Integrator transfer function and arbitrary continuous input signals

Right. The two sided Laplace transform is just a Fourier transform with the change of variable s = jω (omega, that is).

Cheers

Phil Hobbs

It means that instead of just individual numbers, variables consist of pairs of numbers. This is so that both magnitude (length) and phase (angle) can be accommodated. For example (1, 0) is equivalent to the real number 1 and it has a length of 1 unit. It can be written 1 + j0 or just 1 The number (0, 1) also has a length of one unit but it's not on the real number line, it's been rotated 90 degrees counterclockwise. It can be written 0 + j1 or just j The following may be worth watching. It's not specifically about electronics. If you don't want to watch it all, just watch the part at 19:10 Note also that outside electronics j usually becomes i. Except in some of my old handed down textbooks where "operator j" is used and the complex plane is called the Argand diagram.

A couple of points that may clarify issues.

First, the complex numbers are _numbers_. That is, they allow the same arithmetic operations as the reals. The only axiom of the reals that has to be relaxed is the Archimedean order property—you can sort any list of reals by sign and magnitude, but you can’t do that with complex numbers.

Second, even a real-valued function will in general have a complex-valued transform. Constraining the function values to be real is equivalent to constraining the transform to be Hermitian, i.e. if h(t) is real-valued and has transform H(f), then

H(-f) = H^*(f),

the complex conjugate.

Cheers

Phil Hobbs

"Christopher Howard" snipped-for-privacy@librehacker.com wrote in message news: snipped-for-privacy@librehacker.com...

These days you just use LTSpice and you see that your ramps become parabolas as expected. Also look at what happens without R2.

Version 4.1 SHEET 1 880 680 WIRE 208 -96 160 -96 WIRE 352 -96 288 -96 WIRE 160 32 160 -96 WIRE 224 32 160 32 WIRE 352 32 352 -96 WIRE 352 32 288 32 WIRE -304 128 -304 80 WIRE -256 128 -304 128 WIRE -256 144 -256 128 WIRE -64 144 -96 144 WIRE 32 144 -64 144 WIRE 160 144 160 32 WIRE 160 144 112 144 WIRE 240 144 160 144 WIRE 352 160 352 32 WIRE 352 160 304 160 WIRE 400 160 352 160 WIRE 240 176 192 176 WIRE -304 192 -304 128 WIRE 192 224 192 176 WIRE -96 256 -96 144 WIRE -96 384 -96 336 FLAG -256 144 0 FLAG -304 0 +15 FLAG -304 272 -15 FLAG 272 128 +15 FLAG 272 192 -15 FLAG 400 160 Parabola FLAG -64 144 Ramp FLAG 192 224 0 FLAG -96 384 0 SYMBOL OpAmps\\UniversalOpAmp 272 160 R0 SYMATTR InstName U1 SYMATTR Value2 Avol=100k GBW=8Meg Vos=0 SYMBOL voltage -304 -16 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 15V SYMBOL voltage -304 176 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 15V SYMBOL voltage -96 240 R0 WINDOW 123 24 124 Left 2 WINDOW 39 0 0 Left 0 WINDOW 3 76 51 Left 2 SYMATTR Value2 AC 1 SYMATTR Value PULSE(0 1 0 1ms 0ms 0 0 10) SYMATTR InstName V3 SYMBOL res 128 128 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 10k SYMBOL cap 288 16 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C1 SYMATTR Value 10nF SYMBOL res 304 -112 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R2 SYMATTR Value 100k TEXT -464 416 Left 2 !.tran 10ms

No, you can calculate the frequency response by setting s=jw, or the dc responce by setting to 0.

The transfer function is the ratio of the Laplace transform of the system output to the Laplace transform of the systen input. So to find the system output y(t) to an arbitrary input x(t) you:

For example for a heaviside input (x = 1 for t > 0 otherwise x = 0) L[x] = 1/s. For an integrator with a transfer function 1/s the output would therefore be 1/s^2 (ie L[y]). Taking the inverse transform of that would yield y = t (ie a ramp).

That’s one of the many sources of confusion when EEs talk to other technical folk.

WDDNS two- sided Laplace.

Cheers

Phil Hobbs