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I am wondering if there a way to convert a pulse train with a 200 kHz rep rate, a duration of ~100 nanoseconds, and an amplitude of ~4 volts to a DC voltage in such a way that the area under the curves of the pulse train is equal to the area under dc voltage. In other words, if both graphs of voltage in time were integrated out to infinity, the integrals would be equal. These are from a pulsed laser that I am measuring with amplified photodiodes. I am hoping to measure a very small signal that is occurring at a subharmonic frequency using a lock-in amplifier and then compare it to the magnitude of the pulses. It would be really convenient if I could use the auxiliary channel on the back of the lock-in as a voltmeter for this latter part, but it can only measure a stable DC voltage well.

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    \$\begingroup\$ A low pass filter with dc gain of 1 will do what you are asking for. \$\endgroup\$ – The Photon Sep 29 '15 at 4:59
  • \$\begingroup\$ Can you draw these two graphs, because I don't understand what areyou talking about, also the integral with units. You are talking about current pulse, then it becomes a voltage pule, after that an integral voltage*time becomes voltage with two graphs?! \$\endgroup\$ – Marko Buršič Sep 29 '15 at 7:12
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What the photon said; If the modulation for the lockin (LI) is very slow compared to the 200kHz, (say factors of ten or more... 1kHz to 10 kHz.) Then you may do better by low pass filtering the 200 kHz pulses first... with a corner set above the LI modulation frequency, and sending that into the lockin... the LP filter will cause some phase shift.. but should be a real help in terms of dynamic range... (not over loading the LI front end.)

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