[SI-LIST] Re: Impulse vs. Step Response?
- From: Tracy Barclay <Tracy.Barclay@xxxxxxxxxxxx>
- To: "Beal, Weston" <Weston_Beal@xxxxxxxxxx>, David Banas <DBanas@xxxxxxxxxx>
- Date: Fri, 2 Sep 2011 09:53:50 -0700
Hi Weston,
Yes, this is a rather roundabout method to model a step function. My
preference is to avoid using the UIC function.
------------------------------------------------
Tracy Barclay
CAE
Synopsys, Inc.
-----Original Message-----
From: Beal, Weston [mailto:Weston_Beal@xxxxxxxxxx]
Sent: Friday, September 02, 2011 09:04
To: Tracy Barclay; David Banas
Cc: 'si-list@xxxxxxxxxxxxx'
Subject: RE: Impulse vs. Step Response?
The step function is created in a roundabout fashion using the initial
condition. The initial voltage on node -out- is set to 0V. Once the transient
simulation starts, the voltage is allowed to rise up to its defined level of
1V. This method is defined by the .IC statement together with the UIC parameter
on the .TRAN line.
Weston
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Tracy Barclay
Sent: Friday, September 02, 2011 8:00 AM
To: David Banas
Cc: 'si-list@xxxxxxxxxxxxx'
Subject: [SI-LIST] Re: Impulse vs. Step Response?
Hi David,
For your transient analysis, I do not see a transient waveform defined in your
netlist. The source Vin only defines a DC voltage and a AC voltage. I would
expect to see something like:
Vin in 0 DC 1 AC 1 PWL(0 1 0.05n 1 0.06n 0)
Where the PWL definition defines the desired unit step.
------------------------------------------------
Tracy Barclay
CAE
Synopsys, Inc.
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of David Banas
Sent: Thursday, September 01, 2011 17:40
To: 'si-list@xxxxxxxxxxxxx'
Subject: [SI-LIST] Impulse vs. Step Response?
Hi all,
I've got a puzzler. I'm using this HSpice deck:
.TITLE AC Sweep vs. Transient Correlation
.OPTION
+ POST
+ PROBE
+ RUNLVL = 5
Rin in out r=1K
Cout out 0 c=1pF
Rout out 0 10MEG
Vin in 0 DC 1 AC 1
.IC V(out) = 0
.AC DEC 10 1 1G
.TRAN 0.05n 1u 0 0.05n UIC
.PROBE AC V(in) V(out)
.PROBE TRAN V(in) V(out)
.END
to get both the transfer function, H(w), (.AC command) and the unit step
response, u(t), (.TRAN command) of a simple RC LPF.
Then, I take the FFT of the .TRAN output and compare it to the .AC output.
And they match!
I wasn't expecting a match, since all my textbooks say there's a `jw' factor
relating the impulse response and the step response. That is:
U(w) = H(w)/jw
So, when I compared the 2 spectra, I was expecting the FFT of my .TRAN output
to have an extra 6dB/octave roll-off, as compared to the .AC output, due to the
`w' in the denominator, above.
Can anyone shed some light on this?
Thanks,
David Banas
Sr. Member Technical Staff
Altera
+1-408-544-7667 - desk
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