In my project i am using an ultrasonic transducer. The transducer has a high Q-factor so that it's zero input transient responses decay rate is quite low and i am suffering from that because the transient is interfering with the reflected signal.

enter image description here Here is the circuit that i am using, C0 L1 R1 and C1 accounts for the transducer

enter image description here

It has the response enter image description here

where green is Vin and red is Vout

In a linear time invariant circuit,

\$ \frac{d}{dx} \overline{x}(t) =\overline{\overline{A}} \overline{x}(t) + \overline{b} v_{s}(t) \$ where \$ v_{s} \$ is the source and \$ \overline{x} \$ is the state vector then we have,

\$ \overline{x}(t) = e^{ \overline{\overline{A}}t} \overline{x}(0) + \int_0^te^{ \overline{\overline{A}}(t - \tau) } \overline{b} v_{s}( \tau )d \tau \$

, i wonder if for any input \$ v_{s1}(t), v_{s1}(t)= 0~~ for~~t > t_{1} \$ there exists a function \$ v_{s2}(t), ~~v_{s2}(t) \neq 0~~ for ~~t_{1}\leq t\leq t_{2} \$ such that \$ \left\{ \begin{array}{ll} v_{s}(t) = v_{s1}(t) & t\leq t_{1} \\ v_{s}(t) = v_{s2}(t) & t_{1}\leq t\leq t_{2} \\ 0 & o.w. \\ \end{array} \right. \$ \$ ,\overline{x}(t_{2}) = 0 \$

That is to say, i want to apply a signal right after the original signal so that the latter will suppress the transient created by the first. I have found analytical solution for \$ v_{s2}(t) \$ if \$ \overline{b} \$ is an eigenvector of \$ \overline{\overline{A}} \$ but i am seeking a general solution. Here is an example i found with trial and error:

enter image description here

Where the green signal is the input and the red signal is the response of a high Q bandpass filter the signal is a 250Hz pulse with 5 cycles. if we apply a 1 kHz for 2 cycles at 22ms we have,

enter image description here

Observe that the transient response is immediatelly suppressed

I would be glad if you propose a method to find a \$ v_{s2}(t) \$ preferably a square one.

  • 2
    \$\begingroup\$ It looks like you are applying a "counter voltage" that allows the energy in the reactive components to be immediately turned to heat in the lossy part of the device. This is not a new thing but the problem with your question is that it is dealing a lot with formulas and lacks EE content like the equivalent circuit of your device and your method of driving the device. In other words I have no idea what you have done electrically and therefore no way to suggest a better or more proven alternative method. \$\endgroup\$
    – Andy aka
    Jul 23, 2018 at 13:48
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    \$\begingroup\$ Yeah, what Andy said. I have no interest in wading thru the equations to see what they are trying to say. After skipping over them, I'm not sure what you are asking. Ask what you want in words. Use equations to back the words to give details if really needed. However, the concepts shouldn't need that level of detail. It's also hard to see what the graphs are, with them aliased to oblivion like that. Again, there are too many details here obscuring (if you ever said it at all) what you are actually trying to accomplish and what your question is. \$\endgroup\$ Jul 23, 2018 at 14:01
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    \$\begingroup\$ I haven't waded through the equations either, but as @Andyaka mentioned the general problem has been solved by active damping. If you Google "active damping for ultrasonic transducers" you should be able to find some references. \$\endgroup\$
    – John D
    Jul 23, 2018 at 14:34

1 Answer 1


Ok, this is not an issue, though I would like to see user’s response to determine if it is useful or not. Let us solve it in another way. For author’s information:the reason that your transducer starts to ring is because you stop your drive too hard. The switch-off causes a step stimuli and it rings usually longer than you drive it inversely due to the high quality factor. Some of the signal waveform demonstrate the transducer is not driven at its central frequency, so you can see it not following the tuning. To solve your issue, all you need is a soft switch off at the end of the pulse train.

In case you don’t follow my suggestion, you may realise the drive through an analogue switching mechanism where at the end of pulse train, you switch to a capacitive drive branch to produce an damped waveform, this will make your transducer a lot quieter. I understand your work may have already finished, but you don’t need to struggle with the math, only an analogue implementation and the parametric choice of the capacitive drive.

In fact, a lot more questions should be asked first before providing a solution, e.g. your coupling mechanism with the propagation medium which may be either solid or liquid, though that ringing is much different in either case. Looking at the central frequency of the transducer, this analogue switcher is quickest solve because it is too high for digital controlled active damper to sample and process. I believe some commercially available medical or Nondestructive test equipment have similar transient compensation if the ringing is a matter to the reflected wave, or if they want to switch the phase of the drive signal much quicker.


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