# [SI-LIST] Re: Impulse vs. Step Response?

• To: "si-list@xxxxxxxxxxxxx" <si-list@xxxxxxxxxxxxx>
• Date: Fri, 2 Sep 2011 21:32:23 +0000
```David,

Let's neglect with second (large) resistance, so that R=1K and C=1pF

The transfer function of your circuit and also the spectrum of the impulse
response is:

H(p) = 1/(pRC+1), where p = 2*Pi*j*freq

The spectrum of the step response is:

S(p) = 1/[p*(pRC+1)]

You can represent S(p) as a sum:

S(p) = 1/p - (RC)/(pRC+1)

The second summand is similar to H(p) but has additional scaling (-RC) that is
about 1e-9. You should be able to see large difference in magnitude compared to
the AC response. If all computations are correct, then you are right, we should
see the difference between the AC response and FFT from step response.

However, there could be a possibility that FFT method used eliminates the
constant from the step response, if it requires zero at infinity:

S(t) = 1 - exp(-t/RC)

Thus taking the exponential term only, that differs from impulse response by
sign and magnitude:

H(t) = (1/RC)*exp(-t/RC)

Plus, if we compare the AC response to FFT from step response, we may not see
the small magnitude in S(p) because FFT did not multiply the result on the time
step (should be done because of integrating in time domain). Typically, such
normalization should be done outside FFT, but I'm not sure what FFT method FFT
was used.

-----Original Message-----
From: David Banas <DBanas@xxxxxxxxxx>
Date: Fri, 2 Sep 2011 10:20:00 -0700
Subject: [SI-LIST] Re: Impulse vs. Step Response?

Okay, so if we're agreed that, for all intents and purposes, I am driving this
circuit with a unit step, as opposed to a unit impulse, why does the spectral
content of my transient simulation output exactly match the output of the ac
sweep, rather than having an additional 6dB/octave roll-off, as one would
expect?

Thanks,
-db

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