases the length of the return is usually very similar to the transmitted pulse. In the display unit, the pulse (in time) will be converted into a pulse in distance. The range of values from the leading edge to the trailing edge will create some uncertainty in the range to the target. Taken at face value, the ability to accurately measure range is determined by the pulse width.

If we designate the uncertainty in measured range as the range resolution,

1. R RES, then it must be equal to the range equivalent of the pulse width, namely:
2. R RES = c PW/2

Now, you may wonder why not just take the leading edge of the pulse as the range which can be determined with much finer accuracy? The problem is that it is virtually impossible to create the perfect leading edge. In practice, the ideal pulse will really appear like:

• Figure 6
• To create a perfectly formed pulse with a vertical leading edge would require an infinite bandwidth. In fact you may equate the bandwidth, b, of the transmitter to the minimum pulse width, PW by:
• PW = 1/2 b
• Given this insight, it is quite reasonable to say that the range can be determined no more accurately than cPW/2 or equivalently
• R RES = c/4 b

In fact, high resolution radar is often referred to as wide-band radar which you now see as equivalent statements. One term is referring to the time domain and the other the frequency domain. The duration of the pulse also affects the minimum range at which the radar system can detect. Figure 7 Increasing the pulse width while maintaining the other parameters the same will also affect the duty cycle and therefore the average power. For many systems, it is desirable to keep the average power fixed. Then the PRF must be simultaneously changed with PW in order to keep the product PW x PRF the same.

For example, if the pulse width is reduced by a factor of ½ in order to improve the resolution, then the PRF is usually doubled. Pulse Repetition Frequency (PRF) The frequency of pulse transmission affects the maximum range that can be displayed. Recall that the synchronizer resets the timing clock as each new pulse is transmitted.

Returns from distant targets that do no reach the receiver until after the next pulse has been sent will not be displayed correctly. Since the timing clock has been reset, they will be displayed as if the range where less than actual. If this were possible, then the range information would be considered ambiguous.

1. Figure 8
2. The maximum actual range that can be detected and displayed without ambiguity, or the maximum unambiguous range, is just the range corresponding to a time interval equal to the pulse repetition time, PRT. Therefore, the maximum unambiguous range,
3. R UNAMB = c PRT/2 = c/(2PRF)
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When a radar is scanning, it is necessary to control the scan rate so that a sufficient number of pulses will be transmitted in any particular direction in order to guarantee reliable detection. If too few pulses are used, then it will more difficult to distinguish false targets from actual ones.

False targets may be present in one or two pulses but certainly not in ten or twenty in a row. Therefore to maintain a low false detection rate, the number of pulses transmitted in each direction should be kept high, usually above ten. For systems with high pulse repetition rates (frequencies), the radar beam can be repositioned more rapidly and therefore scan more quickly.

Conversely, if the PRF is lowered the scan rate needs to be reduced. For simple scans it is easy to quantify the number of pulses that will be returned from any particular target. Let

• t represent the dwell time, which is the duration that the target remains in the
• N = t PRF
• We may rearrange this equation to make a requirement on the dwell time for a particular scan
• t min = N min /PRF

radar’s beam during each scan. The number of pulses, N, that the target will be exposed to during the dwell time is: So it is easy to see that high pulse repetition rates require smaller dwell times. For a continuous circular scan, for example, the dwell time is related to the rotation rate and the beam-width.

1. t = q/W
2. where q = beam-width W = rotation rate which will give the dwell time in seconds. These relationships can be combined, giving the following equation from which the maximum scan rate may be determined for a minimum number of pulses per scan:
3. W MAX = q PRF / N

Finally, the frequency of the radio carrier wave will also have some affect on how the radar beam propagates. At the low frequency extremes, radar beams will refract in the atmosphere and can be caught in “ducts” which result in long ranges. At the high extreme, the radar beam will behave much like visible light and travel in very straight lines.

• Very high frequency radar beams will suffer high losses and are not suitable for long range systems.
• The frequency will also affect the beam-width.
• For the same antenna size, a low frequency radar will have a larger beam-width than a high frequency one.
• In order to keep the beam-width constant, a low frequency radar will need a large antenna.

Theoretical Maximum Range Equation A radar receiver can detect a target if the return is of sufficient strength. Let us designate the minimum return signal that can be detected as S min, which should have units of Watts, W. The size and ability of a target to reflect radar energy can be summarized into a single term, s, known as the radar cross-section, which has units of m 2,

• If absolutely all of the incident radar energy on the target were reflected equally in all directions, then the radar cross section would be equal to the target’s cross-sectional area as seen by the transmitter.
• In practice, some energy is absorbed and the reflected energy is not distributed equally in all directions.

Therefore, the radar cross-section is quite difficult to estimate and is normally determined by measurement. Given these new quantities we can construct a simple model for the radar power that returns to the receiver:

• P r = P t G 1/4 p R 2 s 1/4 p R 2 A e
• The terms in this equation have been grouped to illustrate the sequence from transmission to collection. Here is the sequence in detail:
• G = r G dir

The transmitter puts out peak power P t into the antenna, which focuses it into a beam with gain G. The power gain is similar to the directional gain, G dir, except that it must also include losses from the transmitter to the antenna. These losses are summarized by the single term for efficiency, r,

Therefore The radar energy spreads out uniformly in all directions. The power per unit area must therefore decrease as the area increases. Since the energy is spread out over the surface of a sphere the factor of 1/4 p R 2 accounts for the reduction. The radar energy is collected by the surface of the target and reflected.

The radar cross section s accounts for both of these processes. The reflected energy spreads out just like the transmitted energy. The receiving antenna collects the energy proportional to its effective area, known as the antenna’s aperture, A e, This also includes losses in the reception process until the signal reaches the receiver.

1. P r = S min, If you solve for the range, you get an equation for the maximum theoretical

radar range: Perhaps the most important feature of this equation is the fourth-root dependence. The practical implication of this is that one must greatly increase the output power to get a modest increase in performance. For example, in order to double the range, the transmitted power would have to be increased 16-fold.

You should also note that the minimum power level for detection, S min, depends on the noise level. In practice, this quantity constantly be varied in order to achieve the perfect balance between high sensitivity which is susceptible to noise and low sensitivity which may limit the radar’s ability to detect targets.

Example: Find the maximum range of the AN/SPS-49 radar, given the following data Antenna Size = 7.3 m wide by 4.3 m tall Efficiency = 80 % Peak power = 360 kW Cross section = 1 m 2 S min = 1 10 -12 W We know from the previous example, that the directional antenna gain, G dir = 4 p / qf = 4 p /(.05 x,07) = 3430 The power gain, G = r G dir G = 2744.

### What frequency is pulse?

Listen to Pulse 1 on FM

FM 97.5 MHz
FM 102.5 MHz

#### How do you calculate pulse width modulation?

Explanation – Pulse-width modulation ( PWM ) allows the BASIC Stamp (a purely digital device) to generate an analog voltage. The basic idea is this: If you make a pin output high, the voltage at that pin will be close to 5 V. Output low is close to 0 V.

What if you switched the pin rapidly between high and low so that it was high half the time and low half the time? The average voltage over time would be halfway between 0 and 5 V (2.5 V). PWM emits a burst of 1s and 0s whose ratio is proportional to the duty value you specify. The proportion of 1s to 0s in PWM is called the duty cycle.

The duty cycle controls the analog voltage in a very direct way; the higher the duty cycle the higher the voltage. In the case of the BASIC Stamp, the duty cycle can range from 0 to 255. Duty is literally the proportion of 1s to 0s output by the PWM command.

• To determine the proportional PWM output voltage, use this formula: ( Duty ÷ 256) x 5 V.
• For example, if Duty is 100, (100 ÷ 256) x 5 V = 1.953 V; PWM outputs a train of pulses whose average voltage is 1.953 V.
• In order to convert PWM into an analog voltage we have to filter out the pulses and store the average voltage.

The resistor/capacitor combination shown below will do the job. The capacitor will hold the voltage set by PWM even after the instruction has finished. How long it will hold the voltage depends on how much current is drawn from it by external circuitry, and the internal leakage of the capacitor.

In order to hold the voltage relatively steady, a program must periodically repeat the PWM instruction to give the capacitor a fresh charge. Just as it takes time to discharge a capacitor, it also takes time to charge it in the first place. The PWM command lets you specify the charging time in terms of PWM duration.

The timing for the units in Duration is shown in in the table above. So, on the BS2, to charge a capacitor for five milliseconds, you would specify five units in Duration, How do you determine how long to charge a capacitor? Use this rule-of-thumb formula: Charge time = 5 x R x C. Since, on the BS2, each unit in Duration is approximately a millisecond, it would take at least 50 units to charge the capacitor. Assuming the circuit is connected to P0, here’s the complete PWM instruction: PWM 0, 100, 50 ‘ Put a 1.96V charge on cap (BS2, BS2e) To charge the same circuit to the same level using a BS2sx, the Duration would require adjustment as follows: PWM 0, 100, 125 ‘ Put a 1.96V charge on cap (BS2sx) After outputting the PWM pulses, the BASIC Stamp leaves the pin in input mode (0 in the corresponding bit of DIRS). In input mode, the pin’s output driver is effectively disconnected. If it were not, the steady output state of the pin would change the voltage on the capacitor and undo the voltage setting established by PWM,

• Eep in mind that leakage currents of up to 1 µA can flow into or out of this “disconnected” pin.
• Over time, these small currents will cause the voltage on the capacitor to drift.
• The same applies for leakage current from an op-amp’s input, as well as the capacitor’s own internal leakage.
• Executing PWM occasionally will reset the capacitor voltage to the intended value.

PWM charges the capacitor; the load presented by your circuit discharges it. How long the charge lasts (and therefore how often your program should repeat the PWM command to refresh the charge) depends on how much current the circuit draws, and how stable the voltage must be.

• You may need to buffer PWM output with a simple op-amp follower if your load or stability requirements are more than the passive circuit can handle.
• The term “PWM” applies only loosely to the action of the BASIC Stamp’s PWM command.
• Most systems that output PWM do so by splitting a fixed period of time into an on time (1) and an off time (0).

Suppose the interval is 1 ms and the duty cycle is 100 ÷ 256. Conventional PWM would turn the output on for 0.39 ms and off for 0.61 ms, repeating this process each millisecond. The main advantage of this kind of PWM is its predictability; you know the exact frequency of the pulses (in this case, 1 kHz), and their widths are controlled by the duty cycle.

1. BASIC Stamp’s PWM does not work this way,
2. It outputs a rapid sequence of on/off pulses, as short as 1.6 µs in duration, whose overall proportion over the course of a full PWM cycle of approximately a millisecond is equal to the duty cycle.
3. This has the advantage of very quickly zeroing in on the desired output voltage, but it does not produce the neat, orderly pulses that you might expect.

All BS2 modules also uses this high-speed PWM technique to generate pseudo-sine wave tones with the DTMFOUT and FREQOUT instructions.

### How is PWM generated?

Hint: Pulse width modulation signal is used to control analog devices with a digital output. In other words, we can drive an analog device by using an output of a modulating signal from a digital device such as a microcontroller. Pulse Width Modulation or PWM is a fancy word that can describe a type of digital signal.

In much simpler words PWM generates analog signals by using a digital source. Complete answer: PWM signal can be generated using a comparator. One input of the comparator is connected to a modulating signal and the other input is fed with a non-sinusoidal wave or saw-tooth wave. The comparator compares the two input signals and generates a PWM signal.

It depends on the value of the saw-tooth signal, if the saw-tooth signal is more than the modulation signal then the output of the PWM wave will be high and if the value of the saw-tooth signal is less than the modulation signal then the output of the PWM signal will be low. Additional Information: The advantages of the PWM signal include preventing overheating of LEDs while maintaining their brightness. PWM signal provides quick response time. Its initial cost is low. Its disadvantages include high switching losses because PWM frequency is quite high.

Also, it induces radio frequency interference. Note: There are numerous applications for PWM technology. The applications of PWM signals include controlling the speed of the motor, driving a buzzer with different loudness, controlling the direction of the servo, providing an analog output, and generating an audio signal.

The behavior of the PWM signal is determined by the duty cycle and frequency of a PWM signal.

## Does PWM frequency Matter?

What effect does the frequency have in PWM? To a first approximation, frequency does not matter at all, provided it’s fast enough to avoid the appearance of blinking. All that’s important is the average power of the LED, which depends only on the duty cycle.

In practice, there are switching losses that increase with frequency. Each transition from high to low requires some amount of energy, such as to charge or discharge the gate capacitance of a MOSFET. There is also a transitional period where the transistor being switched is neither fully off nor fully on, and thus dissipates more power than it would at either extreme.

With increased frequency, this energy cost is paid more times per second, thus losses go up. Furthermore, parasitic components (the inductance of traces and leads, and their capacitance, among other things) becomes increasingly significant with increasing frequency, making design more challenging.

#### What does 100% PWM mean?

PWM 100% Duty Cycle –

Post Essentials Only Full Version
vignesh.J New Member

Total Posts : 9 Reward points : 0 Status: offline

2014/08/12 02:48:50 (permalink) 0 Hi, I’m using dspic30F3011 in the application of 3 Phase AC Drive. Using the MCPWM module in Complementary Mode. I Faced an Issue that, during PWM Output at 100% Duty Cycle, PWM1H is in always high state(Logic 1) and PWM1L is in always low state(Logic 0). The Problem is, IGBT pair (Transistor Pair T1&T2) is always in ON State. This Cause Damage to the Power Transistor, An Interval Time(Dead Time) must be inserted between the both Transistor Switching, Because the power output devices cannot switch instantaneously, some amount of time must be provided between the turn-off event of one PWM output in a complementary pair and the turn-on event of the other transistor In my case if we are using 100% duty cycle,there is no Dead Time occurs in Output Waveform. I checked Dead Time Insertion up to 99% Duty cycle, it works normally and no problem, at 100% duty cycle there is no dead time insertion. How Can i achieve 100% duty cycle? I hope that anyone will Reply & guide me Thanks & Regards #1

Aussie Susan Super Member

Total Posts : 4201 Reward points : 0 Joined: 2008/08/18 22:20:40 Location: Melbourne, Australia Status: offline

Re: PWM 100% Duty Cycle 2014/08/12 19:17:05 (permalink) +2 (2) I must admit that I don’t follow how your argument that 100% duty cycle will cause damage leads to you asking how to achieve 100% duty cycle. A 100% duty cycle means that the output is always high (or the complimentary output always low) so the glib answer is to forget about the PWM and just set the output high! If you want dead time then you cannot achieve 100% duty cycle by definition. Susan #2

ric Super Member

Total Posts : 35416 Reward points : 0 Joined: 2003/11/07 12:41:26 Location: Australia, Melbourne Status: online

Re: PWM 100% Duty Cycle 2014/08/12 19:43:20 (permalink) +2 (2) vigneshwaran.j87[email protected], I Faced an Issue that, during PWM Output at 100% Duty Cycle, PWM1H is in always high state(Logic 1) and PWM1L is in always low state(Logic 0). As Susan just stated, that is the definition of 100% duty cycle! If you don’t want it to stay high, don’t set it to 100%. I also post at: PicForum To get a useful answer, always state which PIC you are using! #3

NorthGuy Super Member

Total Posts : 7385 Reward points : 0 Joined: 2014/02/23 14:23:23 Location: Northern Canada Status: offline

Re: PWM 100% Duty Cycle 2014/08/12 20:19:55 (permalink) +1 (1) [email protected] some amount of time must be provided between the turn-off event of one PWM output in a complementary pair and the turn-on event of the other transistor You need to explain this in some more details. How can a dead time be provided between turn-on and turn-off events when these events do not exist at 100% duty cycle? #4

vignesh.J New Member

Total Posts : 9 Reward points : 0 Status: offline

Re: PWM 100% Duty Cycle 2014/08/13 03:53:17 (permalink) 0 dear susan, Thanks for your reply. In my previous post i tried to state that, In 100% duty cycle there is no dead time in my output waveform. but in case of Power Transistor Switching(Hardware side),we must insert a small amount of dead time b/w two power transistors,else the transistor becomes damage. From your reply, I understood that we cant able to achieve 100% duty cycle with dead time. And if we want to insert dead time, we must stop PWM output up to 99% duty cycle. i’m i correct? Thanks & Regards, vignesh #5

vignesh.J New Member

Total Posts : 9 Reward points : 0 Status: offline

Re: PWM 100% Duty Cycle 2014/08/13 03:56:41 (permalink) 0 hi ric, thank you for reply thanks & regards, vignesh #6

vignesh.J New Member

Total Posts : 9 Reward points : 0 Status: offline

Re: PWM 100% Duty Cycle 2014/08/13 04:08:06 (permalink) 0 hi, In A.C Drive application, a dead time is nothing but a delay time that must be inserted in between two power transistor switching (IGBT pair),else if there is no dead time, transistor leads to damage. the following is my code for PWM initialization in dspic30f3011 void Init_PWM() Thanks & Regards, Vignesh #7

Ian.M Super Member

Total Posts : 13274 Reward points : 0 Joined: 2009/07/23 07:02:40 Location: UK Status: offline

Re: PWM 100% Duty Cycle 2014/08/13 06:45:59 (permalink) +2 (2) At 100% duty cycle the top transistor is on all the time and the bottom one off. Draw us a timing diagram showing where you expect the deadtime, because the consensus here is: Its impossible! – NEW USERS: Posting images, links and code – workaround for restrictions. I also support http://picforum.ric323.com because this forum is sometimes too broken to use! #8

vignesh.J New Member

Total Posts : 9 Reward points : 0 Status: offline

Re: PWM 100% Duty Cycle 2014/08/14 00:14:42 (permalink) 0 I understood that we can’t able to insert dead time between high and low output at 100% duty cycle. Thanks for your valuable comment. Regards, vignesh #9

2022 APG vNext Commercial Version 4.5

### Why is PWM 255?

10-min tutorial: What happens when PWM registers overflow? As you may know, in the Atmega328 that powers the Arduino Uno, several of the pins are capable of Pulse Width Modulation (PWM). These are pins 3, 5, 6, 9, 10, and 11. With PWM, we can approximate analog output programmatically and do things like fade an LED on and off or control the speed of a motor. In the Atmega328, the register that is used by the PWM function has a resolution of 8 bits. This gives us a total of 255 possible “analog” output levels. If we attach an LED to a PWM-capable pin, we can drive it to 255 different brightness levels. And if we attach a motor, we can drive it to 255 different speed levels.

• We can set a PWM value by using the analogWrite(pin, value) instruction.
• So, analogWrite(3, 125) would set pin 3 to value 125.
• The maximum value is 255 since the PWM register is 8 bits wide (2 in the power of 8 is 255).
• What happens if we set analogWrite to a value bigger than 255? Say, 256? Let’s think about this for a minute.

If the PWM value is 255, the binary version is 11111111 is stored in the PWM register. A connected LED would light up in maximum brightness. Let’s add 1 to the register, and make the PWM value 256. The binary version of 256 is 00000001 00000000 since now we need two bytes to represent this value.

But, the Arduino (Atmega) can only fit the first byte in its PWM register, the one in bold. The second byte will overflow and “disappear”. So, what you have stored in the PWM register is 00000000, This is decimal “0”, which means that your LED is practically turned off. In other words, analogWrite(3, 0) and analogWrite(3, 256) would have the exact same effect on an LED or a motor.

Add another “1” to the register, and the PWM value now is 257. The binary version of 257 is 00000001 00000001. The byte in bold is stored in the PWM register, and the rest disappears. In the register now the decimal value “1” is stored. The lesson to take home is that although you can set the PWM value in analogWrite to any decimal you like, only the first byte of this number will fit in the PWM register.

#### What is a good PWM frequency?

Increase the PWM frequency – A higher frequency will cause a shorter cycle time of the PWM; hence the current will have less time to rise. PWM frequencies not less than 50 kHz for brushless dc motors are recommended. PWM frequencies of 80 kHz or more would be even more appropriate for motors with a very small electrical time constant.

## What is bandwidth formula?

Bandwidth in terms of Q and resonant frequency: BW = f c /Q Where f c = resonant frequency Q = quality factor. A high Q resonant circuit has a narrow bandwidth as compared to a low Q. Bandwidth is measured between the 0.707 current amplitude points.

#### How do you calculate pulse width modulation?

Explanation – Pulse-width modulation ( PWM ) allows the BASIC Stamp (a purely digital device) to generate an analog voltage. The basic idea is this: If you make a pin output high, the voltage at that pin will be close to 5 V. Output low is close to 0 V.

What if you switched the pin rapidly between high and low so that it was high half the time and low half the time? The average voltage over time would be halfway between 0 and 5 V (2.5 V). PWM emits a burst of 1s and 0s whose ratio is proportional to the duty value you specify. The proportion of 1s to 0s in PWM is called the duty cycle.

The duty cycle controls the analog voltage in a very direct way; the higher the duty cycle the higher the voltage. In the case of the BASIC Stamp, the duty cycle can range from 0 to 255. Duty is literally the proportion of 1s to 0s output by the PWM command.

To determine the proportional PWM output voltage, use this formula: ( Duty ÷ 256) x 5 V. For example, if Duty is 100, (100 ÷ 256) x 5 V = 1.953 V; PWM outputs a train of pulses whose average voltage is 1.953 V. In order to convert PWM into an analog voltage we have to filter out the pulses and store the average voltage.

The resistor/capacitor combination shown below will do the job. The capacitor will hold the voltage set by PWM even after the instruction has finished. How long it will hold the voltage depends on how much current is drawn from it by external circuitry, and the internal leakage of the capacitor.

In order to hold the voltage relatively steady, a program must periodically repeat the PWM instruction to give the capacitor a fresh charge. Just as it takes time to discharge a capacitor, it also takes time to charge it in the first place. The PWM command lets you specify the charging time in terms of PWM duration.

The timing for the units in Duration is shown in in the table above. So, on the BS2, to charge a capacitor for five milliseconds, you would specify five units in Duration, How do you determine how long to charge a capacitor? Use this rule-of-thumb formula: Charge time = 5 x R x C. Since, on the BS2, each unit in Duration is approximately a millisecond, it would take at least 50 units to charge the capacitor. Assuming the circuit is connected to P0, here’s the complete PWM instruction: PWM 0, 100, 50 ‘ Put a 1.96V charge on cap (BS2, BS2e) To charge the same circuit to the same level using a BS2sx, the Duration would require adjustment as follows: PWM 0, 100, 125 ‘ Put a 1.96V charge on cap (BS2sx) After outputting the PWM pulses, the BASIC Stamp leaves the pin in input mode (0 in the corresponding bit of DIRS). In input mode, the pin’s output driver is effectively disconnected. If it were not, the steady output state of the pin would change the voltage on the capacitor and undo the voltage setting established by PWM,

1. Eep in mind that leakage currents of up to 1 µA can flow into or out of this “disconnected” pin.
2. Over time, these small currents will cause the voltage on the capacitor to drift.
3. The same applies for leakage current from an op-amp’s input, as well as the capacitor’s own internal leakage.
4. Executing PWM occasionally will reset the capacitor voltage to the intended value.

PWM charges the capacitor; the load presented by your circuit discharges it. How long the charge lasts (and therefore how often your program should repeat the PWM command to refresh the charge) depends on how much current the circuit draws, and how stable the voltage must be.

You may need to buffer PWM output with a simple op-amp follower if your load or stability requirements are more than the passive circuit can handle. The term “PWM” applies only loosely to the action of the BASIC Stamp’s PWM command. Most systems that output PWM do so by splitting a fixed period of time into an on time (1) and an off time (0).

Suppose the interval is 1 ms and the duty cycle is 100 ÷ 256. Conventional PWM would turn the output on for 0.39 ms and off for 0.61 ms, repeating this process each millisecond. The main advantage of this kind of PWM is its predictability; you know the exact frequency of the pulses (in this case, 1 kHz), and their widths are controlled by the duty cycle.

1. BASIC Stamp’s PWM does not work this way,
2. It outputs a rapid sequence of on/off pulses, as short as 1.6 µs in duration, whose overall proportion over the course of a full PWM cycle of approximately a millisecond is equal to the duty cycle.
3. This has the advantage of very quickly zeroing in on the desired output voltage, but it does not produce the neat, orderly pulses that you might expect.

All BS2 modules also uses this high-speed PWM technique to generate pseudo-sine wave tones with the DTMFOUT and FREQOUT instructions.

### What is pulse width in flow cytometry?

Pulse area is the most commonly plotted value in FACS analyses and is the total amount of light (e.g., fluorescence) emitted by a particle. Pulse height is the maximum value of the pulse, whereas pulse width provides information about the size of the particle.

### How is pulse width modulation measured?

Calculating RMS voltage of a PWM – Thus for calculating RMS of a varying PWM voltage in response to a sine wave may be acquired by multiplying 0.7 (constant) with the peak voltage. So for a 9V peak we get 9 x 0.7 = 6.3V, that’s the RMS voltage or the average value of a 9V peak to peak PWM simulating a sine wave.