How to know that the set excitation frequency has reached resonance

The first choice is to correct the netizen's "The impact of the moment the circuit is powered on can vibrate the steel string by the excitation coil." A large number of experiments have shown that only a very small number of sensors may vibrate and output signals under the condition of low voltage single pulse, and even then, this vibration is particularly unsatisfactory, resulting in the amplitude of the output signal of the sensor being too low and affecting the signal acquisition.

For most single-chord vibrating wire sensors, high-voltage single-pulse excitation is possible, and the general range of this high voltage is between 80 and 200V. For the other part of the sensor, the high-voltage pulse excitation method is not as good as the low-voltage multi-cycle sweep excitation activation effect, and there are a very few sensors that are not sensitive to the high-voltage single pulse, and the low-voltage sweep method must be used to excite.

How do you know if a steel string has produced vibrations?

This should be done from the signal acquisition end, no matter what method is used to excite, the part of the collected signal has a judgment on the output signal of the sensor, if the signal output by the sensor is regular, to meet the expected requirements, it is considered that the steel string has been vibrate, and the direct acquisition or re-excitation acquisition can be. This process is one or more interactions between the collector and the sensor.

You can use the mature vibrating wire sensor measurement module or reading module: VM301, VM401, VM501, the sensor excitation, signal judgment, filtering, amplification, calculation of these work is automatically completed by the module, and the host computer can only communicate with the digital interface of the module as the sensor frequency. This kind of module saves the trouble of developing your own collector, and you can focus on the product and user interface, and you may not be able to achieve good compatibility and data acquisition performance by doing it yourself.

Hello! Why sweep frequency? The impact at the moment of power on the circuit can vibrate the steel string by excitation coil, and then the resonance frequency signal obtained by the pickup coil is amplified, and a part of it is fed back to the excitation coil to maintain vibration, which we have used a lot.

Today's vibrating wire technology is a bit misguided, complicating a simple problem.

Typing isn't easy, oh!

If the vibrometer outputs a displacement signal from the vibrating body, the displacement signal is fed into the oscilloscope.

The y-axis fails (2 channels), the excitation signal is input to the oscilloscope x-axis (1 channel), and the frequency when the image changes from an oblique ellipse to a regular ellipse is the natural frequency of the vibrating body.

If the vibrometer outputs the velocity signal of the vibrating body, the velocity signal is input into the y-axis of the oscilloscope, and the excitation signal of the front bay is input into the x-axis of the oscilloscope, and the frequency when the image changes from an oblique ellipse to an oblique straight segment is the natural frequency of the vibrating body.

If the vibrometer outputs the acceleration of the vibrating body.

signal, the acceleration signal is input into the y-axis of the oscilloscope, and the excitation signal is input into the x-axis of the oscilloscope, and the traveling frequency when the image changes from an oblique ellipse to a regular ellipse is the natural frequency of the vibrating body.

Because a standing wave can be formed at the resonant frequency, the wavelength of the sound wave can be measured according to the situation of the standing wave, and then the wavelength can be multiplied by the resonant frequency to obtain the magnitude of the sound velocity.

To adjust the resonance, hold the sensor close to the transmitter without touching it, and then adjust the frequency of the drive signal until the maximum waveform received by the sensor.

In an AC circuit with a resistor r, inductance L, and capacitance C elements, the voltages across the circuit are generally different from the current phases in which they are. If we adjust the parameters or supply frequencies of the circuit elements (L or C), we can make them have the same phase and the entire circuit appears purely resistive. The circuit reaching this state is called resonance.

In the case of resonance, the measurement is accurate, and the frequency of the measurement system is gradually adjusted, and when in the resonant state, the signal amplitude increases greatly or multiple resonant spikes can be seen.

Summary. Dear, thank you for your patience According to the teacher's understanding<> the output frequency of the signal generator needs to be the same as the resonant frequency of the ultrasonic probe, so that the probe can produce a strong resonance phenomenon during the working process. Ultrasonic probes are designed according to the principle of resonance, when the external vibration frequency is the same as the resonant frequency of the probe, it can resonate and form a strong harmonic vibration force, which is used to detect defects and anomalies inside the material.

If the output frequency of the signal generator is different from the resonant frequency of the ultrasonic probe, the desired effect will not be achieved, and the probe will not be able to produce strong harmonic vibrations. Therefore, to ensure the best performance of the probe, the output frequency of the signal generator should be the same as the resonant frequency of the ultrasonic probe. <>

The output frequency of the signal generator needs to be the same as the resonant frequency of the ultrasonic probe in order to produce strong resonance phenomena during the operation of the probe. The ultrasonic probe is designed according to the principle of resonance, when the external vibration frequency is the same as the harmonic void closing frequency of the probe, it can resonate and form a strong harmonic vibration force, which is used to detect defects and abnormalities inside the material. If the output frequency of the signal generator is different from the resonant frequency of the ultrasonic probe, the desired effect will not be achieved, and the probe will not be able to produce strong harmonic vibrations.

Therefore, to ensure optimal performance of the probe, the output frequency of the signal generator should be the same as the resonant frequency of the ultrasonic crack probe. <>

Are the wave velocities of sound, infrasound and ultrasound the same under the same conditions? Why.

Different types of waves have different wave velocities under the same conditions. Sound waves, infrasonic waves and ultrasonic waves are three different frequencies of mechanical waves, and their wave speeds are different:- Sound waves are low-frequency mechanical waves in Qin Hall, and the wave speed is generally 343m s (at a temperature of 20, the speed of sound waves in the air).

Infrasound is an ultra-low frequency sound wave that is lower than what is audible to the human ear and has a wave velocity slightly lower than that of sound waves. - Ultrasonic waves are high-frequency mechanical waves, usually referring to waves with a frequency greater than 20kHz, and the wave velocity is much higher than that of sound waves, usually hundreds to thousands of meters. The difference in wave velocity of different types of waves is mainly due to their propagation mode in the medium and the different physical characteristics of the suspect and the hidden sound, such as the wavelength, frequency, vibration direction and intensity of sound waves, infrasound waves and para-leakage sound waves are different, so their wave velocities are also different.

Through your description, I feel that your basic concept of physics is vague. To tell you some common sense1, the unit of current is a, not v. You're confusing these two basic concepts in your experiment.

2. In the vibrating wire sensor technology, what we need is to electronically measure the natural oscillation frequency of the vibrating wire. Excitation of a vibrating string with a frequency that has nothing to do with the natural frequency of the vibrating string makes no sense in measuring the natural frequency of a steel string. The natural frequency of the vibrating string is usually about 1k 3kHz, and you use a 100Hz frequency signal to excite the steel string, I don't know what the point is?

3. You use a 100Hz (or 101Hz) sinusoidal signal to continuously excite the steel string, at this time the steel string will of course vibrate according to 100Hz (note that the steel string is forced to vibrate at this time, not resonance). How do you think the resonant frequency of a steel string has changed?

Excitation: also known as "excitation", is the necessary process of frequency data acquisition of vibrating wire sensor, only when sensor receives suitable excitation signal can produce natural vibration, and only when vibrating wire sensor produces natural vibration can output frequency signal, further, reading circuit can detect and read the natural vibration signal of vibrating wire sensor, can obtain vibration frequency value by calculation. The excitation signal of the vibrating wire sensor (the external signal that can make the sensor produce natural vibration) is generally divided into two categories, one is the high-voltage short pulse, and the other is the multi-group continuous low-voltage pulse signal of a specific frequency.

High-voltage pulse excitation: A process or method that uses a higher voltage (100 200V) to send a short pulse to the coil of the vibrating wire sensor to produce natural vibration of the vibrating wire sensor at any frequency.

Low-voltage swept excitation: The process or method of sending a continuous low-voltage (3 10V) pulse signal to the vibrating wire sensor at a frequency equivalent to (close) to the sensor's natural frequency, so that the sensor produces natural vibration.

In principle, this circuit is correct, but the program in the MCU must take into account the time precedent of reaching the measured frequency. Because it takes time to perform an accurate frequency test and the sweep, amplify, detect, (reamplify), sample, AD loop, and the program executed in the MCU that you design, the frequencies that your loops can really lock on may not be your target frequency.

The correct approach should be to use the algorithm program of infinite frequency approximation to output a set of "swept excitation" signals close to and close to the target frequency, as the excitation signal source of the sensor, and finally use the program to automatically reduce the frequency conversion range of the swept frequency signal to a negligible degree, and the frequency at this time may be regarded as the "natural frequency of the sensor".

This is more difficult to implement.