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I've been asked to estimate the half-power cut off frequency of the filter and sketch the frequency and phase response.

I think the 63% mark is around 1.9 usecs and I need to use the equation f=1/2pit

Is this correct? Any help or workings are much appreciated.

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  • \$\begingroup\$ Im not sure what the "half power" cut-off frequency is(-3db frequency?), but the time to frequency(in Hz) conversion is calculated with f=1/T, so in radians that is 2Pi/T. So converting from time to frequency at the 63% mark sounds right, if this is what you are asking for. \$\endgroup\$
    – Linkyyy
    Sep 17, 2018 at 16:10
  • \$\begingroup\$ yes 63% I'm using the 1.2xPix0.9us so i get 176.839kHz, but i'm not sure if this is the completely wrong way of going about it thanks for the quick answer \$\endgroup\$
    – Ravrag
    Sep 17, 2018 at 16:25
  • \$\begingroup\$ As a side-note: the volts vs. microseconds diagram of FIG.2 is misleading. A first-order filter does not start with an infinite V/us slope. It starts with a slope of \$ \unicode{932} \$ \$\endgroup\$
    – glen_geek
    Sep 18, 2018 at 0:28
  • \$\begingroup\$ hey Glen, whats the relevance of this, is it just to throw the person answering a little bit? \$\endgroup\$
    – Ravrag
    Sep 18, 2018 at 5:50
  • \$\begingroup\$ @Ravrag There's another way to get time-constant besides finding the time-to-0.63V point....that is to extrapolate the initial slope in a straight line (from \$ t_0 \$ up to V). This straight line will cross V after one time-constant. With that graph, its impossible to use this method. \$\endgroup\$
    – glen_geek
    Sep 18, 2018 at 12:23

1 Answer 1

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Equivalence from time to freq. domain for 1st order filter.

Half Power (f) or 50% of V^2/R occurs thus for V,
Vout(f)=0.707 of Vin(f) , where 0.707=1/root(2)

for the exponential time decay towards target step (1-1/e)=~63% input

\$ (1-1/e) * V_{step}=0.63 * V_{step} \$

This is equivalent to exponential V(f) at \$ \omega=\dfrac{1}{T}~where ~t=T= RC , \omega=2\pi f\$

So you are correct. OA's must be RRO types to go to 0V out or bipolar supply and have GBW much greater than f-3dB. (lookup if the abbreviation is foreign)

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  • \$\begingroup\$ if I edit the post adding in the part B table would you be able to help me with that please? I'm a little lost with that \$\endgroup\$
    – Ravrag
    Sep 17, 2018 at 16:35
  • \$\begingroup\$ The full swing pulse BW or FPBW only exceeds your requirement on 2 of the Op Amps. THis is due to current limiting. \$\endgroup\$
    – Tony Stewart EE75
    Sep 17, 2018 at 17:52
  • \$\begingroup\$ thank you very much, is this the only thing that makes op amp 1 & 4 the only suitable ones? \$\endgroup\$
    – Ravrag
    Sep 17, 2018 at 18:42
  • \$\begingroup\$ yes for full step swing you need the FPBW to not affect the rise time by usually a significant amount. such as 2 octaves or 3x f depending on error tolerance. It is like cascading higher LPF effect , a 2nd order effect. \$\endgroup\$
    – Tony Stewart EE75
    Sep 17, 2018 at 19:19

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